An Analytical Calculus: Volume 2: For School and University (v. 2)

AN ANALYTICAL CALCULUS VOLUME II

AN ANALYTICAL CALCULUS FOR SCHOOL AND UNIVERSITY

BY

E.A.MAXWELL Fellow of Queens' College, Cambridge

VOLUME II

CAMBRIDGE AT THE UNIVERSITY 1966

PRESS

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521056977 © Cambridge University Press 1954 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1954 Reprinted 1966 This digitally printed version 2008 A catalogue recordfor this publication is available from the British Library ISBN 978-0-521-05697-7 hardback ISBN 978-0-521-09036-0 paperback

CONTENTS PREFACE

vii CHAPTER VII: THE LOGARITHMIC AND

EXPONENTIAL FUNCTIONS PAGE

1. The logarithm 1 2. First properties of the logarithm 2 3. The graph of log a; 4 4. The use of logarithms in differentiation 10 5. The use of logarithms in integrating simple rational functions 11

PAGE

6. The exponential function ^ d d 7. The relations / = - , / = y dx xdx 8. The integration of ex 9. The reconciliation of logea? and log10a?

REVISION EXAMPLES I I I , 'ADVANCED' LEVEL

18 20 24 27

28

CHAPTER V I I I : TAYLOR'S S E R I E S AND A L L I E D RESULTS 1. 2. 3. 4.

A series giving since A series giving 1/(1 +x) Expansion in series The coefficients in an infinite series 6. Taylor's theorem 6. Maclaurin's theorem 7. Maclaurin's series

38 40 41 43 44 49 49

8. 9. 10. 11. 12. 13.

The series for since and cosy The binomial series The logarithmic series The exponential series Approximations Newton's approximation to a root of an equation 14. Leibniz's theorem

REVISION EXAMPLES IV, 'ADVANCED' LEVEL

50 51 53 54 55 56 62 66

CHAPTER I X : T H E H Y P E R B O L I C FUNCTIONS 1. The hyperbolic cosine and sine 2. Other hyperbolic functions 3. The graph y = cosh a?

84 4. The graph y = sinho? 87 5. The inverse hyperbolic cosine 94 6. The inverse hyperbolic sine

95 96 98

CHAPTER X: CURVES

1. Parametric representation 2. The sense of description of a curve 3. The 'length' postulate 4. The length of a curve 5. The length of a curve in Cartesian coordinates

100 6. The length of a curve in polar coordinates 101 7. The 'gradient angle' tp 102 8. The angle from the radius 103 vector to the tangent 9. The perpendicular on the 104 tangent

105 107 109 110

VI

CONTENTS

Chapter X: Curves—continued PAGE

10. Other coordinate systems 11. Curvature 12. A parametric form for a curve in terms of s 13. Newton's formula

PAGE

111 113

14. The circle of curvature 15. Envelopes 16. Evolutes 116 17. The area of a closed curve 117 18. Second theorem of Pappus

119 123 126 128 132

R E V I S I O N E X A M P L E S V, ' A D V A N C E D ' L E V E L

133

REVISION EXAMPLES VI, 'SCHOLARSHIP' LEVEL

138

C H A P T E R X I : COMPLEX N U M B E R S 1. Introduction 2. Definitions 3. Addition, subtraction, multiplication 4. Division 5. llqual complex numbers 6. The complex number as a number-pair 7. The Argand diagram 8. Modulus and argument 9. The representation in an Argand diagram of the sum of two numbers 10. The representation in an Argand diagram of the difference of two numbers

158 161 163 164 165 166 168 169 171

11. The product of two complex numbers 12. The product in an Argand diagram 13. De Moivre's theorem 14. The nth roots of unity 15. Complex powers 16. Pure imaginary powers 17. Multiplicity of values 18. The logarithm of a complex number, to the base e 19. The sine and cosine 20. The modulus of ez 21. The differentiation and integration of complex numbers

172 173 174 177 178 179 181 183 184 186 186

172

REVISION EXAMPLES VII, 'SCHOLARSHIP' LEVEL

C H A P T E R X I I : SYSTEMATIC

192

INTEGRATION

1. 2. 3. 4.

Polynomials 200 6. Trigonometric functions 214 Rational functions 200 7. Exponential times polynomial Integrals involving *J(ax + b) 203 in a? 215 Integrals involving 8. Exponential times polynomial *J(ax2 + 2hx + b) 203 in sines and cosines of 5. Integrals involving *J(ax2 + 2hz multiples of a? 215 + b): alternative treatment 207 CHAPTER X I I I : INTEGRALS INVOLVING 1. 'Infinite'limits of integration

217

'INFINITY'

2. 'Infinite'integrand

222

REVISION EXAMPLES V I I I , 'ADVANCED' LEVEL

226

REVISION EXAMPLES IX,

227

'SCHOLARSHIP' LEVEL

A P P E N D I X , FIRST STEPS IN PARTIAL DIFFERENTIATION

236

A N S W E R S TO E X A M P L E S

245

INDEX

271

PREFACE Appreciation for help received was expressed in the Preface to Volume I, but I would record how much deeper my indebtedness becomes as the work progresses. E. A. M. QUEENS' COLLEGE, CAMBRIDGE

June, 1953

CHAPTER VII T H E LOGARITHMIC AND E X P O N E N T I A L FUNCTIONS THE particular functions which we have used in the earlier chapters (Volume i) are the powers of x, the ordinary trigonometric functions, and combinations of them such as polynomials. We now introduce an entirely new function, the logarithm. The need for it arises, for example, when we seek to evaluate the integral \xndx for n = — 1. The standard formula xndx =

-xn+l 71+1 becomes meaningless; the integral cannot be evaluated in terms of the functions at present at our disposal.

/•

1. The logarithm. Consider the integral Cdx To make the discussion precise, we shall fix the lower limit, giving it the value unity; the effect of this is merely to remove ambiguity about the arbitrary constant. The integral is a function of its upper limit, which we denote by the letter x, replacing the variable in the integration by the letter t. The function is thus

where (Vol. I, p. 87)

f'(x) = - . x

2

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

The function defined in this way is called the logarithm of x, usually written , logic or loge#, the suffix e being inserted for reasons to be given later (p. 27).

2. First properties of the logarithm. We now prove some of the basic properties to which the logarithm owes its importance. The reader will note the very close connexion with logarithms to the base 10', with which he is presumably familiar. (i)

log 1 = 0.

This follows immediately, since (Vol. i, p. 83)

(ii) For

(

log ay = log x + logy.

log #£

-r*

f

/Vr

Now use the substitution t ==

XU

in the latter integral. We have the relation dt == xdu (remembering that x is CONSTANT here, the variable of integration being t). Also the values xfxy of t correspond to the values \,y of 11. Hence ^ ^ Jx

t

J i XU

Ji U

We therefore have the required relation log xy = log x + log y.

FIRST PROPERTIES OF THE LOGARITHM COROLLARY.

log {xjy) = log x — log y.

For

logo; = log j P ] y\

3

= log(x/y) + \ogy. log(#w) = nlogx.

(iii)

Cx dt

In the relation

logic = 1 —,

h make the substitution We have the relation

t

n

u=t. du — ntn~xdt.

Also the values 1, x of t correspond to the values 1, xn of u. Therefore, since _ 1 ntn ' 1 we have

logo: = Ji

——— n t

=

— (applying the substitution)

log (xn) =

so that

Note, n may have any real value, and is not necessarily a positive integer. (iv) The value of log x increases indefinitely as x does. Suppose that N is aiiy large number and m the largest integer such that 2m

= log sec x. ILLUSTRATION

5. To find /== Lcwlogo;fe

(?i4= — 1).

On integration by partsxn+l (Vol. I, p. 103), we have rxn+l

-

l

logo;— -,-dx n+l r xn B xn+l Jn+1 x log x — r ax n+l 8 Jn+l xn+l

ILLUSTRATION

xn+l

6. To find coso; Ccos CC cos Ccosxdx xdx cosxxdx

,

_

We have

/=

TTT

l

r 2 s—= hi sm a; 2 J cos a; J l~sm u = sin a;,

Let so that

du = GO&xdx.

Then

,,

1 + sin a?

flog2 & l

THE GRAPH OF LOGiC

This may also be expressed in the form

in x\2

,

/I + sin x

= log (sec x + tan x). Another form for the answer is I ILLUSTRATION

+X

7.* To find T—

Notice that the differential coefficient of the denominator is 2x + 4, and express the numerator in the form

Then

f

7=2 JX*+4:X+ I'd

dx

JX*-

dx = 21og (a;2 + 4a; +13) - J tan" 1 ^ - . \ 3 / ILLUSTRATION 8.* To find

Notice that the differential coefficient of the denominator is 2a;-6, and express the numerator in the form f (2a?-6)+23. /

=

_ _L

1

j_ 23

* An important type. Mil

10

LOGARITHMIC AND EXPONENTIAL* FUNCTIONS EXAMPLES I

Find the following integrals: dx

1.

x+1' - dx.

dx

5.

Evaluate the following integrals: d

7. I

-.

8. I —-.

Ja x

9.

J_3 aj+1

3 d

r~ *,

„. r_^_.

Jo

i2. r

Differentiate the following functions with respect to a;: 13. log(3a; + 2). 2

14. logtana;. n

16. a; loga;.

17. x loex.

15. logcoseca;. 18.

Find the following integrals: 20. [l^^dx.

19. flog xdx. J

22. fa; log a;da.

J

2±. [ .

23. [4^-.

J 9^

21. f

J

Jsma;

I v" w ' v / ^ w

Qfi

j v^^/

J si ^/ wW

_

| (^a; + ojaa;

28. f J4. The use of logarithms in differentiation. The differentiation of a fraction (in which the numerator and the denominator may themselves be products of factors) is often made easier by the method known as logarithmic differentiation, illustrated in the following examples. ILLUSTRATION

9. To differentiate the function

THE USE OF LOGARITHMS IN DIFFERENTIATION

11

Take logarithms. Then log y = 3 log x + log (1 + a?2) — 4 log (1 — x) — 2 log (1 + 2x). Differentiate.

Then

1 dy 2/ dx

3 2x x \+x2

4 l—x

4 l-h2#'

and the value of -~ follows at once. dx With a little practice, the two steps may be taken together: ILLUSTRATION

10. To differentiate the function ( 1 - 2x)2 sin 3 x

Take logarithms and differentiate. Then Idy —4 3 cos a; 16a; ydx — l — 2x f EXAMPLES II

Use the method of logarithmic differentiation to differentiate the following functions: cos2#

T+ 2 '

x2(l+x)2 4

3

• (1+ ) ' —I

iTT:

2a; cos 2 a;'

(1+) (1-^)'

5a;(la;)3 ••

(1+a;2)2

ccsina: 2

(1 + COSOJ)2 •

8.

-r-—

•

rr.

9.

5. The use of logarithms in integrating simple rational functions. A rational function of x is an expression of the form u(x) v(x)' where u{x), v(x) are polynomials in x. We shall later (p. 200) give a detailed treatment of the integration of such functions; here we give a preliminary account of the simpler cases.

12

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

If the degree of u(x) is higher than that of v(x), we can divide u(x) by v(x), and obtain an expression of the form p(x) v(x) ^ v + -7-Y, where p(x) is a polynomial, and w(x) is a polynomial whose degree is less than that of v(x). The integration of the polynomial p(x) is immediate. We may therefore confine our attention to the form w(x) where w(x), v(x) are polynomials in x9 the degree of w(x) being less than that of v(x). The method is to express this quotient in partial fractions; details may be found in a text-book on algebra, but, for convenience, a brief account of the calculations involved is inserted for reference. In order to explain what is required, we consider some typical examples. (Different mathematicians use varying methods. Those which follow have the advantage of giving independent checks of accuracy in some of the more complicated cases.) 2x (i)

/ ( )

The denominator consists of the two linear factors (x — 2), (x each occurring to the first degree only. We seek to express f(x) in the form . R x-2x TT7

,

We have

A

+ 3'

B

2x

- +•

Multiply throughout by x — 2. Then A

Blx-2)

2x

x+3

# + 3'

This holds for all values of a;; in particular, for x = 2. Then 2.2 4 ~~ 2 + 3 " 5" Hence

A = -. 5

INTEGRATING SIMPLE RATIONAL FUNCTIONS

13

In practice, these steps are usually telescoped, as we now illustrate in finding B. Multiply throughout by 3 + 3 and then put 3 = - 3. Thus 2(-3) 6 D (-3-2) 5 Hence

f(x) =

(ii)

f(x)

4 5(3-2)

6 5(3 + 3)

3+1

The denominator consists of two linear factors (3 + 2), (3—2), of which (3—2) occurs to degree 3. We seek to express f(x) in the form B C D A •(3-2) 3 + (3-2) 2 + 3-2 >

A

B

SO that

+

C ^

D

x+1

+

Multiply throughout by ru + 2 and then put a; = — 2. Thus -2+1 - 1= 1 "" ( _ 2 - 2 ) 3 " " - 6 4 " " 64* Multiply throughout by (x — 2)3 and then put x = 2. Thus 2+ 2 4 In order to find C, D, we use these values of A, B: C

D

x+l

1

3

+

64(o; + 2) (a? -2) 3 - 123+8-48^-96 64(3 + 2) ( 3 - 2 ) 3 3 3 - 6 3 2 - 4 3 + 24 64(3 + 2) ( 3 - 2 ) 3 '

At this point, we are able to check accuracy for the highest common factor of the denominators on the left-hand side is (x — 2)2.

14

LOGARITHMIC AND EXPONENTIAL

FUNCTIONS

Hence x + 2 and x — 2 MUST be factors of the numerator on the right. By division, we find that

„ Hence

G D x-6 -r^ + -=-—

Multiply by (x — 2)2 and then put x = 2. Thus °

64~~16" (a;-6)

64(a;-2)2

1

64(*-2)'

again checking accuracy by the cancelling of x — 2. D=

Finally,

-^i'

Hence

Ax— 1 The denominator consists of the linear factor (#—1), repeated, and the quadratic factor (x2 + x + 1). We require to express f(x) in theform

A

B

Cx + D

the numerator above the quadratic factor being of the form Cx + D. We thus have A

B

Cx + D

4x-l

+ + Multiply throughout by (x— I) 2 , and then put x = 1. Thus ^ ~ 1+ 1+ 1"

1#

INTEGRATING SIMPLE) RATIONAL FUNCTIONS

^ Hence

B x—\

Cx + D f- -

Ax-1

15

1

„ 5 Oz + D a:-2 Hence -+ the cancelling of the factor x — 1 providing a check of accuracy. Multiply throughout by x — 1, and then put x = 1. Thus 1-2 1 1 + 1 + 1 3' x-2

Hence

1

on cancelling the factor, a?— 1. Hence

and so The following illustrations exhibit some further points about the calculation of partial fractions, and also show how the integration of rational functions is carried out. ILLUSTRATION

11. To find 1=

The numerator is not of less degree than the denominator, so we begin by dividing out: xx

_

2a: 2 -1

16

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

Consider, then, the function

Factorize the denominator, so that 2a: 2 -1

We therefore have to find constants A,B9C9D such that A

B

C

{

2# 2 -l

D

(x — 1 )2 2

A =

(x = 1

and then put | _

(l + l) 2

.

Thus

4'

Hence 5

2a; 2 -1

i>

1

4:(x-l)2

+

Hence, by the usual process, It follows that 1

l h L II

*^ At

3

x "I \ O

I

A I

1

x 1 \

•

A I

.1\O

3 AI

.. 1 \ /

ti'«* /

INTEGRATING SIMPLE RATIONAL FUNCTIONS ILLUSTRATION

17

2. To find / =

f Udx

We must first factorize the denominator. It vanishes when x = 2, so that x — 2 is a factor, and, after division, wefindthat it is (x - 2) (x2 + 2x + 5). We therefore seek to express //

N

13

13

m the form .,, so that

+A Bx + C x-2 +

13

Following a routine which should now be familiar, we have 13 Bx + C

13 1 (x-2)(x2 + 2x + 5) x-2

%-2x-x2

""x Hence

18

LOGARITHMIC AND EXPONENTIAL FUNCTIONS EXAMPLES III

Integrate the following rational functions: 1.

x2 - 9 '

2.

4.

a* s-1"

5.

7.

3.

1

-. 11.

10.

12. 14.

13. 15. 18. 21.

9.

1

4a;+ 7*

3)-

16

-;

19. 22.

x-2

17. 20.

x2+l '

2x

23.

24.

6. The exponential function. Imagine the graph y = logx (Fig. 62) to be turned, as it were, through a right angle and viewed through a mirror, and y the axes then renamed to give the curve shown in the diagram (Fig. 64). Then y is a certain function of x with the property that x = logy. Thus y is an 'inverse' function of log x in a sense similar to that in which sin"1 a; is (Vol. I, p. 38) an inverse function of sin a:.

0

Fig. 64.

THE EXPONENTIAL FUNCTION

We write the relation

19

x = logy

to give y in terms of x, in the form y = exp x, where exp x, whose properties we now study, is called the exponential function. It is defined, as the graph implies, for all values of x, increasing steadily from zero to 'infinity' as x increases from 'minus infinity' to 'infinity'. The exponential function (of a real variable x) is necessarily POSITIVE.

From the relation

logy = x,

we have, by differentiation with respect to x9

1^=1 ' ydx dy ~T~ = V* so that dx * Hence the differential coefficient of exp x is exp x itself. It is convenient to have a name for the value of the function when x = 1, and for this we use the letter c. Thus 1 exp 1 = e, or, in equivalent form,

logc = 1.

From the graph, we have the relation [The value of e, to four significant figures, is 2-718.] We now seek to identify the function expx in terms of the constant e and the variable x. If y = expo;, then

log {exp x) = log y = x.

Also the relation log(^) = nlog# leads, on replacing x,n by the letters e, x respectively, to the relation log(ex)

20

LOGARITHMIC AND EXPONENTIAL

FUNCTIONS

since loge = 1. Hence log{exp#} = log^*). But we have proved (p. 5) that, if the logarithms of two numbers are equal, then the numbers themselves are equal, and so _ exp x = ex. The exponential function exp a; is therefore identified as the number e raised to the power x.

1. The relations -~ = - , ~ = Jv. dx

x dx

(i) THE LOGARITHM.

It follows from the definition of a logarithm that the relation dx

x

yields for positive x the result where C is an arbitrary constant. If x is negative, say __ X — *""" Uy

where u is positive, then dy _dydx __ dy du dx du dx' hence

so that

dy 1 -=— = — du x 1 y = \ogu-\-G

We may therefore conclude that, whether x is positive or negative, the equation , dx leads to the relation

x

y — log | x \ + C,

where\x\is the numerical value of x.

THE EXPONENTIAL FUNCTION

21

In practice, it is customary to use the form

with the tacit assumption that x is positive; but this needs care. The relation y = log x + G may be put into an alternative form by writing C = log a, where a is also an arbitrary constant. Then y = logz + loga = log ax

(ax assumed positive).

By the definition of the exponential function, we then have ax = ev, x = b ev,

or

where b is likewise an arbitrary constant, assumed to have the same sign as x. (ii) THE EXPONENTIAL FUNCTION.

We turn now to the equation dy = y

dx 1 —= dy y we see that interchange of x,y in the above relation x = bev leads to the result ,_ y = bex. dy Hence the equation -~ = y Writing this in the form

CLX

leads to the relation

y = bex,

where b is an arbitrary constant, assumed to have the same sign as y. More generally, the relation

leads to the relation

y = bekx:

22

LOGARITHMIC AND EXPONENTIAL

For the substitution gives

FUNCTIONS

x = ujk dy = dydx=ldy du dxdu kdx9

so that

—= r du k y = beu

Hence

We give two typical examples to show how exponential functions arise in physical applications. ILLUSTRATION 13. We return to Illustration 1 (p. 6) of a circuit with resistance R, self-inductance L and electromotive force E. The equation for the current x at time t is

dx £ — = E — Rx. dt

Hence

E-Rx = u;

Write ,

du __ du dx dt dx dt

= (-BIL)u. Hence

u=

Ae~iB/L)t,

where A is an arbitrary constant, so that If x = 0 when t = 0, then E = Ae° = A, and so or

E-Bx = Ee-{R'L)',

23

THE EXPONENTIAL FUNCTION

ILLUSTRATION 14. To find the variation of pressure with height in an atmosphere obeying the law

pv =* constant, where p, v denote pressure and volume respectively. Consider a vertical filament of air whose cross-sections have area SA (Fig. 65). Let the pressures at heights x9x + Sx be p,p + 8p. Then the element of volume (shaded in the diagram) of height Sx and base SA is in equilibrium under pressure round its sides, which does not concern us, and also P + Si> under the following vertical forces: (i) pSA upwards; (ii) (p + 8p)8A downwards; (iii) pSASx downwards,

j

where p is the weight per unit volume at height x. Hence pSA -(p + Sp) SA - pSxSA = 0, or

8p + pSx = 0,

Now let p09 p0, v0 be the values of p9 p, v at ground level. Since p is the weight per unit volume, the relation pv = povo

is equivalent to and so

qy

rp

- = —, P Po SA Fig. 65.

Po

In the limit, this is and so

dx

p

Vn

,

where A is an arbitrary constant. But p = p0 when x = 0, so that p0 = Ae° = A. Hence the pressure at height x is given by the relation p

24

LOGARITHMIC AND EXPONENTIAL

FUNCTIONS

8. The integration of e*. To

I exdx9

find

we have merely to note that the relation (p. 19)

leads at once to the result ex = exdxy so that the value of exdx is ex itself. eaxdx = - e°*.

COROLLARY. ILLUSTRATION

15. To find

/ = eax sin bxdx. On integration by parts, we have 1 fl / = -eaxsinbx— \-eax.bcosbxdx a Ja 1 b C = - eax sin bx — eax cos bxdx. a aj

Integrating again by parts, we have / = -eaxsinbx—^eax2cosbx-\—^ a

a

a a ;

6

\eax( — bsi aJ

—s/.2

a

Hence

(1 + -^ I / = - eax sin 6# — 5 eax cos 6a;, \ a2/ a a2

so that

/ = -g—^2 (a si n te — 6 cos 6a;).

25

THE INTEGRATION OF ex ILLUSTRATION

16. To prove that, if y = eax sin bx9

then

dx2

dx

-2L =- a eax s u l bx + b eax cos bx dx

We have

= ay+ beax cos bx.

d2y dy . „ , ,„ „„ . . -=~2 = a~- + abeax cosbx -b2eax smbx dx dx

Hence

so that EXAMPLES IV

Find the differential coefficients of the following functions: 1

P2X

3.

2. e*\

6. (l+a: 2 )e-

4. e^cosa. 7. ^(e^ + c-x)2.

8. *e* S in*.

9.

10. (l + e^)si]ia;.

11. e3iCcos4^.

12.

I-* 2 '

enana;.

Find the following integrals: 13. \e2xdx.

14. \e~5xdx.

16. \x2e-x*dx.

17.

r 19. j e

•J

ooB« a, then asinx + bsinSx will have a maximum value for some value of x between 0 and |TT. Find this maximum value when a = 3,6 = 0-5, proving that it is a maximum and not a minimum. 21. Given that y = x5-5x* + 5x2+l, find the stationary values of y. Determine whether these values are maximum or minimum values or neither. 22. The perpendicular from the vertex A to the base BG of a triangular lamina cuts BG at D; CD = q,DB = p (where qa2>--- i n terms of f(x) and its differential coefficients, and later we proceed to a more detailed discussion. First, however, we must say a few words about the meaning of the 'sum to infinity* in general; for a fuller treatment, a text-book on analysis should be consulted. Suppose that we have a series whose successive terms are, say, uvu2,uz, ...,un,.... It may happen that the sum of the first n terms ~ 5

tends to a limit S as n tends to infinity. I For example, if x=$= 1, xn 1 _/y

so that, for this series, 8n = and, if | x \ < 1,

1 X ~~ X

Xx

1

v>

n

,

JL ""^ X

S = lim Sn = -^—.)

In this case we call S the sum to infinity of the series. We write 8

and say that the series converges to 8. On the other hand, if 8n does not tend to a limit as n tends to infinity, the series has no 'sum to infinity', and the expression

has no arithmetical meaning. If the terms uvu2,... of the series depend on x (as in the particular example just quoted) so also does the sum 8n of the first n terms, and the sum to infinity when there is one. We shall be concerned exclusively with series of the form ...+anxn+..., where the coefficients ao,al9... are constants, and we have seen (§§ 1, 2) that such a series may converge for all values of X or for some values only.

EXPANSION IN SEEIES

43

Note. It is important to realize that the word 'converge' implies definite tending to a limit. A series such as for which 8n = 1 when n is odd and 0 when n is even, has a finite sum for all values of n, but does not converge. The series is said to oscillate boundedly. 4. The coefficients in an infinite series. We first assume that expansion in an infinite series is possible, and seek a formula to determine the coefficients: To prove that, iff(x) is a given function which CAN be expanded in 2 ejorm f(x)==a + + ...+a xn +..., n

an = -—r^> n n\ where f(n)(0) is the value offin)(x) when x = 0. We assume without proof that (in normal cases) we can differentiate the sum of an infinite series by differentiating the terms separately and adding the results, as we should for a finite number of terms. Then then

f'(x) = f"(x) = and so on. Putting x = 0, we have successively /(0) /'(0) /"(0) /'"(0) /«v>(0)

= a09 = al9 = 2a2, = 3.2a3, = 4.3.2.a4,

and, generally, /(0) = n(n-l)...Z.2an

= n\an

Hence

Note. This work gives no help about whether the function CAN be expanded; for that we must go to the next paragraph. 4-2

44

TAYLOB'S SERIES

AND A L L I E D

RESULTS

ILLUSTRATION 1. A 'particle falls from rest under gravity in a medium whose resistance is proportional to the speed. To find an expression for the distance fallen in time t. If x is the distance fallen, then the acceleration is x downwards; the forces acting per unit mass are (i) gravity, of magnitude g downwards, (ii) resistance of magnitude kx upwards. Hence

x = g — kx.

The formula just given, when adapted to this notation, is t2 .. X =

t3 ...

XQ-JT tXQ + -jry XQ + — XQ + . . . ,

where xQ, x0, x09... are the values of x9 x, x,... when t = 0. From the initial conditions, x0 = 0, xo = 0, so that

x0 = g — kx0 = g.

By successive differentiation of the equation of motion, we have x o = -Jcxo = -Jcg, XQ

=

KXQ

= /c g9

and so on. Hence

The series converges for all values of t. EXAMPLES II

Use the formula of § 4 to find expansions for the following functions: 1. sin#.

2. cos#.

3. !/(!+«).

5. Taylor's theorem. We come to a somewhat difficult theorem on which the validity of expansion in series can be based. Suppose that f(x) is a given function of x, possessing as many differential coefficients as are required in the subsequent work.

T A Y L O R ' S THEOREM

45

To prove that, if a, b are two given values of x, then there exists a number £ between a, b such that, for given n9

Write

(ft-*)* (a?)

~

where Rn is a number whose properties will be described as required, and k is a positive integer to be specified later. We propose to use Rolle's theorem (Vol. I, p. 60) for F(x) exactly as we did (Vol. I, p. 61) for the mean value theorem, of which this is, indeed, a generalization; we therefore want the relations F(a) = F(b) = 0. By direct substitution of b for x in the expression for F(x), we have

F(b) = 0.

In order to obtain the relation F{a) = 0, we substitute a for x on the right-hand side and equate the result to zero; thus

0 = ftb)-f(a)- (b-a)f'(a)-^^f"(a)

-...

We must therefore give to En the value

Rn-Ab) -/(a) - ( 6 - a ) / » J±

We have now ensured the relations = F(a) = 0, and so, by Rolle's theorem, there exists a value £ of x between a, 6 at which F'{x) = 0.

46

TAYLOR'S SERIES

AND A L L I E D

RESULTS

The next step is to evaluate F'(x). This will involve a number of terms of which

is typical, and the differential coefficient of this term is

Hence, remembering that Rn is a constant, we have F'(x) =

-f[x)-{{b-x)f"{x)-f(x)}

(n-2)\J ~

(n-l)\J

J

{X {X)+

{b-af

(X) +

j

(b-a)*

"n

Kn>

after cancelling like terms of opposite signs. But there is a number £ between a, b for which F'(g) = 0, and so M(6-q)(6-fl

Equating this to the value of Rn obtained above, we have

The value of k is still at our disposal. If we put k = n (as we might have done from the start, of course, had we so desired) we have (6-q)»

TAYLOR'S THEOREM

47

and the result enunciated at the head of the paragraph follows at once. This gives us the most usual form for Taylor's theorem, and we have adopted it in the formal statement; but there are advantages in keeping k more general, as we see below. The theorem is therefore established. The expression

is called the remainder after n terms. It is, in the first instance, what it says, namely a remainder, the difference between

Kb) lh — n\n-l

(h — n\%

and

/

The theorem just proved enables us, however, to express this remainder in the suggestive form (with k = n) >(0 by choosing £ suitably. This expression is known as Lagrange's form of the remainder. By giving k other values, we obtain various forms for the remainder. In particular, when k = 1, we have =

"~

(6-q)(6-g)n-i J (n-1)!

( }

*

This may be expressed alternatively by writing f, which lies between a, b, as a + 6(b—a), where 6 lies between 0 and 1. Then

This is called Cauchy's form of the remainder. Though less 'in sequence' than Lagrange's form, it enables us to deal with some series for which Lagrange's form does not work.

48

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

ALTERNATIVE TREATMENT. There is an alternative treatment of Taylor's theorem, based on integration by parts, which leads to yet another form of the remainder. The method is essentially a continued application of the formula

Cb(b-t) JO V^

the term

/(fc) (a +1) vanishing for t = b when & > 0.

This formula may be expressed more concisely by writing

bk

uk = T*f{k) {) +

so that Hence

ux = bf (a) + u2

r6 Moreover,

/ ' (a +1) dt = f(a + 6) —/(a). Jo Equating the two values of uv we obtain the 'Taylor' relation (with our original C6' now replaced by ea + 6')

where

%=

T A Y L O R ' S THEOREM

49

EXAMPLE I I I

1. Prove that, if Rn(%) is the function otx defined by the relation

( + ... +* ~ _ ^V then

i?»

-

^

^

and deduce that Rn{x) = f*^"^ Jb

f{n\t)dt.

(n—1)1

6. Maclaurin's theorem. If we write 6 = a + h, Taylor's theorem (with the Lagrange remainder) becomes

where f is a certain number between a,a + h. A convenient form is found by putting a = 0 and then renaming h to be the current variable x:

f(x) =

|J

^^

where f is a certain number between 0,x. This important result is known as Maclaurin's theorem. Compare p. 43. With the Cauchy form of remainder, the corresponding result is

where ^ is a certain number between 0,1. 7* Maclaurin's series. The remainder Rn in Maclaurin's theorem appears in the form or

50

T A Y L O R ' S S E M E S AND A L L I E D R E S U L T S

where £ lies between 09x and 6 between 0,1. Then f(x) = /(0) + xf (0) + . . . + ~ ^

/Yl +

+

+

It may be proved that this expansion for ex is valid for all values ofx. [See Examples V (1) below.]

THE EXPONENTIAL SERIES

55

COROLLARY. The work of this paragraph enables us to fill a gap in the discussion of the exponential function (pp. 18-20) by obtaining an expression for the number e. Putting x = 1, we have

EXAMPLES V

1. Use the method given for the sine series to prove the validity of the exponential series for all values of x* By direct calculation of the differential coefficients and substitution in Maclaurin's formula, obtain expansions for the following functions as series of ascending powers of x: 2. e2x. 5. l/(l+a;). 2x

8. e~ .

3. log (1 + 2x). 2

6. 1/(1-a;) . 9. yl{l + 2x).

4. sin2#. 7. cos4x. 10. log (l-3z).

12. Approximations. If the successive terms in the expansion

become rapidly smaller, a good approximation to the value of the function f(x) may be found by taking the first few terms. ILLUSTRATION 2. To estimate 7(3-98).

Writing the expression in the form

we may expand by the binomial series (p. 51) to obtain

= 2{1 - -0025 - K-000025) + ^ (-000000125)...}, and a good approximation is 2(1--0025) = 1-9950. A limit may be set to the error by means of Taylor's theorem, as follows:

56

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

Since, for a general function f(x),

f{x) =

|J

the error cannot exceed the greatest value of

rt

or, here,

2x

(-005)2

2!

1

-00000625

' 4(1 - -0050)*—- (1 - -0050)*'

In the most unfavourable case, with 0 = 1 , this gives a value less than -0000063 for the error. EXAMPLES VI

Estimate the values of the following expressions:

1. V( 4 '02).

2

- V( 8 * 97 )-

3. ^(8-02).

4. ^(26-98).

5. ^(31-97).

6. l/V(9-03).

13. Newton's approximation to a root of an equation. Suppose that we are given an equation f(x) = 0 and know that there is a root somewhere near the value x = a. Newton's method, which we now describe, shows that, under suitable circumstances, a better approximation to the root is a

7W

If the correct root being sought is | == a + h, then

so that, by Taylor's theorem for n = 2, with Lagrange's form of the remainder,

NEWTON'S APPROXIMATION

57

for a value of 77 between a and a + h. If h is reasonably small, then the term involving h2 may be regarded as negligible for practical purposes. We therefore have f(a) + hf'(a) = 0, or so that the root is approximately

a

"7T«T

This crude statement, however, should be supplemented by more careful analysis, and the following graphical treatment shows the precautions which ought to be taken. We begin with an examination of the curve

y=f(x) near the point £, taking first the case where the gradient is positive near and the concavity 'upwards' near that point, so that f(x)J"(x) are both positive.

Fig. 66.

(i) Let X be the point (f, 0) on the curve, and let A be the point for which x — a, where a > £; draw AM perpendicular to Ox, and let the tangent at A meet Ox in P.

58

Then

TAYLOR'S SBBIES AND ALLIED RESULTS

OM = a, AM=f(a), AM -p^ = tan^=/(a),

so that and

PM = {jy\, OP = a-

But under our assumptions that f{x)J" (x) are positive near X (so that the gradient is positive and the concavity upwards) the tangent AP lies between AM and the curve, so that P lies between X and M. Hence, under these conditions, OP is a better approximation than OM to OX. That is, a — ^V^ is a better approximation rr /'(a) than a to the root. It is assumed that /'(a) is not zero, and, indeed, that/'(#) is not zero near the required root. (ii) Suppose next that, with the same diagram, B is the point for which x = b, where b < £, so that B lies 'below' Ox; draw BN perpendicular to Ox, and let the tangent at B meet Ox in Q. Then

BN = — f(b)

since f(b) is negative,

QH = tan fo =f'(b), so that

and

OQ^b-jj^-y

But now we cannot be sure that Q is nearer to X than JV, for Q may be anywhere to the right of X according to the shape of the curve. Hence we cannot be sure whether b — ~-L is, or is not, a better approximation than b to the root.

NEWTON'S APPBOXIMATION

59

Similar argument applied to the accompanying diagram (Fig. 67) shows that the results (i), (ii) are also true if (the concavity still being 'upwards') the gradient is negative near X; P is closer than M to X, whereas Q may or may not be closer than iVtoX.

Fig. 67.

We have therefore proved, so far, that, iff" (x) is positive near £, the approximation a — J±JL {S better than a itself when f (a) is positive. It is easy to verify in the same way that, if /"(#) is negative near £, the approximation a — JTT\ is better than a itself when f(a) is negative. (The whole diagram is merely 'turned upside down'.) In other words, we can be sure of a better approximation if f"{x) retains the same sign near x = a, that sign being also the sign off(a). In more complicated cases, it is wise to draw sketches such as we have shown in order to determine how the tangent at A cuts the X-axis. It is, however, clear from the graphs that, for ordinary functions, the approximation a — jrrk is ALWAYS better than a itself once we come sufficiently close to the correct answer. In other words, once it is ascertained that an approximation is reasonably good, it may confidently be expected that Newton's approximation will make it better still. 5-2

60

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

ILLUSTRATION 3. An example where the answer is apparent from the outset may help to make the principle clear. Consider the equation

f so that

It is obvious that x = 0 is one root, but let us attempt to reach that value by approximation from (i) x = 1, (ii) x = — 1. When x = 1,

/(!) = ¥ , /'(I) = H, /"(!) = ¥ , so that /(I), / " ( I ) have the same sign; moreover/"(#) remains positive near x = 1 (in fact, down to x = — f, where/(#) is negative, indicating a point on the other side of the root). Hence we expect Newton's formula to give a better approximation; and since _/0^== /'(I)

1 9 _ 14 33 33'

this is actually the case. On the other hand, when x = — 1,

Here, again,/( —1),/"(- 1) have the same sign; but f"(x) changes sign from negative to positive at x = — f, which is near — 1; we are therefore in a doubtful region; and since

the approximation is actually worse. ILLUSTRATION 4. To find an approximation to that root of the equation ^-Zx+l =d which lies between 0,1.

We have

f(x) = x3 - 3x + 1, f'(x) = 3x*-B, f"(x) = 6x.

NEWTON'S APPROXIMATION

61

Since f"(x) is positive when x lies in the given interval 0,1, it is advisable to begin with an approximation which makes f(x) also positive. Now & We should therefore like an approximation between \ and \ which keeps f(x) positive, and inspection shows that \ appears very suitable. We then have

so that and the corresponding approximation is H TV = ft ='3472. The correct root is '3472..., so we have already obtained four correct figures. 5. / / 77 is a small positive number, to find an approximation to that root of the equation ILLUSTRATION

sin x = rjx which lies near to x = 7r. (The intersection of the graphs y = sin x, y — rjx shows that there is a root near to TT.) Since sin n = 0 and 77 is small, the approximation x = IT is reasonably good. Write f(x) = sin x — -qx, so that

f'(x) = cos x-7], f"(x) = —sin a;.

Then

/(TT^-T^,

/ ' ( 7 7 H - I - 7 7 , / ' » = 0.

Although /"(TT) is actually zero, so that the curve (Fig. 68) y = sinx-rjx has an inflexion at x = TT, the concavity is 'downwards' in the interval O,TT and /(IT) is negative; moreover the gradient /'(TT) is also negative. The accompanying sketch shows that the tangent lies between the ordinate x = 77 and the curve, so that Newton's method will improve the approximation.

62

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

The corresponding solution is

— 7

7

y

z

r

— 7

T

—

i

•

(-1-7,) 1+r, l+7j To the first order in -q this is, on expansion of (1 + ij)"

o

Fig. 68.

Note. This solution is less than 77, as we expected from the diagram. 14, Leibniz's theorem. The theorem which follows is useful in calculating the higher differential coeflRcients necessary for a Maclaurin expansion. To prove that, iff{x) is the product of two functions u, v, so that f(x) = uv9

then f(n){x)

where ncp is the binomial coefficient nCp nCp

n\

~~p\(n-p)V

We use the method of mathematical induction, assuming the result to be true for a certain integer N9 so that

LEIBNIZ'S THEOREM

63 (N+1)

Now differentiate this expression to obtain f (x). differential coefficient of a product such as u{N~p)vip) is M(N-p+l) v(p)

_|_ u(N-p)

The

v(P+Dt

As we write down these terms for the series on the right, we put the answer in two lines, the top line consisting of terms such as U(N-P+DV(P) a n ( j the lower of ^w-^dH-D; also we displace the lower line one place to the right, thus:

Now the coefficient of u{N~p+1)v{p)

N\ p\(N-p)\

is

N\

Hence

It follows that, if the theorem is true for any particular value N9 then it is true for N + 1 , N + 2, and all subsequent values. But it is easily established when N «= 1, being merely the result

It is therefore true generally. Note. The expression is symmetrical when regarded from the two ends, and will equally well be written in the form ln)

ff

(x) = uv{n) + ... + n

64

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S EXAMPLES VII

Use Leibniz's theorem to find the following coefficients: d4 . !• -m (#8sina;).

d4 2. -r—= i

(XX

3^

differential

iXjJU

ifix QQg Qx),

4.

6. To ap2% Leibniz's theorem in finding a Maclaurin expansion for

We have f {x)

'

=

on simplification. Hence (ar«+1) {/(*)}• = 1 . Differentiate. Then (o:2+ l).2/'(^)/"(^) + 2 4 / ' ( ^ = 0, or

(x2+l)f"(z) + xf'(x) = 0.

Differentiate n times, using Leibniz's theorem. Write the expansion from (x2+ l)f"{x) on the first line, and from xf'(x) on the second; note that the expansions terminate since (x2 +1)'" = 0, x" = 0. We obtain (x2 + 1 )/

simplifying your answers as much as you can. Prove that the first function has a maximum at x = 1 and that the second function (whose differential coefficient vanishes at x = 0) has neither a maximum nor a minimum at x = 0. 6. Find the first four differential coefficients of sin4 x. Show that the function has turning values at x = 0 and at x = JTT, and determine whether they are maxima or minima. 7. Prove that, if y = sin2(a;2), then

Prove that the result of changing the independent variable in this equation from x to | , where £ = x2, is

where a, b are constants to be determined. Prove also that, if A9B are any constants, the function y = | + A cos (2a;2) + B sin (2x2) satisfies the first differential equation. 8. Differentiate /

l\ w

(cos x + sec x)n

with respect to x, and find the nth differential coefficient of

K)' for all positive integral values of n, distinguishing the cases n < 2 and n > 2.

68

T A Y L O R ' S S E R I E S AND A L L I E D

RESULTS

9. Show that the polynomial 4 16 -(2ir-5)x* + —ATT-3)x5

f(x) = x

77

77"

and its differential coefficient/'(a;) have the same values as sina? and its differential coefficient respectively, at the values x — 0, x = ± \n.

J

%

i7T

0

f(x)dx as an approximation for

I sinxdx is less than 0*1 per cent. Jo 10. (i) Given that —- = ex} -j- = since, ax ax and that u = 1, v = 0 when x = 0, show that d2(uv) _ ~dtf~ = when # = 0. (ii) Given that y = axn + bx1"71, prove that x-7- + (n-l)y = (2n-l)axn, ax (in d t J

and form an equation in x,y9-~-9 -=-^ which does not contain a or 6. ax ax 11. Differentiate with respect to x:

Prove that

-r—

1=

, —r,

and find the nth differential coefficient of x 12. Find the nth derivatives with respect to x of

W (ii)

i

and

sin a; and

x>—v x sin #.

REVISION EXAMPLES IV

69

2

13. (i) Find the derivative of y = sin a; with respect to z = cos x by evaluating the limit of 8y/8z. (ii) Differentiate with respect to x: sin# (iii) Prove that

dy2

dx2l \dxt

14. Find from first principles the differential coefficient of \jxz with respect to x. What is the differential coefficient of ljxB with respect to #2? Differentiate with respect to x: logecosa;,

xj{\-x%

(1^2)2.

15. Find the nth differential coefficient of y with respect to x in each of the following cases: (i) y = l/x2, (ii) y = sin2#,

(iii) y = e2*sin2#.

There are three values of k for which the function y = xke~x satisfies the equation

Find these values. 16. Differentiate with respect to x: l-2x

3 tan2 X

ci-n~l-

Prove that, when y = at2 + 2bt + c, t = ax2 + 2bx + c, and a,b,c are constants,

17. Differentiate

c tanaJ ,

Show that, if y = sin 6, a; = cos 0, then

70

T A Y L O E ' S S E K I E S AND A L L I E D R E S U L T S

18. A particle moves along the axis of x so that its distance from the origin t seconds after starting is given by the formula x = a cos ht + \dkt, where a, k are positive constants. Find expressions for the velocity and acceleration in terms of t, and prove that the particle is not always moving in the same direction along the axis. Find the positions of the particle at the two times t = TT/6& and t = 13TT/6&; also find the position of the particle at the instant between these times at which it is momentarily at rest, and deduce the total distance travelled between the two times. 19. A particle moves in a plane so that its coordinates at time t are given by x = el cos t9y = e'sintf. Find the magnitudes of the velocity and of the acceleration* at time t and prove that the acceleration is always at right angles to the radius vector. Draw a rough sketch of the path of the particle, from time t = 0 to t = 77, and indicate the direction of motion at times 0, | T T , JTT,

f 77, 77.

20. A particle moves in a plane so that its position at time t is given by x = a cos pt,y — b sin pt. Find expressions for (i) the magnitude v of the velocity at time t; (ii) the magnitude / of the acceleration at time t; and prove that there is no value of t for which / = dvjdt. Prove also that the resultant acceleration makes an angle 6 with the normal to the path, where 2ab tan 0 = (a2- 62) sin 2pt. 21. A point moves in a straight line so that its distance at time t from a given point 0 of the line is x, where x= Find its velocity at time t, and prove that the acceleration is then — t2 Bint —2t cost+

2smt.

Determine the times (£>0) at which the acceleration has a turning value, distinguishing between maxima and minima. * If v,f are the velocity and acceleration respectively, then v2 = x2+y2,

BEVISIOST EXAMPLES IV

71

22. A particle moves along the axis of x so that its distance from the origin t seconds after starting is given by the formula x = a cos pt. Prove that the velocity of the particle changes direction once, and only once, between the times t = 0, t = 2TTJP and that the change of direction occurs at the point x = — a. The distance from the origin of a second particle is given by the formula . 1 n . x = a cos pt + \a cos 2pt. Write down expressions for its velocity and acceleration at time t. Show that between t = 0 and t = 2TTJP the velocity of the particle changes direction three times, and find the values of x at which these changes occur. 23. Prove that the equation of the tangent to the curve given by at the point where t = tan a is y — #tana+.tan 3 a + t a n a + l = 0. Show that the curve lies on the positive side of the line x = 1 and is symmetrical about the line y = — 1, and prove that the area bounded by the line x = 4 and the curve is ^ . 24. Prove that there are two distinct tangents to the curve

y = x*-x + 3 which pass through the origin. Find their equations, and their points of contact with the curve. Give a rough sketch of the curve. 25. A curve is defined by the parametric equations where a is a positive constant. Prove that (i) the curve passes through the points A,0,B whose coordinates are (0,6a), (0,0), ( - 3a, 0). (ii) the point t is in the first quadrant when - 1 < t < 1 and in the third quadrant when 1 < t < 2. Make a rough sketch showing the part of the curve corresponding to values of t between — 1 and + 2. Find the equation of the tangent to the curve at O, and prove that the area bounded by the arc AB and the chord AB is 27a2/2.

72

TAYLOR'S

S E R I E S AND A L L I E D

RESULTS

26. Prove that the slope of the curve whose equation is is always positive. Show that the curve has a point of inflexion where x = — 1, the slope there being \% Prove also that the tangent at the point (0,1) meets the curve again at the point (— 3, — 2). Sketch the curve, indicating clearly the point of inflexion and the tangents at the points (—1,4) and (0,1). 27. A,0,B are three fixed points in order on a straight line, and AO = p, OB = q. A fixed circle has centre 0 and radius a greater than p or q, and P is a point on this circle. Show that the perimeter of the triangle APB is a maximum when OP bisects the angle APB9 and find the corresponding magnitude of the angle POA. 28. Prove that the maximum and minimum values of the function y = x cos 3x occur when 3 tan 3x — Ijx, and discuss the behaviour of the function when x = 0. By considering the curves y = tan 3x, y = \jx, show that maximum values occur near the values x = § kir, and minimum values near x = J(2&+ 1)TT, the approximation becoming more exact as x becomes larger. 29. Given that a 2 # 4 + 62?/4 = c6, where a, b, c are constants, show that xy has a stationary value c3/>/(2a6). Is this value a maximum or a minimum ? 30. (i) A right circular cylinder is inscribed in a given right circular cone. Prove that its volume is a maximum when its altitude is one-third that of the cone. (ii) A right circular cone of height h stands on a base of radius h tan a. A cylinder of height h — x is inscribed in the cone. Prove that 8, the total surface of the cylinder, is equal to 2TT{X2 (tan

2

a — tan a) + xh tan ex},

and prove that, when tan a > ^, S increases steadily as x increases. 31. P is a variable point in the circumference of a fixed circle of which AB is a fixed diameter and 0 is the centre. Prove that,

REVISION EXAMPLES IV

73

when OP is perpendicular to AB, then the function AP + PB has a maximum value and the function AP3 + PBZ has a minimum value. 32. If/(#) is a function otx and/'(#) is its differential coefficient, show that, when h is small, f(a + h) is approximately equal to f(a) + hf'(a). Without using trigonometrical tables, find to three significant figures (i) the value of cos 31°, and (ii) the positive acute angle whose sine is 0*503. Give your answer to (ii) in degrees and tenths of a degree. 33. Calculate (i) 0, then

EEVISION EXAMPLES IV

75

By induction, or otherwise, prove that

where n is a positive integer. 43. If y = e^loga;, show that

Find the equation obtained by differentiating this equation n times. 44. Prove that, when y = eax sin bx,

and that

-p = t/(a + 6 cot bx). CbX 00

Prove that, if then

eaxsmbx

£

= T. —%xn,

c n + 2 - 2acn+1 + (a2 + b2)cn = 0,

and find the values of cl9 c2, c3. 45. If cos y = cos oc cos #, where #,^, a lie between 0 and \TT dii

du

radians and a is constant, find the values oty,-^-, -~ when x = 0. Taking a; to be so small that xz and higher powers of x are negligible, use Maclaurin's theorem to show that y as a + ^#2 cot a. Hence calculate y in degrees, correct to 0-001°, if a = 45° and a; = 1° 48'. [Take IT = 3-142.] 46. By using Taylor's theorem obtain the expansion of tan (x + \IT) in powers of x up to the term in #3. Hence calculate the value of tan 44° 48' correct to four places of decimals. [Take TT = 3-142.] 6-2

76

T A Y L O B ' S S E R I E S AND A L L I E D R E S U L T S

47. Prove that, if y = loge I (i)

:—I,

then

| -

and (ii) the expansion of y in a series of powers of x as far as the term in #4 is 2x + \xz. Find, correct to four significant figures, the value of y when x = 1° 48', taking TT = 3-142. 48. Prove that, if y = (sin"1 #)2, then

The Maclaurin expansion of 2/ in powers of x is taken to be

Given that y = 0 when a; = 0, prove that a0 = 0 and ax = 0. Prove also that aw+2 = w2an when n > 0, and hence show that

49. If y =

w"dM, prove that

3-•*•>£• Find the values of the first four differential coefficients of y when # = 1, and, by using Taylor's expansion in the form 11 deduce that the value of f ' uudu is approximately 0«1053.

h 50. If y = tana:, prove that

and find the third, fourth, and fifth derivatives of y.

BEVISION EXAMPLES IV

77

Hence find the expansion of tana; in a series of powers of x up to x5. 51. Prove that, if y = sin^ttsin^a;), then

Show that the first two terms in the expansion of the principal value of y in ascending powers of x are mx + Jm( 1 — m2)x3. 52. Find the indefinite integrals: x e~x* dx,

-

2~ fa* x* e* dx-

If u, v are functions of x, and dashes denote differentiation with respect to x, show that (uv"f + u'"v)dx — uv" — u'v' + u"v + constant. 53. Show that

\f(x)dx*= \f(a-x)dx. Jo Jo

ta,n2xdx, I- : ——)dx= Jo \l+sin2aj/ Jo and evaluate the integral. Deduce that

54. Find the indefinite integrals: f\xlogxdx, i *

C dx f\- x*~.dx. J \——,

Prove that, when the expression e^sina; is integrated n times, the result is 2-tn ex s i n (£ _ in

wj

where Pn_ x is a polynomial of degree w — 1 in a;.

78

TAYLOR'S

SERIES

AND A L L I E D

RESULTS

55. Find the indefinite integrals of

56. Find the indefinite integrals of 32cos4#,

27x2(logx)2,

~-^j->

a*.

JL — X

57. Integrate with respect to x: 4

Evaluate t h e definite integrals: C\

f*» dx Jo 1 + BOOBV

)*****'

58. Find the indefinite integrals:

59. Find the indefinite integrals: f dx J(l-3x)3

r J

2O

_

f_ J

,

1

,

f cos J2~cos 2 o;

60. Find the values of sin5^^ Jo

cos 4 a;^, Jo

ex sin 2 x dx. Jo

61. (i) Find t h e indefinite integrals of x{x+l)2' (ii) By substitution, or otherwise, prove that f1/V2 Jo

irsin-1^^ = |,

f1 x*J{x*+ l)dx = Jo

62. Find the indefinite integrals of cos3 3x,

-— , 3 —2cosa;

Xs sin x.

REVISION EXAMPLES IV

79

Use the method of integration by parts to integrate

twice with respect to x. 63. By means of the substitution x = oc cos2 0 -f j8 sin2 0, or otherwise, prove that Cfi dx Ja^{{x-oc){p-x)}^7Tt Prove also that

[

- a) (j8 -

64. P r o v e t h a t | c o s 2 Odd = JTT,

Jo

|COS4

Jo

§77.

Find the area bounded by the curve r = a(l + cos0) and determine the position of the centre of gravity of the area. 65. Find the equation of the normal at (the point (f, sin £) to the curve whose equation is y = sin x. Prove that, if £ lies between 0, TT, the normal at P divides the area bounded by the #-axis and that arc of the curve for which 0 ^ x ^ 77 in the ratio (2 - cos £ - cos3 £) : (2 + cos £ + cos3 £). - 2 + v2= l a2 b2 is rotated through two right angles about the #-axis. Prove that the volume generated is ^Trab2. (i) Prove that, if a and 6 are varied subject to the condition a + b = £, then the greatest volume generated is 2TT/81. (ii) The volume is cut in two by the plane generated by the rotation of the y-axis. Prove that the centre of gravity of either part of the volume is at a distance fa from the plane of separation. 66. The ellipse r

67. The complete curve x2/a2 + y2\b2 = 1 is rotated round the 2/-axis through two right angles. Find the volume generated by the area enclosed by the curve.

80

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

The semi-axes a and b of the curve are each increased by €. Prove that, if e is small, the increase in volume is approximately 68. OA is a straight rod of length a in which the density at a point distant x from 0 is b + ex, where b and c are constants. Find the distance of the centre of gravity of the rod from 0. 69. The gradient at any point (x, y) of a curve is given by dx and the curve passes through the point (2,0). Find its equation and sketch the graph, indicating the turning points. Find the distance from the y-axis of the centre of gravity of a uniform lamina bounded by the curve and the positive halves of the x and y axes. 70. A lamina in the shape of the parabola y2 — 4ax bounded by the chord x = a is rotated (i) about the axis of y, (ii) about the line x = a. Prove that the volumes generated in the two cases are 71. Integrate with respect to x: sin2#, sin3 x9 x2 cos x. The portion of the curve y = sin a; from x = 0 to x = \n revolves round the axis of y. Prove that the volume contained between the surface so formed and the plane y = 1 is JT^TT2 — 8). 72. The coordinates of a point on a curve referred to rectangular axes are {at2,2at), where t is a variable parameter which lies between 0 and 1. Make a rough graph of the curve. Calculate (i) the area enclosed by the curve and the lines x = a, y = 0; and (ii) the area of the surface obtained by revolving this part of the curve about the #-axis. 73. Find the area contained between the #-axis and that part of the curve x = 2£2 -f 1, y = t2 — 2t which corresponds to values of t lying between 0 and 2. Find also the coordinates of the centroid of this area.

REVISION EXAMPLES IV

8J

74. Prove that the tangent to the curve

Jx + Jy = Ja at the point (x,y) makes intercepts ^{ax)^(ay) on the axes. A solid is generated by rotating about the #-axis the area whose complete boundary is formed by (i) the arc of this curve joining the points (a, 0), (0,a), and (ii) the straight lines joining the origin to these two points. Prove that, when this solid is of uniform density p, its mass M is x^Trpa3. Prove also that the moment of inertia of the solid about Ox is 75. Find the area bounded by the curve x = 2a(t3 — l),y = Sat2 and the straight lines x = 0, y = 0. Prove that the tangent to the curve at the point t = 1 meets the curve again at the point t = — | , and find the area bounded by the parts of the tangent and of the curve that lie between these points. 76. Prove that the parabola y2 = 2ax divides the area of the ellipse 4#2 + 3y2 = 4a2 into two parts whose areas are in the ratio 77. The portion of the curve y2 = 4ax from (a, 2a) to (4a, 4a) revolves round the tangent at the origin. Prove that the volume bounded by the curved surface so formed and plane ends perpendicular to the axis of revolution is ^-Tra3, and find the square of the radius of gyration of this volume about the axis of revolution. 78. Find the coordinates of the centre of gravity of the area enclosed by the loop of the curve whose equation is r = acos20, which lies in the sector bounded by the lines 6 = ± \TT. Find also the volume obtained by rotating this loop about the line 0 — ^77. 79. Find the coordinates of the centre of gravity of the loop of the curve traced out by the point x = 1 — t2, y = t — ts. Find also the volume obtained by rotating this loop about the line x = y. 80. A plane uniform lamina is bounded by the curve y2 = 4a# and the straight lines y = 0, x = a. Find the area and the centre of gravity of the lamina.

82

T A Y L O R ' S S E R I E S AND A L L I E D R E S U L T S

The lamina is rotated about the axis Ox to form a (uniform) solid of revolution. Find the centre of gravity of the solid and, assuming the density of the solid to be p, find its moment of inertia about the axis Ox. 81. A lamina in the shape of the parabola y2 = 4=ax, bounded by the chord x = a, is rotated (i) about the axis of y, (ii) about the line x = a. Prove that the two volumes thus generated are in the ratio 3 : 2. 82. Prove by integration that the moment of inertia of a uniform circular disc, of mass m and radius a, about a line through its centre perpendicular to its plane is |ma 2 . The mass of a uniform solid right circular cone is M, and the radius of its base is a. Prove that its moment of inertia about its 2 . axis is 83. Evaluate

Jo Jo

cosn0d0 when n = 1,2,3,4.

The area bounded by the axis of #, the line x = a, and the curve x = asin0, y = a(l — cos 0), from 0 = 0 to 0 = \n, revolves round the axis of x. Prove that the volume generated is ^7ra3(10 — 3TT). 84. The portion of the curve x2 = 4:a(a — y) from x = — 2a to x = 2a revolves round the axis of x. Prove that the volume contained by the surface so formed is ffrra3, and find its radius of gyration about the axis of revolution. 85. Sketch the curve r = a(l + cos#), and find the area it encloses and the volume of the surface formed by revolving it about the line 0 = 0. 86. Find the three pairs of consecutive integers (positive, negative, or zero) between which the roots of the equation lie, and evaluate the largest root correct to two places of decimals. 87. Show that the equation x*-3x-7 = 0 has one real root, and find it correct to three places of decimals.

BEVISION EXAMPLES IV

83

88. Show that the equation has one real root, and find it correct to three places of decimals. 89. For a function / having an Nth differential coefficient, Taylor's theorem expresses f(a + x) as a polynomial of degree JV— 1 in #, together with a remainder. State the form taken by this polynomial, and one form of the remainder. Prove by induction, or otherwise, that dn -=— (e* sin 3^3) = dX

Hence find the coefficients an in the Maclaurin series Hanxn of the function e^sina;^. By means of Taylor's theorem, show that when x>0 the JV-1

difference between exsinx^j3 and 2 anxU *s n o ^ greater than n - 0

(2x)Nex

[A proof showing that the difference is not greater than (2x)Nex is acceptable if the form of remainder which you have quoted leads to the result.] 90. If show that

y (l + x)2-^+(l+x)y=

1.

dX

Deduce the first four terms in the Maclaurin series for y in powers of x. 91. Show that y = {x + ^ ( l + x2)}k y'y](l+x2) = ley, satisfies the relations Deduce the expansion

Verify that this agrees with the series derived from the binomial series when h = 1.

CHAPTER IX THE HYPERBOLIC FUNCTIONS 1, The hyperbolic cosine and sine. There are two functions with properties closely analogous to those of cos x and sin x. They are called the hyperbolic cosine of x, and the hyperbolic sine of x, and are written as cosh x and sinh#. We define them in terms of the exponential function as follows: cosh# = \{ex-\-e~x), sinhz =

\{ex-e~x).

We establish a succession of properties similar to those of the cosine and sine: (i) To prove that

cosh2 a; — sinh2# = 1.

The left-hand side is

l{{e2x + 2 + e~2x) - (e2x - 2 + e~2x)}

= 1. (ii) To prove that cosh (x + y) — cosh x cosh y 4- sinh x sinh y It is easier to start with the right-hand side: (e* + e~x) (& + e~y) + {ex - e~x) (e^ - e~y)} == i{(ex+y + ex~y + e-x+y + e~x-y) + (ex+y - ex~y -

= cosh (x + y).

THE HYPERBOLIC COSINE AND SINE

85

(iii) Similarly sinh (x + y) = sinh x cosh y + cosh x sinh y. (iv) As particular cases of (ii), (iii), cosh 2x = cosh2 x + sinh2 x, sinh 2# = 2 sinh a; cosh a\ COROLLARY

(i). cosh 2 x = |(cosh 2x + 1), sinh2 a; = ^(cosh2# — 1).

COROLLARY

(ii). Since cosh a;—1 = 2 sinh 2 -

and sinh2 ~ is positive, it follows that cosh x is greater than unity for all (real) values ofx, i.e. Note.

cosh x ^ 1. coshO = 1, sinh 0 = 0.

(v) To prove that

cosh a; = cosh (— x).

The right-hand side is = cosh a;. (vi) To prove that For

sinh x = — sinh (— x).

sinh ( - x) = l{e~x - e-) = i(e-* - ex) = —sinh a:.

iV'o^e. Sinh a; is positive when a; is positive, and negative when x is negative. For example, if x is positive, then ex is greater than e~x, since e is greater than 1. Hence \(ex — e~x) is positive. (vii) To prove that -r~ (cosh a;) = sinh a:, CLX

•j- (sinh x) = cosh a;. ax

86

THE HYPERBOLIC FUNCTIONS

We have

-=- (cosh x) = - -=- (ex + e~x) dxK

'

2dx

and

= cosh a;, (viii) To prove that coshrfa; = sinha;,

= cosh#. These results follow at once from (vii). (ix) By theformulce of (vii), we have the relations -=-s (cosh a;) = cosh#, cLx

-r-g (sinh#) = sinha;. dx Thus cosha:, sinha; both satisfy the relation dx*

y

(x) The following expansions in power series are immediate consequences of Maclaurin's theorem: 1

x2

+

x* xB +

+

The series converge to the functions for all values of x.

OTHER HYPERBOLIC FUNCTIONS

87

2. Other hyperbolic functions. The following functions are defined by analogy with the corresponding functions of elementary trigonometry: sinh x cosh a;' KJCVXIXX U/

cotha;

cosh a? sinh a;'

cosecha;

1 sinh x9

seen a;

1 ~~ cosh a;"

sech 2 # + tanh 2 # = 1,

The relations

are found by dividing the equation cosh2 x — sinh2 x = 1 by cosh2#, sinh2 a; respectively. Note the implications sech2 xl, there are TWO values of cosh"1 a;, equal in magnitude but opposite in sign. We can express cosh"1 a; in terms of logarithms as follows: y = cosh"1 a;,

If then

x = cosh y

so that

e*v-2xev+l = 0.

Solving this equation as a quadratic in ev, we have and so, by definition of the exponential function, y = log{x±J(x2-!)}. These are the two values of y. Moreover their sum is

= 0.

Hence the two roots are equal and opposite. The positive root is log {x + 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 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

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