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EXAMPLES AND SOLUTIONS 
 
 OF THE 
 
 DIFFERENTIAL CALCULUS. 
 
 JAMES HADDON, M.A., 
 
 Second Mathematical Master of King s College School. 
 
 Author of " Rudimentary Algebra," " Rudimentary Arithmetic," " Rudimentary 
 Book-keeping," &c &c. 
 
 LONDON: 
 VIETUE AND CO., 26, IVY LANE, 
 
 PATERNOSTER ROW. 
 
H3 
 
 5TAT 
 INTRODUCTION. 
 
 THE Doctrine of Limits is now very generally adopted as 
 the basis of the Differential and Integral Calculus. 
 
 Of the methods which were formerly in use it may be 
 advantageous to the mathematical student to glance at some 
 of the most prominent. 
 
 By inscribing successively in a circle, regular polygons of 
 four, eight, sixteen, thirty-two, &c. sides, we may at length 
 suppose a polygon to be inscribed whose area shall be less 
 than that of the circle by a quantity so small as to be unas 
 signable. In this manner the area of the circle may be said 
 to be exhausted. Hence, the method which was based upon 
 this mode of operation was termed the Method of Ex- 
 Jtaustions. 
 
 In the early part of the seventeenth century a work was 
 published, in which all quantity was assumed to be composed 
 of elements so small that it would be impossible to divide 
 them. An infinite number of points in continued contact 
 were supposed to form a line, an infinite number of lines to 
 form a surface, and an infinite number of surfaces to form a 
 solid. Now, since a line has magnitude, namely, lengthy 
 and a point has no magnitude, it is obvious that a line 
 cannot properly be considered to be made up of a series of 
 
 02848 
 
IV INTRODUCTION. 
 
 points. The method founded upon these suppositions is 
 consequently objectionable. Cavalerius, the inventor of it, 
 called his work " Geometria Indivisibilibus ;" and hence this 
 method was styled the Method of Indivisibles. 
 
 Sir Isaac Newton considered all quantity to be generated 
 by motion j a point in motion producing a line, a line in 
 motion producing a surface, and a surface in motion pro 
 ducing a solid. This motion or flowing of a point, a line, 
 and a surface, gave rise to the terms fluents and fluodons : 
 the quantity generated by the motion being called influent 
 or flowing quantity, and the velocity of the motion, at any 
 instant, the fluodon of the quantity generated at that instant. 
 The method founded upon these considerations has been 
 long known as the Method of Fluxions. 
 
 As applications of this method are continually met with 
 in mathematical works, it may not be inappropriate to give 
 a few instances of its notation, compared with that proposed 
 by Leibnitz, and now generally adopted by writers on the 
 Differential Calculus : 
 
 n . n 
 
 ii, it, if, u, u, suicT, {(x 2 l) m } . 
 du, d 2 u, d s u, d*u, d n u, dsinz, d n (x 2 -l) m . 
 
 The fluxional symbols in the first line are placed exactly 
 over the corresponding differential symbols in the second. 
 
 Leibnitz considered every magnitude to be made up of an 
 infinite number of infinitely small magnitudes. His mode of 
 reasoning was as follows. Any quantity u consists of an infi 
 nite number of differentials, each equal to ph + qk 2 + rh 3 + &c., 
 and h being infinitely small, each term in the series is infi 
 nitely less than the next preceding term, and consequently 
 the sum of the terms after the first is infinitely less than 
 
INTRODUCTION. V 
 
 that first term. Hence ph is the only term necessary to be 
 retained to represent the series. 
 
 Lagrange, in his " Calcul des Fonctions," endeavoured to 
 simplify the subject by rejecting the consideration of infi 
 nitely small differences and limits, referring the Differen 
 tial Calculus to a purely algebraic origin. He defined the 
 differential of a quantity to be the first term of the series 
 ph+qk 2 + r/t 3 + &c. This is the foundation of his theory. 
 
 Each of these methods has found numerous advocates 
 among mathematicians, a fact which excites no surprise 
 when we consider the extraordinary genius of the great men 
 whose names are associated with the origin of these various 
 and most interesting theories. 
 
 In our own day several highly talented men have directed 
 their attention to this subject, and it seems now to be very 
 generally admitted that the method best adapted to ele 
 mentary instruction is that founded on the Doctrine of 
 Limits. 
 
 Among the valuable works which have recently enriched 
 this subject may be mentioned those of Whewell, Hall, 
 O Brien, De Morgan, Thomson, Young, Price, and Walton, 
 in our own language, and Duhamel, Cauchy, Moigno, and 
 Cournot, in the French. 
 
 Let us suppose a certain magnitude u to be dependent 
 for its value upon some variable magnitude x, so that the 
 value of u may be represented by some expression into which 
 x enters, then u is a function of x. "We will assume, for 
 instance, that u=aP, and, in this simple example, supposing 
 a- to undergo a change of value, we will trace the corres 
 ponding effect produced upon the function u. 
 
 Let x take the increment h, that is, let x change its value 
 A 2 
 
VI INTRODUCTION. 
 
 and become x-\-h, then if we represent the corresponding 
 value of u by u t , we shall have 
 
 .*. u t u=3 x 2 h -f- 3 xh 2 + h 3 = corresponding increment of u, 
 -1 = 3x 2 -\-3xk + k 3 = ratio of increment of function to 
 
 increment of variable. 
 
 Now the first term of this expression for the ratio being 
 3# 2 , it is obvious that h may undergo any change of value 
 whatever, without affecting this first term. 
 
 Let h then continually decrease in value until it is=0, 
 then the expression for the ratio will be simply 3# 2 . 
 Hence this first term is the limit towards which the ratio 
 approaches as h is diminished, and which limit the ratio 
 cannot reach until h = Q. 
 
 Now if u=x 3 , du = 3x 2 dx, -=3x 2 , where du is 
 
 dx 
 
 the differential of u, dx the differential of x, and -- the 
 
 dx 
 
 differential coefficient derived from the function, that is the 
 coefficient of dx. Thus the limit 3 x 2 is equal to the differen 
 tial coefficient. 
 
 These remarks are offered to the reader in this place, not 
 only with a view to remind him of what the Method of 
 Limits is, and to regard it in its connexion with the methods 
 above alluded to, but also in the hope of inducing him con 
 stantly to recollect that, when he is performing that very 
 common operation in the Differential Calculus of ascertaining 
 the differential coefficient, he is virtually seeking the limit of 
 the ratio of the increment of the function to the increment 
 of the variable. 
 
CONTENTS. 
 
 Page 
 
 Algebraic functions of one variable, differentiation of, examples . 1 
 Asymptotes to curves, to determine the, examples 
 Bernoulli s formula for the expansion of functions . 
 
 method of evaluation of indeterminate functions . 30 
 
 Binomial theorem, examples 21 
 
 Co-ordinates, polar 
 
 Curvature of curved lines . . . .129 
 
 radius of described 
 
 examples 130 
 
 Curves, asymptotes to, examples 82 
 
 normals to, examples 79 
 
 subnormals to, examples 79 
 
 subtangents to, examples 80 
 
 tangents to, examples ....... 
 
 tracing of, examples 100 
 
 De Moivre s formula of the expansion of functions . . .27 
 Differentiation of functions of one variable 
 
 of two or more variables . . .57 
 
 successive 15 
 
 elimination of functions by . . . .64 
 
 Elimination of constants and functions by differentiation, examples 64 
 Envelopes to surfaces and lines, examples of . . . .147 
 Euler s deductions of Taylor s theorem . . . . . 19 
 theorem for the integration of homogeneous functions, 
 
 examples . . . ^ 62 
 
 Evaluation of indeterminate functions 
 
 Bernoulli s theorem 30 
 
 Evolutes to curves, examples 
 
 Exercises, miscellaneous 
 
 Expansion of functions. Bernoulli s formula .... 
 
 binomial theorem . . . .21 
 
 . De Moivre s formula ... 27 
 
 Euler s formula 19 
 
 , . exponential theorem .... 22 
 
 Maclaurin s theorem . . . 17, 20 
 
 Taylor s theorem . . . . 16 
 
 Exponential theorem formula 22 
 
 Functions, development of, Maclaurin s theory *. . . .20 
 Machin s formula .... 23 
 
IV CONTENTS. 
 
 Pago 
 Functions, elimination by differentiation, examples . . .64 
 
 expansion of, Taylor s theorem 23 
 
 differentiation of 1, 57 
 
 maxima and minima of ..... 35, 68 
 
 successive differentiation of, examples . . .15 
 
 algebraic, differentiation of, examples ... 1 
 
 homogeneous, theorem for the integration of . . 62 
 
 indeterminate, evaluation of 30 
 
 transcendental, differentiation of, examples . . 8 
 
 Homogeneous functions, Euler s theorem for the integration of, 
 
 examples 62 
 
 Implicit functions, expansion of 28 
 
 Indeterminate functions, evaluation of ..... 30 
 
 Bernoulli s system 30 
 
 Lagrange s notation of Taylor s theorem 17 
 
 Leibnitz s theorem of successive differentiation of functions 
 
 enunciated 16 
 
 Lines and surfaces, envelopes to, examples . . . .147 
 
 Machin s formula for the development of functions, examples . 23 
 Maclaurin s theorem for the development of functions, formula 
 
 and examples 20 
 
 theorem of the expansion of functions . . .17 
 
 Maxima and minima of functions of one variable . . . 35 
 
 of functions of two or more variables, examples . . 68 
 
 of implicit functions of two or more variables, examples 55 
 
 term defined 35 
 
 Miscellaneous exercises 158 
 
 Normals to curves, examples 79 
 
 Polar co-ordinates, examples 95 
 
 Kadi us of curvature 129 
 
 to find the . 130 
 
 Singular points 100 
 
 Spirals, equations to, examples 95 
 
 Subnormals to curves 79 
 
 Subtangents to curves 80 
 
 Successive differentiation of functions, examples of . . .15 
 
 Leibnitz s theorem 16 
 
 Surfaces and lines, envelopes to 147 
 
 Tangents to curves, to find the equations of . . . .78 
 Taylor s theorem of the expansion of functions described . . 16 
 
 examples and formula? 16 
 
 Euler s formula and deductions of . . . .19 
 
 Lagrange s notation of 17 
 
 Tracing of curves . . 100 
 
 Transcendental functions, differentiation of .... 8 
 Vanishing fraction defined 30 
 
EXAMPLES 
 
 ON THE 
 
 DIFFERENTIAL CALCULUS. 
 
 CHAPTER I. 
 
 DIFFERENTIATION OF FUNCTIONS OF ONE VARIABLE. 
 
 Ex. (1.) Let u=ax. Then =a. 
 
 dx 
 
 (2.) Let u =a + x. Then =4. 
 
 (3.) Let y=3ax 2 + l 2 . Then -=2 x 3ax= 
 
 (jLOC 
 
 (4.) Let u= S&=&. Then = 
 
 dx 
 
 (5.) Letw=-^ - 5 . Then = 
 a 2 x 2 dx 
 
 (6.) M= 
 
2 ALGEBRAIC FUNCTIONS 
 
 (7.) =. 
 
 (8.) W=: 
 
 (9.) u=(a + bx 
 ~= n(a + b 
 
 (10.) = 
 da . f 
 
 m ~*=l m n (a 
 
 2 V^f 2 
 
 (11.) M = 
 
 o <*" 
 
 2w = 
 
 ?\/l + -. 
 
 X 
 
 Squaring, we have 
 
 .= ! 
 
 /IHH? 
 
 i+- " 
 
 + ^ 2 2 vl + ^ 
 
OF ONE VARIABLE. 
 
 (12.) K Multiplying both numerator and 
 
 va 2 + x 2 x 
 
 denominator by ^a 2 + x 2 + x, we have 
 
 dx 
 
 (13.) =:(a + a? 
 
 rfM 
 
 - 
 
 (14.) == 
 
 
 m(l +x n ) m - nx n 
 + (1 +x n } m . n (1 + m ) w " 1 . 
 =mn(l +x m )*- 1 . (1 +a; n ) m - 1 . {(1 +ar in 
 
ALGEBRAIC FUNCTIONS 
 
 (15>=V fa b 7 =+V(c 2 -x 2 ) 2 \ 3 =L ^+( c 2_ 
 
 V V x i \ v x 
 
 
 , I* x 
 
 \2x\/x ~ 3 t/c 2 x 2 
 
 3b 
 
 a 
 
 V iC 
 
 V CC 
 
 "ccx/l-ic 2 
 
 du 
 
 Tx 
 
 (17.) u=/\/ a + x+\ 
 
 &c. in inf. 
 c. in inf. 
 
OF ONE VARIABLE. 
 
 =, 
 
 dx dx 
 
 du 1 
 
 But v M 2 u=a-\-x, 
 
 Tl 2 
 
 (18.) =1 + -^ 
 1 + 
 
 i - t f 
 
 1 -f- &c. in im. 
 
 --- 
 
 -2= ~2- 
 ,u 1 
 
 (19.) 2wC + w 2 &ic 2 =0. This is an implicit function. 
 
 -r-= > T~ - ; 
 
 dx dx au + x 
 
 B 2 
 
ALGEBRAIC FUNCTIONS 
 
 But v bx 2 ux= 
 a bxu_u 
 " au + x x 
 
 (20.) u= 
 (21.) w= 
 
 (22.) u=c-2x*. 
 (23.) u=2x 2 -3x+G. 
 (24.) ^=4^- 
 (25.) u= 
 (26.) w= 
 
 m \ ? - 
 
 ~\l 
 
 ^28 \ 7 
 (29.) M= 
 
 du 
 dx 
 
 f du u 
 " dx~ x 
 
 (30.) a? 
 
 (31.) =(! + *) 
 
 
 dx 
 
 
 dx 
 
 dx 
 
 
 
 rfo; 
 
 dw 
 
 ^ 
 du_ 
 dx~ 
 
 1 3sc 
 
OF ONE VARIABLE. 
 
 (33.) u= 
 
 Vx 2 +l + 
 
 du 2x 2 + 1 
 
 ..! 
 dx 
 
 (34) u= 
 (35.) u=(a 
 
 dx 2(1+ 
 
 du 
 
 du 
 
 , 
 (36.) w= a3v ax x*. = - 
 
 3a dx 3a 2 Vax-x 2 
 
 dx 
 
 (38.) = 
 (39.) = 
 (40.) u= 
 
 (xa)-* dx 2 V x a 2 
 Vs^+lx du 2% 
 
 dx \/x 2 +l.(\/x 2 +l+x) 2 
 du 3a 2 +4:ax-x 2 
 
 v a x dx 6 (a 2 
 
 /,! x / o o\ / o ox 1 ^ 
 
 - 1 (41.) u=x(a 2 + x 2 ) (a 2 x 2 )^. = 
 
 dx V 
 
 (42.) u =s\/2x-I\/2x-I ^2x 1 &c. 
 
 in inf. 
 
 (43 .) = 
 
 1 &c. in inf. 
 
TRANSCENDENTAL FUNCTIONS 
 
 (44.) tt =- ~^_ r 
 
 l + (l-4a;)*-2a: 
 
 CHAPTER II. 
 
 TRANSCENDENTAL FUNCTIONS OF ONE VARIABLE. 
 
 du 
 
 Ifu = SlTLX } = COSC. 
 
 -= sin x. 
 
 au 
 u=ta,nx; ^^ 
 
 u=cotx; = (l-j-cot 2 #) = cosec 2 o; = r j 
 
 du 
 
 u=secx: -=sec x. tana;. 
 dx 
 
 du 
 
 u=cosecx =cosecx. cotx. 
 dx 
 
 du 
 
 u=v. sin x : =sm x. 
 dx 
 
 du a 
 
 du 
 
 Ex. (I.) Lettt=sin 2 #. Then^=2smo^^^=2sina.cosa:. 
 
 dx dx 
 
OF ONE VARIABLE. y 
 
 (2.) tt=cosm#. i.e. the cosine of the product of 
 m and x. 
 
 du dcosmx dmx 
 
 = = sm mx . j = m smmx. 
 
 dx dx dx 
 
 (3.) 
 
 du dcosx 
 
 f- cos x. 
 
 dx dx dx 
 
 -cos#. 3sin 2 #. cos x 
 sin 4 ir = sin 2 a7 (3 cos 2 .r sin 2 ^) 
 = sin 2 # (3.1 sin 2 a7 sin 2 ^) = sin 2 # (33 sin 2 a; sin 2 # ) 
 
 = sin 2 ^ (34 sin 2 ir) . 
 
 (4.) u=e x . cos x, e being the base of the Napierian 
 system of logarithms. 
 
 du r c?cos# de x 
 
 -=e^. - -- j-co&r--^ 
 ax dx dx 
 
 =e x . (sin x) + cosx. e^=^(cosa; sin^r). 
 (5.) u=x.e cosx . 
 
 du >de cosx dx dcosx 
 
 - . - 
 
 dx dx dx dx 
 
 (6.) = 
 
 COS 7 * X 
 
 ~ ] d?( siiuc) 
 
 . _L. 
 
 ^ ^ n 2w. _. " 
 
 cos n ~ l x COS M+ 
 
10 TRANSCENDENTAL FUNCTIONS 
 
 7. M=cos~ 1 .rvl a 2 . This is an inverse function. 
 
 /y A/ "I /y*2 /v T^VlATt ?/ ^ /"*rG! 1 X * * r*ACl -?/ 
 
 Ju v JL "~~ fct- ^^ ^w JL ilt?JJL W 1 OUS J5 ^ OC/> it 
 
 rf__ .J^_ 1 1_ 1 
 
 dz dz sin ?i vl cos 2 ^ vl 
 
 1 
 
 
 But Y *=# v1 a; 2 , :. = 
 
 1 
 
 -f- \l 
 
 1-tf 2 
 
 du du dz 
 
 Hence = 
 
 dz dx 
 
 (8.) u=a(s : mx 
 du 
 
 , \ o i 
 
 = a (cos x + sm a?) . Squaring, we have 
 
 2 si 
 
 (9.) 
 
 =a 2 (l +2 
 = 2 l 2 si 
 
 Put ^ for log x n , then M=^ TO , /. -= 
 
 dz 
 
 dz nx n ~ l n 
 and v *=!<**, .-._ = _;-=... 
 
 </M du dz n 
 
 Hence -7- = -; --- =mz m - l --= 
 dx dz dx x 
 
 x 
 
OF ONE VARIABLE. 11 
 
 (10.) log= 
 
 du 
 
 dx u X* 
 
 du 
 
 dx x 1 ^/ \-\-x 1 
 (11.) ua^^\ 
 
 ~ 1 ^. log e 
 
 - 1 ^, v log 0=1, 
 
 _ cos^r) 
 
 - " a2 + l 
 
 Since the denominator is constant, and since the differen 
 tial coefficient of e ax is 
 
 :.= 
 
 
12 TKANSCENDENTAL FUNCTIONS 
 
 /io\ i ^+ N/tf 
 
 (13.) W =log -- - 
 
 Va v x 
 
 _ >v x 
 
 dx (vaVx} 2 
 
 "/a v #+ V# + v^ 2 v a 
 
 2 v ^ ( v^a v 7 ^ 
 (14.) M =a< i+ *. 
 
 log w=c* s+<r . log a. 
 log (log M) =(^ + ar) log c + log (log a). 
 
 Put.?=logM. then = = , 
 
 du u a * + * 
 
 log z=(x 2 +x) log c + log (log a), 
 -f =logc(2#+l). But^=c* a 
 
 /.^=^. logc (2a?+l)=loga. logc. 
 =~=loga. log*. a* >+ 
 
 \ i / r du / \ / 
 
 .) w=i^ma?lv 1 cos#. =^os#x l-j-sm#. 
 
 n a \ f \ du / \ 
 
 (io.) ?=cos(sm^). -= cos x sin (sin x). 
 
 (17.) =sin-C 
 
 (18.) =, 
 
 du 
 
 ~j 
 
 dx 
 
 _ -- 
 
 V (t j 
 
OF ONE VARIABLE. 13 
 
 xl-* 2 du 2 
 
 o (19.) tt=sm 1 T 5- -=r - 
 
 l+x 2 dx \-\-x- 
 
 /x 2 a 2 du x 
 
 .) w=siii- 1 A/ -- = 7= - =- 
 
 V 2_ a 2 ^ v/(^_ a 2) ^2_^2) 
 
 (20.) 
 
 (21.) tt=cosa?+cos2#+cos3#+ &c. 
 
 2 sin 2# + 3 sin 3# 4- &c.) . 
 o 
 
 (22.) =cot-> 
 
 /no \ f\x 
 
 (23.) u=taiL~ l A/ -- 
 V 
 
 du 
 
 i +af dx 
 
 /-< \ i fbx + a du vba 
 
 (24.) w^tan- 1 A/ - 1 = - 
 
 V ^_^ ^ 2l^v/ 
 
 /oer \ i 
 
 (25.) =i_ 
 
 ~~ 2 
 
 (27.) w= 
 
 /og \ _ /i \ du_\ 
 
 dx x 
 
 /on \ i x du I 
 
 \"v-) W = lOg 
 
 ax x 
 du 1 
 
 (30.) u= 
 
 * This expression means the i th logarithm of x, not the w th power of 
 the logarithm of x. Log (logic), which means the logarithm of the 
 logarithm of x, might be written (log) 2 a;. 
 
 C 
 
14 
 
 TRANSCENDENTAL FUNCTIONS. 
 
 Jl.) u=\ogx log(rt Va 2 x 2 ). ~7~: 2 :, 
 
 , du 
 
 t r (32.) wr=logvsm#+logv cos#. =cot2^ 
 
 (33.) = 
 
 (34.) =log 
 
 (35.) = . 
 
 (36.) u=^"t 
 
 (37.) u=^. 
 
 (38.) u=x 
 
 (39.) u= 
 
 (40.) M=iT8 to 
 
 (41.) =*: 
 
 dx 
 
 C?tfc 
 
 e/^7 
 
 du a lo % x . log a 
 
 c&6 /sna? , \ 
 
 ._&/ -- |-cosa?. logir 
 c^r \ a? 
 
 c?w . N 
 
 - = e a * (a cos rx r sin rx] 
 dx 
 
 c?w 
 d^ : 
 
 du 
 
 (42.) f*s=* r (4BBr*), = 
 
 (43.) u =ed& n *. 
 (44.) ^=, 
 
 c^tt 1 - te\ 
 ~dx~~x* XX S W/ 
 
 (d sin rx + mr cos rx] 
 
 
 
 -!(** (log) *... (log)- 1 * 
 
 being a function of ^. 
 
SUCCESSIVE DIFFERENTIATION. 15 
 
 (45.) u=z vV , z, v, and y being functions of x. 
 
 du r, dy y. dv 1 dz 
 
 - r =z vJ . MJtar*. logv -+ - log;? H r 
 
 dx \ do? c <fe 
 
 CHAPTER III. 
 
 SUCCESSIVE DIFFERENTIATION. 
 
 Ex. (1). Let ?/=^ n . 
 
 Differentiating, we obtain the/rstf differential coefficient, 
 
 du . 
 =na?-\ 
 dx 
 
 Differentiating, we obtain the second differential coefficient, 
 
 Differentiating as before, we have the third, 
 
 __ -....--. 
 
 It must be borne in mind that d^u, d^u, d 4 u, &c. c? r w are 
 merely symbols ; and that dx 2 , dx*, dx 4 , &c. dx r are powers 
 of dx. 
 
16 LEIBNITZ S THEOREM. 
 
 (2.) u=logx. 
 
 du 1 d 2 u 1 d s u 
 
 dx x* dx^ a? 2 dy? a 
 
 d 4 u_ 2-3 ^ 5 ^_2-3-4 
 
 dx^ x^ dx^ x^ 
 
 (3.) ua x . 
 
 3 .? ( 4 -) w = si 
 1 LI ( 5 -; u=e a *. 
 
 o? vi 7.3...^ w - 
 
 (6 - } w= w -i^- ^ = (i^^- 
 
 Leibnitz s Theorem, whicli is useful in finding the diffe 
 rential coefficient of the product of two or more simple 
 functions, may be thus enunciated, u and v being both func 
 tions of x y 
 
 d r (uv] d r u dvd r - l u r(r -1} d*v d r ~*u 
 
 CHAPTER IV. 
 TAYLOR S THEOREM. 
 
 This theorem may be thus enunciated. 
 If u=f(x), and x take the increment //, 
 
 du. d^u h 2 d*u // 3 d n u h n 
 
 tt + _A + ^_ + ^- s . TT3T5 +..._ TT1 
 
TAYLOR S THEOREM. 17 
 
 This theorem, written according to the notation of 
 Lagrange, is 
 
 In using it, if we take n terms of the series, the error we 
 shall commit by leaving out the terms beyond the n i} \ will lie 
 
 h n 
 between the greatest and least values of f {n) (x+ 6/1) - - > 
 
 which values will depend upon giving to 6 various values 
 between (0) its least value, and (1) its greatest. 
 
 Maclaurin s Theorem is easily deducible from this. 
 
 Ex. (1.) Expand cos (x + ft) in a series of powers of h. 
 
 du d 2 u d^u 
 
 Let?^:=cos#, then = sm#, -^= cos x, Q =sm#, &c. 
 dx da? doP 
 
 Whence, substituting these values of u, -> &c. in Taylor s 
 
 CLOO 
 
 theorem, we have 
 
 h? * 
 
 :. ii cos^- 
 
 Cor. By making #=0, we have 
 A 2 A 4 
 
 (2.) Expand sin" 1 (x + h), according to ascending powers 
 ofh. 
 
 Let %=sin~ 1 #, then = - -- = (1 a 2 )"*, 
 
 dx v 1 
 
 d 2 U 1 - 9x-^/ n \ /i 9\- X 
 
 _=--(! -^ *(- 2 ,)=,(l-^) .=__ 
 
 c 2 
 
18 TAYLOR S THEOREM. 
 
 Whence, by substitution in the theorem, 
 
 h 
 
 &c 
 
 2- 3(1 -a?)* 
 (3.) Expand log (# + /*) by Taylor s theorem. 
 
 _ du 1 d 2 u 1 d^u 2 
 
 Let u=.\ogx, then -=-, -j-^= -- ^, - rT =-, <fcc. 
 dx x doc 2 x 1 dx* cc 3 
 
 Whence, by substitution, 
 
 h h 2 A 3 
 
 log^ + - o-2 + o-3-&C. 
 X AJf/ O3U 
 
 .) Iiu=f(ai) t show that 
 
 JM x 2 d 2 u x* 
 
 
 r 4 
 
 7.2 ^ 13 
 
 TT , ~\9 
 
 =/(*), /(*+/*)= 
 
 Substituting these values in Taylor s theorem, we have 
 
 x \ du x 2 d 2 u x 4 
 \ _ ^ t I t 
 
 d*U X* 
 
 (5.) If/(^)=tan~ 1 a ; and we put -^=sin 
 
EULER S FORMULA. 19 
 
 k 
 or tan-%=~y, then, tair^ + A^tair^ + siiiy siny - 
 
 2t 
 
 h 2 & 
 
 sin 2y sin 2 y + sin 3y sin 3 y y &c. 
 
 Now, since h may have any value whatever, put h=x, 
 y being an arc in the first quadrant ; then 
 \MQT\X + /O = tan- 1 = 0, 
 
 . . tan~ 1 iP= siny siny ^ + sin 2y sin 2 y + sin 3y sin 3 y + &c. 
 
 TJ- cos ? 
 
 But tan- 1 ar= -y, and ^=coty=- 
 
 /. ^=y + siny cosy + - sin 2y . cos 2 y -f - sin 3y cos 3 y + &c. 
 2, m o 
 
 Similarly, putting A=-(* +1) = - -^- , -e have 
 
 TT siny 1 sin2.?/ 1 sinS.y 
 
 - ~> -- T 77 - ^ 
 
 - ^ 
 
 > 
 
 2 cosy 2 cos-y 
 And, putting h= \/l + # 2 , 
 
 - sn y + - sn 
 2 
 
 Hence, by differentiation, 
 
 + cosy + cos 2y + cos 3y + &c. = 0. 
 
 2i 
 
 These formulae are deductions of Euler s. 
 
 Taylor s theorem may be applied to find approximate roots 
 of equations of the higher degrees. 
 
 (6.) Show that Taylor s theorem comprehends the Bino 
 mial theorem. 
 
 (7.) Expand sin (x + h} by Taylor s theorem. 
 
 h IP 
 
 T sinar- 
 
 i J 
 
20 MACLAURIN S THEOREM. 
 
 (8.) Show, by Taylor s theorem, that 
 
 + &C. 
 
 (9.) Show that tan (x + h) = tan# -f sec 2 ^ - 
 
 h Jfi 
 
 + 2sec 2 #tan#: i -^ + 2secXl + 3tanV) + &c. 
 
 1 J 1 A 6 
 
 (10.) If u=cot~ l x, show that 
 
 h h 2 
 
 cot~ l (x -f- A) = u sin u sin u - -f sin 2 w sin2 ? &c. 
 1 2i 
 
 h* 
 
 (11.) If/(^) = , prove that 
 l x 
 
 CHAPTER V. 
 
 MACLAURIN S THEOREM. 
 
 This theorem, which is used for the development of a 
 function according to the ascending powers of the variable, 
 may be thus enunciated, 7" , U l} U z , U s , &c. representing 
 
 du d 2 u d 3 u 
 
 the values of z, - , - r, -=-T, &c. when a?=0, 
 <te dr dtr 
 
 This theorem was first given in Stirling s " Lineae Tertii 
 
MACLAURIN S THEOREM. 21 
 
 Ordinis ISTewtonianse." It is, however, generally attributed to 
 Maclaurin, and is improperly styled " Maclaurin s Theorem." 
 Ex. (1.) Expand (a + x) n , n being any number whatever, 
 
 positive or negative, integral or fractional, rational or 
 
 irrational. 
 Let u(a-}-x) n , whence if#=0, U =a n . 
 
 &c. 
 
 AA irnlnpH nf 77... TT. . Ar.o. fnr 11.. } 
 
 Substituting these values of V , U l} &c. for u, , &c. 
 
 in Maclaurin s theorem, we have 
 
 + &c., which is the Binomial Theorem. 
 (2.) Develop a*. 
 
 Let u = a*, whence if x = 0, U a = 1 
 
 =",!* ...... U = A 
 
 - 
 dx 2 
 
 * A is here put for the hyp. log. of base a, that is, for the expres 
 sion (a- 1) - - (a- 1) 2 + i (a- 1) 3 - &c. 
 
22 MACLAURIN S THEOREM. 
 
 Whence, by substitution in Maclaurin s theorem, 
 
 which is the Exponential Theorem. 
 :A=loga, :.a*= 
 
 When a?=l, a = l+loga + ~(\oga) 2 + 
 
 an expression for any number #, in terms of its Napierian 
 logarithm. 
 
 If for a we write the Napierian base e } we have, since 
 Iog0=l, 
 
 a? a? 
 
 _-_ 
 And, when #=1, 
 
 L &c.=2-71828 &c. 
 
 (3.) Expand tan"" 1 ^ by the method of indeterminate coef 
 ficients, 
 
 u=tao.~ l x, whence if ^=0, C/y^tan^OrrO. 
 
 - =- - -=1 op+ofi ^ 6 + &c., by actual division. 
 dx 1 +a? 
 
 But (Maclaurin s Theor. Cor.), 
 
 Equating coefficients of like powers of #, we have 
 *7 1= 1, U 2 =0, U 3 =-2, U 4 =0, *7 5 =:2.3.4, &c. ; 
 
MACLAURIN S THEOREM. 23 
 
 2^ 2.3.4^ 
 whence by substitution, ux o + Q ,< g~"^ c - 
 
 ^ G <U O 4: D * 
 
 " 1 ;? =tf - +- =r + &C. 
 O / 
 
 3 + 5 7 
 
 which is an expression for the arc, in terms of its tangent. 
 
 By help of this and Machin s Formula, we may find an 
 approximate expression for the length of the circumference 
 of a circle. 
 
 1 1 
 
 .Let tan&=-, A = ka. then J.=4 tan^ 1 - : 
 o o 
 
 4 4_ 
 
 4tana 4tan% 5 125 120 
 
 1- + ~ 119 
 
 120 
 
 , N tan.4-1 119 1 
 
 Now ^-46-) = __= ^-= 
 
 119 + 
 
 .-.^-45 -tan" 1 
 
 239 
 or 
 
 239 
 
 =4 fi- J L _!_ 
 
 \5 3(5)^5(5)5 7(7) 7 
 
 _ /_!_ 1 , ! _ A \ 
 
 \239 3(239) 3 " t "5( r 239 N ) 5 J 
 
 3(239) 3 ""5(239) 
 This is Machin s Formula. 
 
24 MACLAURIN S THEOREM. 
 
 a very convergent series, by which, taking seven terms in 
 the first row, and three in the second, we obtain. 
 
 7r=3 -141592653589793, 
 
 which is the approximate length of the semicircle, the radius 
 being unity. By talcing three, terms in the first row, and 
 one in the second, we obtain 7r=3 1416, an approximation 
 sufficiently near for ordinary purposes. 
 
 (4.) Expand sec#, in ascending powers of #. 
 
 Put ?< = sec#, whence if &=(), sec#=l, U =l. 
 
 - : 
 
 (LOG 
 
 = sec x tan x + 2 sec x . 2 tan x( 1 + tan V) + 2 taii 2 .r sec x tan x 
 
 = 5 sec x ( 1 -|- tan 2 ?) + 5 tan x sec x tan x 
 
 + 6 sec x . 3 tan 2 ^ ( 1 + tan 2 ^) + 6 tan 3 x sec x tan x 
 
 Whence, by substitution, 
 
 a? 
 
 1+ 
 (5.) Expand cos 3 .r. 
 
 Put u=cos*x, whence if #=0, cos 3 #=l, . U Q =\. 
 
 - = 3cos 2 #( sin^r)=:3sin 3 ^ 3sin^, . . ^=0. 
 ctoo 
 
 U= 3. 
 
MACLAURIN S THEOREM. 25 
 
 d^u 
 
 - 2 
 
 dor 
 
 dor 
 
 + 18cos 2 ^.cos^, 
 
 30* 21^ 3^2 
 
 
 (6.) Develop (1-f-e*) 71 according to ascending powers of x. 
 Let M = (l+e*) n , whence if ^=0, 
 
 . make ^= 
 
 dx 
 
 make ir=0, 
 
 n-l)2 n - 2 .2 + w(w-l)(^-2). 
 l)2 n - 1 + n(n l)(>-2)2 n 
 
 "Whence, by substitution in the theorem, 
 
 ! 2 1-2 2 3 
 
 D 
 
26 MACLAURIN S THEOREM. 
 
 y2 y2> yA 
 
 (7.) Prove tliat log(l + a?) = x i-H-r ~+ &c - 
 
 Let M = log ( 1 + a?), whence if x = 0, Z7 = log ( 1 ) = 
 
 = - - = 1 x + x 1 ofi + o 
 x 1 + x 
 
 But (Maclauriii s Theorem. Cor.) 
 
 = - - = 1 x + x 1 ofi + ofi <fec. by actual division. 
 ax 1 + x 
 
 And, equating coefficients of like powers of a?, 
 
 rr , rr , #. ! ^4 ^ 
 
 Z7 1= =l, U,= -l, = l, 273= -1, ^3: 
 
 ..Z7 1= l, ZT 8 =-1, ^=2, Z7 4 =-2.3, Z7- 
 
 "Whence, by substitution in the theorem, 
 
 x* ^ x* 
 log (1 +*)=*- - + ~ + &c " 
 
 Cor. Writing # for a? we have 
 
 o O <J 
 
 log (l-aO = -a?- ~ ~ 
 
 (8.) Show, by help of the last example, that 
 
 - 
 
 x 1 111 1 
 
 Put - -=!#, then 
 
 log (1 +*)=*- 2 + 3 -&c. (Ex. 7.) 
 
 x x x+\ 1 
 
 But z - 7 1 = - r~ 
 
 x 1 # 1 a7 
 
 1 111 1. 
 
 l ^1 " 2 ^1 
 
MACLAURIN S THEOREM. 27 
 
 (9.) If a n and b n respectively represent the coefficients 
 of x n in the expansions of u=f(x\ and log w; show that 
 
 Assume u=a + a l x + a. 2 x * . . . . +a n x n , then 
 =a 1 + 2 2 # 
 
 ^ 
 
 =t ^ _ _ _ +nb ^ 
 
 . . . +a n x n 
 
 ^ multiplying by the denominator, and equating coef 
 ficients of like powers of x, we have 
 
 na n =b l a n _ l + 2b. 2 a n _s + 3b. 3 a n _3+ . . . +nl n a . 
 (10.) Develop siiio? and coso; in ascending powers of #. 
 
 l- +I - 
 (11.) Prove Euler s formulae, 
 
 2 
 (12.) Prove De Moivre s formula, 
 
 cosm#+ \/ 1 sin mx = (cos x+ v / 1 
 
 (13.) Prove that (tan^) 4 =^ + |^ 6 + | 
 
 o 
 
 (14.) Ifw=sin~ 1 tf, show that 
 
 sin 3 w 3 2 sin 5 ? 3 2 . 
 __ + ___ 54 _ 
 
28 MACLAURIN S THEOREM. 
 
 (15.) Develop w=cot# by the method of indeterminate 
 coefficients. 
 
 1 x a? 2^ 
 
 cot *=^- 3-3^5-33^7 
 (16.) Prove, by Maclatirin s theorem, that 
 
 (17.) Show that 008-^=^ x ^-^ ^^ T- &c. 
 
 A A o A o 1 O 
 
 (18.) Show that sin( 
 
 3 . 
 
 (19.) Prove that -^- =*1 - ^ - ^ - &c. 
 (20.) If cos^+ sino: -/ I =4*^, and ^ take the parti 
 cular value 2* prove the two formulae of John Bernouilli, 
 
 namely, 
 
 1), and 
 
 Implicit Functions. 
 
 Ex. (1.) Given u 3 3 u + x=Q, to expand w in a series of as 
 cending powers of x. 
 
 #=0, u*-3u=0, :.u=Q, . T =0. 
 
 dx 
 
MACLAURIN S THEOREM. 29 
 
 du 
 
 u 
 
 -1 
 
 9 (u 2 - I) 3 
 
 (u 2_i 
 2 v 
 
 27(u*-rf U *~27 
 ( u 2-iy.2u 
 
 20 _4w 3 -2^ 48 2w 
 
 81 (w 2 -!) 7 " 81 (^2- 
 
 d*u_ 40 22u*+\9u 2 +l _ 40 
 
 ~S?~~~243 (w 2 -!) 9 * 5 ~243 
 
 "Whence, by substitution in Maclaurin s theorem, 
 
 (2.) 2w 3 ux 2=0; expand u in a series of ascending 
 powers of x. 
 
 x 3? 
 
 + 2T3 ~ 2^ + &C 
 
 (3.) i- ==6 ; show that z=2 + a?- +- 
 
 .r 4 3^ 7 
 (4.) ifix 8u Sx=Q ; show that ux ^ -- ^ -- &c. 
 
30 EVALUATION OF 
 
 (6.) u 3 a 2 u + aux a?=Q ; show that 
 
 a* a* a* 
 
 u &c. 
 
 & a 6 4 
 
 (7.) sin u =x sin (a + u}, show that 
 
 /V /v2 /& 
 
 u = TTT + sin a - + sin 2 a -^-^ + 2 sin a (3 4 sm 2 a) -r ^ + &c. 
 
 J. 1 A L O 
 
 CHAPTER VI. 
 
 EVALUATION OF INDETERMINATE FUNCTIONS. 
 
 f 
 
 When the two terms of any fraction contain a common 
 
 H/ 
 
 factor, as x a, and the particular value a be given to x, 
 then, since x a will be equal to 0, the fraction will assume 
 
 the form > and be indeterminate. 
 
 Such a fraction is improperly termed a vanishing fraction ; 
 since its values may be finite, infinite, or nothing. 
 
 When the common factor is obvious by inspection, it may 
 of course be removed by division. 
 
 The method of John Bernouilli is to differentiate the 
 numerator and denominator, separately, until they do not 
 vanish simultaneously by making x=a, and thus to deter 
 mine the true value of the fraction in that case. 
 
 P(ft Cl} m 
 
 If the fraction be of the form -^-. (-* and in or n be a 
 
 Q(x-a} n 
 
 fraction, this method of successive differentiation will not 
 apply, since, however often we differentiate, we shall never 
 eliminate the common factor. 
 
INDETERMINATE FUNCTIONS. 31 
 
 In this case we may put a L h for x, expand both terms 
 of the fraction in a series of ascending powers of k, and 
 then put k = Q. 
 
 The process of evaluation of indeterminate functions 
 enables us to find the sum of a series for a particular value 
 of the variable. 
 
 Ex. (1.) Find the real value of the fraction 
 ax 1 
 
 Here P=ax 2 -2acx+ac 2 , Q=bx 2 -2bcx+lc 2 , 
 
 :. =2ax 2ac=Q if x=c 
 dx 
 
 dx 
 
 d 2 P _ 
 
 rr=2 i 
 
 2a a 
 - = - 
 
 (2.) Let u - -r = -- Find u, when xl. 
 
 X"^ 1 
 
 Here P=a?+2x 2 -x-2, Qx^-l ; 
 
 dP 
 
 , . 1 = 6 if x=l "} 
 
 dx \ 6 . 
 
 JK ( "^=0=2. 
 
 ^$09 0-^1 3 
 
 -=3x 2 =3 if x=\ J 
 
 dx 
 
 gX gSina; 
 
 (3.) u : =1, when ^=0. 
 
 ^ = ^r_ -x ( = 1 1 =0 . f 
 
 dx ^ & C Sir l X 
 
 * ^u -I -i -I r\ / s\ 
 
 dx~ ~ 
 
32 EVALUATION OF 
 
 d 2 P 
 
 .. =0 if #=0, 
 
 d 2 O 
 2 
 
 d 3 P 
 
 ~ 
 dar 
 
 -^=008^=1 if#=0, . . u==l. 
 dap 1 
 
 (4.) M=(l ^r) tan -= - =- whena?=l. 
 
 2 . 7T^7 7T 
 
 cot y 
 
 7T 
 
 dP dQ 2 
 
 -7= ! -7= make#=l, then 
 
 dx dx . 9 7r 
 
 smy 
 
 7T 7T 
 ( IQ 2 "2 7T 12 
 
 dx . 2 71 " 12 7T 7T 
 
 o o 
 
 ( Oj - / 7*^ j^ l // ^_ /^ 
 
 (5.) e= p - Find M, when xa. 
 
 Put x=ah, then 
 
INDETERMINATE FUNCTIONS. 33 
 
 k*(2a-h)* + k (2 g -. 
 
 ~ 
 
 Now, putting /?=0, we have u = 
 
 (6.) u =- =^, whenar=0. 
 
 dP 
 
 dx sec 2 **? cos^ 1 1 cos 3 .r 1 cos 3 .*? 
 
 ~dx 
 
 since the factor ^-=1 when ^=0 ; 
 
 3cos 2 .r.sina? sin# 
 
 ; = v ooAsBi. when #=0 : 
 
 cos^r 1 tan x si 
 =-^_=-, hence =-, when ^=0. 
 
 (7.) Find the real value of - - when x = \. 
 
 ^.4_6. c 2 + 8^ 3 
 
 Ans. oo . 
 
 a (a 2 tf 2 )* 1 
 
 (8.) If%= 5 ^ "^> when #=0, u= 
 
 3 Ct^- T^ 
 
 (9.) u= -j=. = > when #= a, w=3. 
 
 v.ax a 
 
 /1A\ * . .. 
 
 (10.) ^= - - - - - , when x=a, u = Q. 
 
 - * 
 
 ! =^ 
 
 cotar + coseca? 1 ?r 
 
 (IJ.) ?= -> when ar=Tt U=a:l. 
 
 cot x cosec# + l 2 
 
 7T 
 
 a; sm a? - 
 
 (13.) z= 3 when#=-> M= 1. 
 
 cos x 2 
 
34 EVALUATION OF 
 
 ( a ._ fl )i + a .i_ fl * 
 (14.) u=- - - when x=a, u= 
 
 
 _ 
 
 (15.) If in - x \, show that n ~ l 
 
 lx 
 
 ,-.m 1 _ ,>n 
 
 x i *v \ S\3C & \v I V */ T n 
 
 (17.) ? = T) - *" when =0, ?<=4. 
 
 (18.) M = 
 
 tan?r^ 
 
 (19.) u=- sr -y 
 
 2 
 
 (20.) u= 
 
 . 
 log.a? 
 
 (21.) u=* 
 
 x a+ \2ax 2 2 
 (22.) z<= - L = = - = 1, when xa. 
 
 va?a 2 
 
 x n 
 (23.) u> when #=<x>, z = 0. 
 
 v 
 
 
 (25.) u=xe~ x =., when ^=co. 
 
 1 7T 
 
 -- 2 
 
 (26.) w= - > when ^=0, ^=-x 
 
 7T^ 
 
 (27.) If ye* xQ; show that when # approaches oo the 
 limiting values of e~ x and y are identical; and that the limit 
 ing value of y is zero. 
 
INDETERMINATE FUNCTIONS. 35 
 
 log (tan 2 x\ 
 (28.) u=~^- / -- / whena?=0, u=l. 
 
 log (tan x) 
 
 (29 .) tt= 
 
 x n a 
 
 log(l+na;) 
 
 (33.) ue x , when #=0, ^=e w . 
 
 . v OiAA C^/ VyV^kJ^L/ 1 r\ C\ 
 
 (34.) w= > when as=Q, u=2. 
 
 1 cos^7 
 
 (35.) If the fraction - -. 7- assume the form GO oc 
 
 /W 0W 
 when ^= ; show that this illusory form oo GO , and also 
 
 x oo are each identical with the form - 
 
 CHAPTER VII. 
 
 MAXIMA AND MINIMA. 
 ONE VARIABLE. 
 
 If a quantity increases to a certain extent, and then 
 decreases to a certain extent, its values at these limits 
 respectively are a maximum and a minimum. 
 
 If it repeatedly increases and decreases alternately, it 
 has several maxima and minima. 
 
36 MAXIMA AND MINIMA. 
 
 If it increases continually or decreases continually, it has 
 no maxima or minima. 
 
 Let u=f(x) } then, to determine the values of x which 
 
 du 
 
 render u a maximum or minimum, put -^-0 or oo , and 
 
 ax 
 
 substitute the possible roots of the resulting equation in 
 
 - > then, if =a negative quantity, the value of x which 
 dx l ax* 
 
 d 2 u 
 is substituted renders u a maximum ; if -a positive 
 
 it L&^ 
 
 quantity, the value of x which is substituted renders u a 
 minimum. 
 
 A maximum or minimum can exist only when the first 
 differential coefficient which does not vanish is of an even 
 order. 
 
 nt 
 
 If u = a maximum or minimum, then a u and are 
 
 a 
 
 maxima or minima. Hence, before differentiating, we may 
 reject any constant positive factor in the value of u. 
 
 If ^=:a maximum or minimum, then u n is a maximum or 
 minimum if n is positive ; but when w a maximum u~ n is a 
 minimum, and when u = a minimum u~ n is a maximum. 
 Hence, before differentiating, we may reject a constant 
 exponent. 
 
 If U=B, maximum or minimum, logu is a maximum or 
 minimum. Hence, when the function consists of a product 
 or quotient of powers or roots, we may use the logarithms. 
 
 Ex. (1.) Find when a? 5 # 4 + 5 .r 3 + 1 is either a maxi 
 mum or a minimum. 
 
 Let u=a?> 5sc* + 5a? + l, then 
 
 15 a?, and putting this=0, 
 
MAXIMA AND MINIMA. 
 
 37 
 
 a 2 -4.r=-3, a; =3, x=\, 
 
 =20 # 3 60^ 2 + 30^, and substituting successively 
 c/^ 2 
 
 the values of #, (0, 1, 3) in this expression, 
 
 11^=0, from which we can infer nothing, 
 daP 
 
 _1 = 20 60 + 30= 10, which indicates a maximum, 
 da? 
 
 =540 540 + 90= + 90, which indicates a minimum. 
 dx 2 
 
 Hence, when #=1, ^t=2, a maximum, 
 and, when # = 3, w= 26, a minimum. 
 1 (2.) If u= \/a 2 x 2 2aa?, ascertain those values of x 
 which make u a maximum or minimum. 
 
 Rejecting the radical and the common factor 2 a, put 
 
 o* =4aa?-r-3a? 8 =4a-3as a;=0, 
 
 4 
 
 d 2 u 
 
 __!!=4;a_G#=-f 4a. 
 Hence x=-^ makes u 
 
 ^7 
 
 a maximum, 
 
 3v x 3 
 
 ic=0 makes M = 0, a minimum. 
 
38 MAXIMA AND MINIMA. 
 
 y (3.) Determine the maxima and minima values of the 
 
 \o& < 
 
 function n=-> - - - 
 
 Y\* 2 /-f*?- 
 
 1 1 -(- x 2 
 
 Putting u=-> we shall have v = - * 
 
 c x 
 
 dv x2x-(l+x 2 } x 2 -! 
 
 Tx= - ^ - = = *= + !> =-i. 
 
 <l 2 va?.2xa?l 2x2x 2 
 
 :. - = +-> which indicates a minimum, 
 dx 2 1 
 
 d 2 v 2 
 
 - ;= -> ...... maximum, 
 
 dx 2 1 
 
 1 1 
 :.u=- r=-^ a maximum, 
 
 -1 1 
 
 -= -- a minimum. 
 
 1 + 1 2 
 
 (4.) Divide a number a into two such parts that the pro 
 duct of the m ih power of the one and the w th power of the 
 other shall be the greatest possible. 
 
 Let x, and a x be the parts, then 
 u-=.x m (a x) n . 
 
 = x m ~ l (ax) n ~ l {xn + (ax)m} 
 
 ma 
 
 x= 
 
 m -f n 
 Or thus, log u = m log x+n log (a x), 
 
 die 1 in n du 
 
 dxu x ax dx \ (a x) 
 
 a u _ lammx n x\ 
 dx \ (a x] x i 
 
MAXIMA AND MINIMA. 39 
 
 flu am (m + n] x 
 
 :.=x"(a-x)* -- , V , ; =0, 
 dx (ax)x 
 
 am 
 
 x=- 
 
 HI -\- n 
 
 the values and a may be rejected, since there can 
 be no division of the line if =0 or a. 
 
 Hence, differentiating again, and substituting - in 
 the second differential coefficient, we have 
 
 - = (m 4- n) . which indicates a maximum, 
 dx 2 
 
 am an 
 
 :.x= - and a x= - are the parts. 
 m + n m -f n 
 
 U 
 (5.) If =sin 3 ccosic, show that u is a maximum when 
 
 x=60. 
 
 du 
 
 = sin- 3 ;*; sin x + cos x 3 sin 2 a; cos x 
 dx 
 
 = 3 sin 2 x cos 2 # sin 4 * = 0, 
 .*. 3 sin 2 # co$ 2 x = siu 4 x, 3 cos 2 o? = sin 2 ic = 1 cos 2 a?, 
 
 . . 4coa 2 aj=l, cos,^=-> . .a?=60, 
 
 ^SH 
 
 -=3 sin 2 ic. 2 cosx ( sin a?) + 3 cos 2 a;. 2 sin a? cosx 
 
 dx 
 
 4 sm 3 # cos x = 6 sm 3 # cos a; + 6 sin x cos^x 
 4 sin 3 a; cos x = 1 sin 3 o? cos x 4- G sin # co$?x. 
 
 ^s ow sin x = > 
 
 rf 2 w 30v/3 1 6^3 1 24 /- 
 
 -,= -- _._ + __.__ -x/3, a negative residt, 
 
 3v/3 1 3 /- 
 
 /. M = - -= v 6, a maximum. 
 o U 10 
 
40 MAXIMA AND MINIMA. 
 
 (6.) Divide a number n into two such factors that the 
 sum of their squares shall be the smallest possible. 
 
 fl 
 
 Let x be one factor, the other ; then 
 
 x 
 
 2 n 2 du_ Zn 2 _ 
 
 x 2 dx x> 
 
 ;.x= > x*=n 2 , x=\/n, 
 
 x 3 
 
 d 2 a Qn 2 6n 2 
 
 - = 2 -\ =2-\ = 2 + 6 = + 6, a positive result, 
 
 dx* x* n* 
 
 :. u is a minimum. Hence the sum of the squares will be 
 the smallest possible when the factors are equal, each being 
 the square root of the given number. 
 
 (7.) Into how many equal parts must a number n be 
 divided that their continued product may be a maximum 1 
 Let there be x equal parts, then 
 
 /%* 
 
 - is the magnitude of each, and 
 x 
 
 =:( ] is their continued product, 
 log u = x log f - ) = x (log n logo:), 
 
 \00/ 
 
 du 1 / 1\ 
 
 =x> +losr/2 logx = 14- log n 
 
 dx u \ xi 
 
 =u{ l+logn logx} =0, 
 
 du 
 
 Tz 
 
 =- ,7 a negative result 
 
 in\* - 
 :. M=|-| =.e", a maximum. 
 
MAXIMA AND MINIMA. 41 
 
 (8.) Show that - -- : is a maximum when x=4=5. 
 1 + tana 
 
 du ( 1 + tan x) cos a? sin x ( 1 -f t an 2 ;) 
 <ta~~ (1+tana;) 2 
 
 cos x + sin a? sin x sin a? tan 2 & 
 (1 + tanz;) 2 
 
 cos x 
 
 tan 2 #=l, 
 
 d*u 3 /- 
 
 := - v z, a negative result, 
 
 C/^T 2 4 
 
 sin^ 1 /- 
 .*. u=- -- = v z, 
 1 tan# 4 
 
 a maximum. 
 + tan 
 
 (9.) If # be the hypothenuse of a right-angled triangle, 
 find the length of the other sides when the area is a maxi 
 mum. 
 
 Let x be one of the other sides, then 
 v a 2 x 1 is the remaining side. 
 
 And area =-x\/a 2 x 1 . 
 *2i 
 
 Now, rejecting the constant > we may take 
 
 - = 2a 2 l2x 2 = 2a 2 6a 2 =4:a 2 , a negative result, 
 
 Cl JO 
 
 :. u is a maximum, and the area is a maximum when the 
 
 two sides are each = 
 
 v/2 
 
 E 2 
 
42 MAXIMA AND MINIMA. 
 
 (10.) What fraction exceeds its w th power by the greatest 
 number possible ? 
 
 Let x be the fraction, then u=x x n , lnx n ~ l =.Q, 
 
 ax 
 
 - = n (n 1 ) x n ~ 2 = n (n 1 ) which is negative, 
 
 .*. u is a maximum. 1 
 
 Ans. 
 
 v n 
 
 (11.) Within an angle BAG a point P is given, through 
 which it is required to draw a straight line so that the 
 triangle cut off by it shall be the smallest A 
 
 possible. 
 
 Let PN=a, A N = b, A D = x, then 
 ND = x-b, ND : PN :: A D : A E or 
 
 ax 
 
 # b \ a . . x \ AE, 
 
 - - 
 
 X O 
 
 Now area A D AE=-A D-AE sin A=~x -- 7 sin -4, 
 
 2 2 x h i 
 
 X 
 
 ~ 
 
 a positive result, . . the area is a minimum. 
 Since AD = 2 AN, :.DE = ZDP, :. the line must be 
 so drawn as to be bisected by the given point P. 
 
MAXIMA AND MINIMA. 43 
 
 * (12.) From two points A, B, to draw two straight lines to 
 a point P in a given line ON, so that AP + BP shall be a 
 minimum. 
 
 Let be the origin of co-ordinates, and the given line the 
 axis of x, 
 
 Let OPx, and let the co-ordinates of A be a, I, and 
 those of B be ft,. Then 
 
 BP= 
 
 :.u=AP+I>P= V l 2 + (x af+ v l^-\-(a, #) 2 , a minimum, 
 du x a a, x 
 
 dx v 7 ^-)-^ a} 2 vb?+ (ax} 1 
 x a a, x 
 
 (13.) If the length of an arc of a circle be 2 a, find the 
 angle it must subtend at the centre so that the correspond 
 ing segment may be a maximum or minimum. 
 
 Draw CD bisecting the arc, and let x be the 
 
 radius, then -= Z A CD. 
 x 
 
 Now area segment ADB= sector AGE A AGE 
 
 =. rad x arc - x 2 sin A CB 
 
 1 . 
 =# ^ ^ sm cos 
 
 . 
 . . tt = a# a, sin cos? 
 
44 MAXIMA AND MINIMA. 
 
 du 
 
 au 9 . a , i , a\ i a\ a a i a 
 
 =axl sin - sm : x* cos - cos 
 
 dx x\ xl \ x 1 ! x x\ x 2 
 
 .a a 
 
 2.rsin- cos- 
 
 a . a a 
 
 -- .rzsin-cos- 
 x x xx 
 
 a a a 
 
 =a a + 2acos^ -- ^2sm-cos- 
 
 x xx 
 
 at a . a\ 
 = 2 cos- cos -- ^rsm- =0. 
 x \ x xi 
 
 a a TT 2 a 1 
 
 lake cos- = ; /.=-> x=. > and the segment is a -Q 
 # ^ 2 TT 2 
 
 = maximum. 
 
 sm- 
 
 m , a . a x a a 
 
 lake acos- = .sin-> .". - = tan-=-> 
 
 x x a x x 
 
 cos 
 x 
 
 a . . 
 
 :. - = 0, /.#=oo and z = minimum. 
 x 
 
 (14.) Within a given circle to inscribe the greatest isos 
 celes triangle. 
 
 Let radius OA=a, AB = AC=x, C=2y, 
 
 AB-AC.BC o*y 
 BD=y. Then A = =_ 
 
 BC.AD - , 
 
 Also A = - - - = -- - -- y Vx 2 y 2 , 
 
 =2 a v/^V, 
 
 =4: cftx- 4 a 2 y 2 , 4 a 2 ^ 2 = 4 d-x 1 x 4 , %ay=.x v&a- u -, 
 
 ;. y = _^ V / 4^-^ 
 
MAXIMA AND MINIMA. 45 
 
 Now A = - x 2 u = - x 1 - x v/4^ 2 x 2 . a maximum. 
 
 2 a 2 a 2 a 
 
 Put M=a^(4a 2 --tf 2 ):=4aV 5 a 8 , 
 
 ^=24aV-8tf 7 =:0, . . 8^=24 A 5 , 
 ax 
 
 x 2 =3a 2 , :.x=a^3, 
 2y=a\/3v4:a 2 3a 2 =av3, and A is equilateral. 
 
 (15.) Of all equiangular and isoperimetrical parallelo 
 grams, show that the equilateral has the greatest area. 
 
 The perimeters of the figures being all equal, the perimeter 
 of each may be considered as one line, and the proposition 
 then resolves itself into the following. " To divide a given 
 straight line into two such parts that the rectangle contained 
 by those parts shall be the greatest possible." 
 
 Let a be the line, x one part, then a x is the other, 
 x(o, x) is the rectangle, and u-=.ax # 2 , a maximum. 
 
 du a 
 
 =a-2x=0, :.a7= ? . 
 ax 2 
 
 .*. the line must be divided into two equal parts, and the 
 parallelogram will be equilateral. 
 
 -* (16.) Of all triangles on the same base and having equal 
 vertical angles the isosceles has the greatest perimeter. 
 
 Let a be the base, a the vertical angle, x and y the two 
 sides, then u =.a-\-x-\-y=-& maximum. 
 
 , ., 
 
 ax dx 
 
 y- + 2y cos a=2.r + 2^-7-; 
 
 O 00 Ci\C 
 
 :. (xy) cosa=:r y, :. xy 
 :. x=y, and the A is isosceles. 
 
46 MAXIMA AND MINIMA. 
 
 (17.) The segment of a circle being given, it is required 
 to inscribe the greatest possible rectangle in it. 
 
 Let BAD be the segment, radius = a, r ^^J^. 
 
 AM=.x, draw A C through the centre per- ,/J "| j\ 
 
 pendicular to PM or BD. Let A C= b. 
 
 Then PM 2 =(2a-x}x, Euc. B. iii. p. 35. 
 
 /. PM- */2aaT^, MG- b -x, 
 
 Area rectangle = M C - 2 P M = 2(bx) \/Zax x 2 . 
 
 ?v,iu=(b-x} 2 -(2ax-x 2 \ 
 
 . . (b x) (a x) = 2 axx 2 , abaxbx + x 2 = 2 axx 2 , 
 
 
 4 
 
 > (18.) To cut the greatest parabola from a given right 
 cone. 
 
 Let BD=a, AD=b, BC=x, CD=a-x, A 
 
 Then -:BNDMia a circle, and MC=NC, 
 :. MC 2 =BC- CD, MG- Jx(a-x\ 
 
 9 B" 
 
 Also BD : AD :: EG : PC, :.PC=-~=> 
 
 a 
 
 2 lx 
 
 Area parabola -=-PC-MN=- ^vaxx 2 , a maximum. 
 o 3 a 
 
 Put u = x 2 (ax x 2 ) = as? x 4 , 
 
 - ,= -a 2 , which indicates a maximum. 
 dx 2 4 
 
MAXIMA AND MINIMA. 47 
 
 (19.) Within a given parabola to inscribe the greatest 
 ? parabola, the vertex of the latter being at the bisection of 
 the base of the former. 
 
 Let BA C be the given parabola, L its latus rectum. 
 AD a, JBD=b, DN=x, PN=y. 
 
 2 2 
 
 Area parabola =-.2PN-ND=-.2yx. if 
 o o 
 
 Now Y the square of any ordinate to the axis = the rect 
 angle under the latus rectum and abscissa, 
 
 vi 4 l / - 
 :. area parabola = -- -=xv a x. 
 
 3 -s/a 
 
 Put M=^ 2 (a x)=-ax 2 a?, 
 
 =2ax-3x 2 =Q, :.3^=2^r, *=fa. 
 dx 3 
 
 > (20.) Inscribe the greatest cylinder within a given right 
 cone. 
 
 Let ABC be the cone, AD=a, BD=b, DN=x, PN=y, 
 AN-a-x. 
 
 Volume of cylinder =-.(2PN) 2 . ND=Trfx. 
 AD : BD :: AN : PN, :.PN=~AN, or B ^ 
 
 u =-(a x). :. cylinder = ?r -r (a x] 2 x . 
 a a 2 
 
 Put u = (a a?) 2 .r = 2 ^ 2 a # 2 -f ir 3 , 
 
 vl t* 04 . r o rw .noj o 
 
 -" *= OT 3- 
 
48 MAXIMA AND MINIMA. 
 
 > (21.) If the volume of a cylinder be a, find its form when 
 its surface is the least possible. 
 
 LetA = x, C=y. A " 
 
 Surface = convex surface -f- 2 area of base 
 
 Volume =a=BC^^-AB=~xf, :.x= 
 4 4 Tr 2 
 
 TT * W . W 9 * * i ^ 9 
 
 Hence ?< = TT- -y + -^ = + -?/ J , 
 </? 4 4 
 
 or altitude = diameter of base. 
 
 d 2 u Say 8 a 
 
 = ~ -\-7r= 5+7r=+o7r, a positive result, 
 
 dy y y 
 
 :. the surface is a minimum. 
 
 > (22.) The latitude of a place and two circles parallel to 
 the horizon being given ; to determine the declination of a 
 heavenly body, whose apparent time of passage from one 
 circle to the other shall be a minimum. 
 
 Let P be the pole, Z the zenith, S, S t the positions of the 
 heavenly body on the parallel circles, the polar distances 
 PS, PS, being equal, 
 
 / ZPS=P, ZPS,=P t , polar distance PS or PS t =x, 
 arc ZS=a, ZS t = a t , latitude =/, declination = 3 ; then 
 Y the passage along the arc SS, is the shortest possible, 
 /. the angle SPS t = a minimum, 
 
MAXIMA AND MINIMA. 
 
 49 
 
 dx 
 
 dx 
 
 dx dx 
 
 dP cotS 
 
 But -7-= : 
 
 </# sin ^ 
 cot/S cot$, 
 
 dx 
 
 sn # 
 
 Again cos S= 
 
 sin x sin x 
 
 sin/ cos a cos# 
 
 sin a sin # 
 sin/ cos cos # sin/ 
 
 cosS,= 
 
 sin/ cos a i cos # 
 sm, sina? 
 
 , cosx 
 
 sin a 
 
 eos#= 
 
 I 
 
 cos- 
 
 sin/. 
 
 cos -(a, a) 
 
 And v the declination is the complement of the polar 
 
 1 
 cos (,-f a) 
 
 distance, .*. sin^= sin/. 
 
 oo* 5 (,) 
 
 Cor. 
 
 If a=.> and ,= +2J, this expression becomes 
 
 sin = tan d sin/; and if the heavenly body be the sun, 
 and 2d=l& nearly = his depression below the horizon 
 when twilight begins in the morning or ends in the evening, 
 we are enabled to determine the time of shortest twilight by 
 means of the analogy rad : sin lat :: tan 9 : sin 3, where 
 the negative sign indicates that, if the latitude be north, the 
 declination will be south, and vice versa. 
 ^ (23.) The centres of two spheres (radii r 19 r a ) are at the ex 
 tremities of a straight line 2 a, on which a circle is described. 
 
50 MAXIMA AND MINIMA. 
 
 Find a point in the circumference from which the greatest 
 portion of spherical surface is visible. 
 
 Let x and y be the distances of ^E 
 
 the point from the centres of the ^ 
 two spheres ; draw tangents JSA, 
 EB, ED, EF; join AJB, DF. 
 
 Then, of the sphere C the portion 
 visible is the convex surface of the 
 segment A HB S f whose area = height HS x circumference 
 
 of the sphere. 
 
 r 2 
 
 Now x : t\ :: r l : C/S, :.CS=> . .height of segment 
 
 x 
 
 r 2 
 
 = r 1 -=fftS, circumference of sphere = 2 7rr 15 
 
 CO 
 I T\ 
 
 :. 27rr 1 (r 1 Invisible portion of sphere C; and similarly 
 
 (V \ 
 r a )=visible portion of sphere c. 
 
 t7 
 
 Hence 2ir! r-\r\ ) +rJ r a ) i=whole visible surface. 
 
 (V xl \ y 1 \ 
 
 r 3 r 3 
 
 Put u=rf + r? -> then 
 
 x y 
 
 /y 
 
 7 * r f~r * 
 
 du _r{> "X_ m rf_ r 
 
 ~, ^" \ o *J } ~n 
 
 dx x 2 ?r x- y 
 
 Hence 
 
MAXIMA AND MINIMA. 51 
 
 * (24.) Of all ellipses that can be inscribed in a rhombus 
 whose diagonals are 2m and 2n, show that the greatest is 
 
 . m n 
 
 that whose major and minor semi-axes are = and j=. 
 
 v2 /2 
 
 respectively. 
 
 ABCD the rhombus, OCm, OB=n, 
 a and b the semi-axes of the ellipse. 
 
 Let ON=x, NP=.y. Then by the 
 properties of the ellipse 
 00 -ON a 2 , OB-NP=b 2 , 
 
 on 99 997919 M " V v 
 
 ..m~x 2 =a 2 .a 2 , n 2 y 2 =b 2 .b 2 , = ? 70= -5 
 
 a 2 m 2 b 2 n 2 
 
 ;.-}- = --] ==1: . . . (1), where a and b alone 
 
 a 2 b 2 m 2 n 2 
 
 must be considered as variables. 
 
 But, area ellipse =7rab= a maximum. 
 
 Rejecting the constant TT, and differentiating this and 
 equation (1), we have 
 
 m 
 
 db n 
 i == 0, 
 dct 
 
 a b db 
 
 m 2 n 2 da 
 
 
 1 l 
 n 2 a 
 
 a 2 b 2 
 
 m 2 n 2 
 
 2 a 2 2b 2 
 
 b 1 
 
 m 
 
 n 
 
 (25.) If w=tf 4 -ar 3 + 22# 2 -24#+12 ) find the values of x 
 which render u a maximum or a minimum. 
 
 Ans. When #=3, M is a minimum, 
 
 #=2, u is a maximum, 
 
 a?=l, u is a minimum. 
 
 (26.) Find when x 3 6^ 2 -f 9a;+10 is a maximum, and 
 
 when it is a minimum. , TT1 _ 
 
 VViien $=6, u is a minimum, 
 
 #=1, u is a maximum. 
 
52 MAXIMA AND MINIMA. 
 
 (27.) Find the maxima and minima values of the function 
 
 . I 2 
 
 When # = u is a minimum. 
 oa 
 
 I 2 
 # == -> u is a maximum. 
 
 (28.) ^ = 7 - 5 ascertain when u is a maximum and 
 (a - a) 2 
 
 when a minimum. a 
 
 When x=- a, u=j> a minimum, 
 
 =+, u = cc , a maximum./ 
 
 (29.) ?^= < r r j find when % is a maximum. 
 
 #=6=2 -71828 &c. 
 
 (30.) w=^ - r^J determine when w is a maximum and 
 
 when a minimum. \^jr== 2, ti = o^a^maximuni, 
 
 #=0, M=6|, a minimum. 
 
 (31.) M=ar+ v^ 2 26^ + or 2 ; when is ^ a maximum 1 ? 
 
 a 2 a 2 
 
 When x=> w = ---f b a maximum. 
 26 26* 
 
 (32.) u= - - ? show that ^^ is a minimum 
 
 when #= 
 
 ; show that u is a minimum when 
 
 (34.) In a given triangle to inscribe the greatest paral 
 
 lelogram. 
 
 Ans. Side of parallelogram = ^ side of triangle. 
 
 (35.) A column a feet high has a statue on the top of it, 
 the height froifci the ground to the top of the statue is I feet ; 
 
 l^ (k~ *)(* --) - - 
 
 U) ".A.- ^ 
 
MAXIMA AND MINIMA. 53 
 
 find a point in the horizontal plane at which the statue sub 
 tends the greatest an^le. /Tf i f ,1 i 
 
 Ans. v ab feet from the base. 
 
 > (36.) Show that the difference between the sine and 
 versed sine is a maximum when the arc is 45. 
 
 (37.) Let AC and BD be parallel, and join A 
 AD] it is required to draw from C a straight 
 line so that the triangles EOD, A 00 together 
 shall be a minimum. 
 
 Let AC a, ADl>, A0=cc ; then x= A/T. 
 
 (38.) The base and vertical angle of a triangle being given, 
 show that when it is isosceles its area is a maximum. 
 
 (39.) A farmer has a field of triangular form, which he 
 wishes to divide into two equal parts by a fence ; find the 
 points in the sides of the field from which he must draw the 
 line, for his fence to be the least possible expense to him. 
 
 Ans. If a, b, c be the sides, the distance of each point 
 
 from the angle is A / -~-> and the length 
 
 c ^ f 
 of the fence 
 
 /( 
 is A/ 5 
 
 (40.) If the greatest rectangle be inscribed in an ellipse, 
 the greatest ellipse in that rectangle, again the greatest rect 
 angle in that ellipse, and so on continually ; show that the 
 sum of all the inscribed rectangles is equal to the area of any 
 parallelogram circumscribing the given ellipse. 
 
 (41.) Prove that the greatest area that can be contained 
 by four straight lines is that of a quadrilateral inscribed in a 
 circle. 
 
 (42.) Inscribe the greatest ellipse in a given isosceles 
 
 triangle. (^ ^ >/ f* *>^ . * ^- ^^ -^ ^ -/*>- 
 Ans. Major axis = altitude of triangle. 
 
 fct , 
 
54 MAXIMA AND MINIMA. 
 
 (43.) A tree, in the form of a frustram of a cone, is 
 n feet long, and its greater and less diameters are a and b 
 feet respectively ; show that the greatest square beam that 
 
 7? CL 
 
 can be cut out of it is ; 7^ feet long. 
 6(a b) 
 
 (44.) Describe the least isosceles triangle about a given 
 circle. The triangle is equilateral. 
 
 (45.) To inscribe the greatest right cone in a given sphere, 
 whose radius is r. 
 
 Distance of base of cone from centre of sphere = 
 
 o 
 
 (46.) If the polar diameter of the earth be to the equato 
 rial diameter as 229 : 230 ; show that the greatest angle 
 made by a body falling to the earth, with a perpendicular to 
 the surface, is 14 58", and that the latitude is 45 7 29". 
 See fig. ex. 9. page 84. 
 
 (47.) In a parabolic curve, whose vertex is A, and focus 
 S, find a point P, such that the ratio A P : SP shall be a 
 maximum. AP : SP \: 2 : \/3. 
 
 (48.) Inscribe the greatest parabola in a given isosceles 
 
 Altitude of parabola = altitude of triangle. 
 
 (49.) If in a circle, whose radius is r, a right-angled tri 
 angle be inscribed ; show that, when a maximum circle is 
 inscribed in the triangle, the area of the triangle is r 2 . 
 
 (50.) Inscribe the greatest cylinder in a given prolate 
 spheroid. 
 
 (51.) Required the maximum and minimum values of u 
 in the equation u s a?x + x 3 =Q. 
 
 e coax 
 
 (52.) u== ) find the maximum and minimum values 
 
 cos n # 
 
 of u. 
 
MAXIMA AND MINIMA. 55 
 
 (53.) Show that the greatest paraboloid that can be in- 
 
 2 
 
 scribed in a given right cone is -- of the height of that cone. 
 
 o 
 
 (54.) u=x l "~ losx j show that when u is a maximum, 
 
 (55.) Find that sphere which, being put into a conical 
 vessel of given dimensions, will displace the greatest possible 
 quantity of fluid. 
 
 (56.) Two circles of given radii intersect each other ; find 
 the longest straight line which can be drawn through either 
 point of intersection, and terminated by the circumferences. 
 
 (57.) If a tangent to a great circle of a sphere measure 5 J, 
 and a perpendicular to a tangent meeting the great circle 
 measure 4 feet ; show that the volume of the sphere is to 
 the volume of its greatest inscribed semispheroid as 27 : 16. 
 
 (58.) Find what values of x make (x 2) (# + 3) (5^-) a 
 maximum or minimum, and distinguish the one from the 
 other. 
 
 (59.) Inscribe the greatest cone in a given hemisphere 
 ABC, the vertex of the cone being at A. 
 
 For other examples and solutions see chap. xi. 
 
 IMPLICIT FUNCTIONS OF TWO VARIABLES. 
 
 If u-=.^(xj y), u being an implicit function of the two 
 variables x and y, by putting - = 0, we shall find the 
 values of x which render y a maximum or minimum. 
 
 By substituting the particular value of x in [-7-3 -r--r-f i 
 
 \dx* dyi 
 
56 MAXIMA AND MINIMA. 
 
 if the result be positive, y will be a maximum ; if negative, 
 a minimum. 
 
 Ex. (1.) Let u=x? 3 a 2 x+ifi=Q ; determine the maxi 
 mum and minimum values of y. 
 
 Differentiate with respect to x, considering y constant. 
 
 U Q 9 O 9 rv 9 9 
 
 -=<> d<jr==U, .".# - =a-% #=-(-#, x=a. 
 
 d 2 it 
 
 ~i~2 = 6 # Differentiate the given function with 
 
 respect to y, considering x constant. 
 
 = 3?/ 2 . Substitute the values of x in M. 
 
 -- ; -- = - = - ^=H -- a positive re- 
 dx* dx 3y 2 3 a 2 . 2f a 
 
 suit, *.y==av2 is a maximum. 
 
 rfy 6^r 6 a VI 
 
 - = - = - - = -- ? a negative re- 
 dx 3 2 3 
 
 suit, y = v 2 is a minimum. 
 
 (2.) u=aP 3a^-y4-y 3 =0 ; show that when o^=0, y=0, 
 a minimum ; and when xa V 2, y=a v4, a maximum. 
 
 (3.) 4#y y* ^*= 2 ; show that when x=- +1 or 1, 
 y=-fl or 1, neither being a maximum or minimum. 
 
 (4.) f 3= 2x(xy + 2); show that when x=l,y= \, 
 
 neither a maximum nor a minimum j but when x > 
 
 & 
 
 = a maxmum. 
 
57 
 
 CHAPTER VIII. 
 
 FUNCTIONS OF TWO OR MORE VARIABLES. 
 
 If uf(x^ y\ x and y being two variables independent 
 of each other, then 
 
 d 2 u _ d 2 u d 3 u d 3 u d 3 u d A u 
 
 dydx dxdy dy 2 dx dxdy 1 dy dx 2 da? dy 
 
 d n+r u d n+r u 
 
 and generally = 
 
 dy r dx n dx n dy r 
 
 In a function of any number of variables, the order of 
 differentiation is indifferent. 
 
 The total differential of two variables is equal to the sum 
 of the partial differentials ; or if u=f(x, y), 
 
 d n u = - dx n 4- n - - r^~ dx n ~ 
 dx n dx n ~ l dy 
 
 w+ a 
 
 d 2 u 
 
 Ex. (1.) Lefjttsstf 8 ^ 2 : find du. and 7 
 
 dxdy 
 
 To find the partial differential coefficient ( ), consider 
 
 \dx/ 
 
 y constant, and differentiate with respect to x ; and to find 
 
 ( ), consider x constant, and differentiate with respect 
 
 \<y/ 
 
 to y. 
 
58 FUNCTIONS OF TWO OR 
 
 m i d^u cl 2 u (du\ . 
 
 To find - - or - - > differentiate I - considering j 
 dydx dxdy \dxi 
 
 constant, or differentiate (-) considering y constant. 
 \dyl 
 
 du= ~ dx+ dy 
 
 dydx dxdy 
 
 dydx 
 i<hi\_(a?-f}. s > l x- 
 
 \dJ~ ^ 2 - 
 
 . 2 a^- 2 -2 
 
 ^ 
 
 (3.) w=sin~ 1 -; find 
 y 
 
MORE VARIABLES. 59 
 
 sinz=-> consider x constant. 
 
 y 
 
 du x 
 
 COSW- = 5 > 
 
 dy y 2 
 
 /du\_ x x _ x 
 
 "\dyl~~ y 2 cos u ^2 \/l sin 2 ^ y V -y 2 
 
 rdu\ 7 (du 
 Hence Jw = }dx + - ? - 
 
 1 a? _ , _ ydxxdy 
 
 x 2 yJylx 2 " y^y^x 1 
 
 dydx ytx 1 (y 2 x dxdy 
 
 yZjj 
 
 (4.) M= , j find rfw, and show that 
 a 1 * a 
 
 dxdz (a 2 z 2 ) 2 dzdx 
 
 d 2 u _ Zx 2 z _ d 2 u 
 (a 2 z 2 ) 2 dzdx 
 
 4a^y^ _ d 2 u 
 
 = -^-r 2^ = -r 
 
 ) and 
 
 dxdydz (a 2 z 2 ) 2 dzdydx dydxdz 
 
 First differentiate considering y, z constant ; then consi 
 dering x, z constant ; and lastly considering x, y constant. 
 
 du %x du x 2 du 
 
 x 2 
 
 9 9 
 
 a 2 z 2 
 
60 FUNCTIONS OF TWO OR 
 
 -r- = -^ L - s Consider x. z constant, and differentiate. 
 dx a 2 z 2 
 
 dxdy a 2 z 2 
 
 = -^ r, Consider z constant, and differentiate. 
 
 dy a 2 ,? 2 
 
 d 2 u 2x TT d 2 u 2x d 2 u 
 
 -, r-=) - Hence 5-= -5 r,= - 7 r" 
 
 dydx a^z- dxdf/ a 2 z 2 dydx 
 
 Again = 2 __ ^ 2 - Consider x } y constant, and differentiate. 
 
 dxdz~ (a 2 z 2 ) 2 ~(a 2 -z 2 ) 2 
 
 = 7-J5 ^ Consider y, z constant, and differentiate. 
 dz a 2 z^ 
 
 d 2 u _ 
 
 ?e ~ 
 
 7 9 
 
 Again = - - - Consider x constant, and differentiate. 
 dy a 2 z 2 
 
 (a 2 -* 2 ) 2 
 
 (a 2 z 2 } 2 (d 2 z 2 } 2 
 
 \ / \ / 
 
 Consider x, z constant, and differentiate. 
 
 d 2 u 2 3?z d 2 u $ 
 
 \-r-n srs- Hence = - 
 
 dydz (a 2 -z 2 ) 2 dzdy 
 
 Now r-=~5 - r,- Consider x, y constant, and differentiate. 
 dxdy a"z* 
 
 dxdydz~ (a 2 -z 2 ) 2 "(a 2 -^ 2 ) 2 
 
 =T-S ^jTo Consider ?/, z constant, and differentiate. 
 dzdy (a*z*y 
 
 dzdydf (a 2 -z 2 ) 
 
MORE VARIABLES. 61 
 
 l = f__ . Consider x constant, and differentiate. 
 
 dydx a 2 z 2 
 
 Hence 
 
 dxdydz~ (a 2 -z 2 ) 2 (a 2 -z 2 ) 2 
 d s u 
 
 dxdydz (a 2 z 2 ) 2 dzdydx dijdxdz 
 (6.) ux 2 y^ > nnd du, and show that 
 
 d 2 u 
 
 dydx dxdy 
 
 (7.) u~> du=- l 
 
 (8.) uxy-, du=xv (-dx+logxdy, and 
 
 d 2 u 
 
 ,/!,;/! \ 
 xy \ | logff = 
 \x x i 
 
 x \ . 
 
 dydx \x x i dxdy 
 
 , x . d 3 u 2 . x x x 
 
 - 
 y 
 
 , x . u . 
 
 (9.) u=$m-> , , 9 =-5Sin -cos-= -r^-j 
 
 dydx 2 y 6 y y 4 y dx*dy 
 
 (10.) u =y sin ^r + x sin^/ ; show that 
 
 d 2 u d 2 u 
 
 -r - = UU v* -J- UU U - -= - J~ 
 
 dydx dxdy 
 
 (11.) n = sin (x 2 y) j show that 
 
 d 2 u , , 9 x 9 / o \ i ^ 2 ^ 
 
 =2^ (cos (ofy) -Jfljf sm (o; 2 y) } = 
 
 C12 ) u^L. , show that 
 
 dydz 2 ~dzdydz 
 
 dzdxdz 
 
62 EULER S THEOREM. 
 
 (13.) u = - > find du. and show that 
 
 x+y 
 
 d 2 u _ . xy d 2 u 
 ~ 
 
 dy dx~ (oc + yf ~ dx dy 
 (14.) u={(a-x) 2 + (b-y) 2 + (c-z) 2 }~ *- } show that 
 
 1? - 77 
 
 (15.) ^=sin"" 1 - > find du, and show that 
 
 x 
 
 dydx y (^xy)^ dxdy 
 
 x 2 y 2 
 
 (16.) ?=sin~ 1 } -$> show that 
 
 * 2 
 
 2 
 
 CHAPTER IX. 
 EULER S THEOREM FOR THE INTEGRATION OF HOMOGENEOUS 
 
 FUNCTIONS OF ANY NUMBER OF VARIABLES. 
 
 If u be a homogeneous algebraic function of n dimensions 
 of any number of variables x, y, z, &c., then 
 du du du 
 
 Ex. (1.) tt = T + ;/T ; here n= - 
 
 2 
 
 dx 
 
EULERS THEOREM. 
 
 63 
 
 du du 
 
 (2.) ysain- 1 -j here w=0. 
 x y 
 
 o- -i i 
 Similarly = 
 
 dy \/2y (x+y) Jxy 
 du du _ xyxy 
 
 dx dy 
 
 : = 0. 
 
 y 
 
 (3.) w= v# 2 -{-y 2 ; here w = 
 
 du . du n d 2 u d 2 u 
 
 (4.) w= 
 
 ^=- 
 here n=2. 
 
 du du du 
 
 - r +y- r + z- r 
 dx dy dz 
 
 (5.) u= ^ .1 here n=3. 
 
 x+y 
 
 du du 
 
 -r+y-r 
 dx * dy 
 
64 ELIMINATION 
 
 CHAPTER X. 
 
 ELIMINATION OF CONSTANTS AND FUNCTIONS BY 
 DIFFERENTIATION. 
 
 Ex. ( 1 . ) Let y ax? + & = ; eliminate the constants a and I. 
 
 , 
 dx dx 2# 
 
 Substituting this value of in the given equation, 
 y -- -~+b=Q, an equation from which a is eliminated. 
 
 To eliminate b, take the equation -~= 2 #, and proceed 
 to the second differential coefficient. 
 
 1 , , 
 
 9 an equation from which a and 6 are 
 
 both eliminated. 
 
 (2.) y 2 axbx 2 =Q ; eliminate a and 5. 
 
 :.a=2y-2bx ..... (1) 
 
 Differentiating again, we have 
 d*y dy dy 
 
 Substituting from (1), (2), the values of a and b in the 
 given equation, there results 
 
 
 
 y 2 =%xy~ -- x 2 */ x 2 \-r-\ , an equation from 
 
 ^ dx * dx 2 \dxl 
 
 which a and b are eliminated. 
 
BY DIFFERENTIATION. 
 
 65 
 
 (3.) 
 
 + 5 + 4 y= 
 
 - 
 ** 
 
 dx* 
 
 (4. ) y = otf 1 + ae*"^ ; eliminate e 
 
 Substituting this value of in the given equation, 
 
 y=x n + (4- no?- 1 } > my=mx n +-/- - 
 
 \dx I m dx 
 
 First, consider y constant, and differentiate with respect to x. 
 
 o 2 
 
66 ELIMINATION 
 
 Again, #/*-j +$(xy)> Considers constant, and differen 
 tiate with respect to y. 
 
 TT ^2 2 
 
 Hence ar - y 2 :=0. 
 da? dy 1 
 
 (6.) Let y=mx s ; eliminate the constant m, and show that 
 
 _ 
 
 (7.) Let y = v miP + n ; eliminate m and n, and show that 
 
 (dy\ 2 ._ d 2 i/ 
 \Tx) ~~ ~ y ^ 
 
 (8.) Let a-\-c(cx y) = ; eliminate c ; and show that 
 
 (9.) Let ^ 2 + -- t y 2 = ; eliminate the constants a and >, 
 
 TO , <y 7 
 
 and show that coy ~ +x (} y~=Q. 
 
 dx 1 \dxi dx 
 
 (10.) Let (a 1) (x+y) xy + # = ; eliminate a, and 
 show that 
 
BY DIFFERENTIATION. 67 
 
 (11.) Let ctanm# ysecm#-f #=0 ; eliminate a and c, 
 and show that 7^z = m V* 
 
 (12.) Let y e x cosx ; eliminate the circular and exponen- 
 
 dy 1 d*y 
 
 tial functions, and show that # = ~7 t> 175* 
 
 <z# JB ar 
 
 (13.) Let y=w cos(r# + a) ; eliminate a and n, and show 
 that ^= ^ 2 y- 
 
 (14.) Let y = sin(log#) ; eliminate the functions, and 
 
 2 <% , dy , n 
 show that ^2 ^^7 ry 111 ^ 
 
 (15.) Let y = a<? 2 *sin(3#+&) \ eliminate a and b, and 
 show that 2 ~^~j h!3y = 0. 
 
 (16.) Let (x a) 2 + (y /3) 2 =r 2 ; eliminate a and ft and 
 
 show that 
 
 \d2! 
 
 (17.) Let y= e x _ x ; eliminate the exponentials, and 
 
 & a 
 
 show that / 2 =l 
 
 (18.) Let - = 0(^ y 2 ); eliminate the arbitrary 
 o:+y w 
 
 function 0, and show that y~ + x-j-=mz. 
 
 1 y 
 
 C19.") Let -xzfa- ; eliminate the function <i>. and show 
 n r x 
 
 that ir-^4-v^ + ^=0. 
 
68 MAXIMA AND MINIMA. 
 
 (20.) Let -=6- ; eliminate the function d>, and 
 
 x a r x a 
 
 show that (yV)-! h (z y) -7-== tf a . 
 
 dy ^ dx 
 
 CHAPTER XI. 
 
 MAXIMA AND MINIMA. 
 FUNCTIONS OF TWO OK MORE VARIABLES. 
 
 If u be a function of two variables x and y } then putting 
 du du . n d 2 u d 2 u / d 2 u \ 2 d 2 u . d 2 u 
 
 T-= T-=; rf -7-5- TT > i r 3^ and T^ 
 
 rf^ y i/^ J dy* \dydxl dx* dy* 
 
 having both the same algebraic sign, u will be a maximum 
 when that sign is negative, and a minimum when it is 
 positive. 
 
 If, on substituting the particular values of x and y, de 
 termined bv putting =0. =0. in the second differen- 
 dx dy 
 
 tial coefficients, these should vanish, then the third diffe 
 rential coefficients must also vanish, or the function will not 
 be a maximum or minimum. 
 
 If u =f (oc, y, z), then we must put = 0, - = 0, 
 
 dx dy 
 
 = 0, and we must have the condition fulfilled that 
 dz 
 
 d 2 u d 2 u / d*u \ 2 ) 
 
 "J-5 T~S "" \-J~r) f exceeds 
 ifo 1 6/y \dxdz) ) 
 
 d 2 u 
 
 tu u __ u_ u \ 
 \dydz dx 1 dxdy dxdzi 
 
TWO OR MORE VARIABLES. 69 
 
 I Ex. (1.) Let uxt + y* kaxy 2 , find x and y when u is 
 a maximum or minimum. 
 
 Differentiate, first considering y constant, and then 
 x constant. 
 
 _ = 4#3_ 4dw 2 =0, __ == 4 ? y3 
 
 das ay 
 
 . .=av / 8. 
 
 7O 
 
 = 
 dxdy 
 
 d 2 u d 2 u d 2 u 
 
 " TT, T-TT > -; ?-> and since the algebraic sign of 
 dx* dy 2 - dxdy 
 
 d 2 u ,d 2 u . - 4r - 
 
 and is positive, sc=_a v 2, and y=a: v8, give 
 
 u= a minimum. 
 
 7ft 
 
 If we take the values #=0, v=0, then =0, and 
 
 dx 2 
 
 d 2 u 
 
 j=0, and also the third differential coefficients 
 dy 2 
 
 Hence also ^=0, 2/=0, give u= a minimum. 
 
 (2.) To determine the greatest right cone that can be cut 
 out of a given oblate spheroid. 
 
 Let AEDE be the ellipse which generates the spheroid, 
 0, I its semi-axes, CN=x, NP=y= radius of base of cone. 
 
70 
 
 MAXIMA AND MINIMA. 
 
 Then y* =. (<$ #2) equation to 
 
 a* 
 
 ellipse j and 
 
 V altitude of cone =AN=< 
 and try 2 = area of base, 
 . . its volume v = J Try 2 , (a + #), a maximum, 
 . . ?/ 2 . ( + ^r) = a maximum, 
 
 fa I 
 
 2 (a + x) a 
 
 But, differentiating the 
 
 equation to the ellipse, y=va 2 x 2 , we have 
 
 a 
 
 .-.,=-, 
 
 ^ a; J = 
 
 Hence 
 
 v=-7r?/ 2 . 
 
 
 
 , /^ (3.) Let M=^ 4 + 2/ 4 ~~2(^ 2/) 2 ; find the values of x and y 
 which render u a maximum or minimum. 
 
 du_ ,_ 
 
 </Z7 
 * 
 
 <^y 
 
TWO OE MORE VARIABLES. 71 
 
 d 2 u d 2 u d 2 u d 2 u , . 
 =4, :. - -T^>I T- and smce tne 
 2 2 
 
 , , . . . d 2 u . <Z 2 M . 
 algebraic sign 01 and is positive. 
 ax 2 " dyj* 
 
 :. x= + A/2, and y=+ v2, give M= a minimum. 
 
 JT (4.) Let w=a{sin^4-sin?/+sin(^4-2/)}; show that u is a 
 maximum when ^=r?/=60 . 
 
 =:[ cos o; + cos (# + 2/)} := 0, = a [ cos ?/+ cos (^ +?/)] =0, 
 
 . . X = y, COS #+ COS ( + ?/):=: COS #+ COS 2 # 
 
 = cos^? + 2 cos 2 ;r 1 = 0, 
 
 d 2 u 
 
 j,=a{ sin^? sin(^4~2/)} = a {sin 60 + sin 120} 
 
 = sin!20= a r 
 
 2 
 
 . t? 2 ?^ rf 2 M d^U d 2 U 
 
 "~rr> ~r~o>~; T- 9, nc * V the algebraic sign of and 
 dx 2 dy* dxdy dx 2 
 
 d 2 u . 3 a /- 
 
 is negative, .*. u = - \/3= a maximum. 
 
 (o.) A cistern, which is to contain a certain quantity of 
 water, is to be constructed in the form of a rectangular 
 parallelopipedon ; determine its form, so that the smallest 
 possible expense shall be incurred in lining its internal 
 surface. 
 
72 MAXIMA AND MINIMA. 
 
 Let a s = its content, x = length, y = breadth, then 
 
 = depth. 
 
 y ^ / 
 
 a 3 a 3 
 :. surface =uxy+Z h 2 a minimum. 
 
 * y 
 
 / 
 
 
 
 du 2 a* du 2 a* 
 
 ~ry -- 2-=> -r= x -- ^ 
 
 dx x 2 dy y 2 
 
 a s a* 2*a 
 
 =-5 = -- Hence the base must be a square. 
 
 xy 2%a 2 2 
 
 and the depth equal to half the length or breadth. 
 
 d 2 U d 2 U I d 2 U \ 2 
 
 ~r~r> * ~7~r> > ? r~ ) * Hence u is a minimum. 
 
 das* dy* \dxdyl 
 
 (6.) In a given circle to inscribe a triangle whose peri 
 meter shall be the greatest possible. 
 
 Let r be the radius, and 6 and <p two of 
 the angles of the triangle ; draw ED J_ 
 A C the base : then, Euc. B. 6. prop. C, 
 
 7? T) 
 
 2r, :.a=2r -- = 
 
 c 
 
 c sin0 
 - i 
 
 a sm0 
 
 sinfl 
 
 Hence u = a -f c + b = 2r { sin + sin -f- sin (0 + 
 
TWO OR MORE VARIABLES. 73 
 
 -^= 2 r { cos <i> + cos (0 + </>)}= 0, 
 dty 
 
 ;.cos0=cos0, 6=<f>, + 0=20, 
 
 /.cos0 + cos20=0, cos0 + 2cos 2 0-l = 0, 
 
 v , WIJVy _ . . 0=60=0. 
 
 22 2 
 
 Hence the Z s are all equal, and the A is equilateral. 
 d 2 u 
 
 \/3 / 
 
 =-2r. =-rv / 3, 
 
 .-."LJl.^-lZr 2 , and ^j~ 
 
 do 2 dtp* dO^ d$* 
 
 Hence the perimeter is a maximum. 
 
 (7.) To determine the least polygon that can be described 
 about a given circle. 
 
 Let 0j, 2 , 3 , . . . n , be the successive angles contained 
 between the lines from the centre to the angular points of 
 the polygon and the radii of the circle ; then if the radius 
 be r, and the first of those lines be I, the area of the right- 
 angled triangle whose angle at the centre is d l will be 
 
 11 r 2 
 
 -r^sin0 1 =-r.r s,ecd 1 . sm6 l = tan 0! ; 
 22 2 
 
 and similarly of all the n triangles successively, into which 
 the polygon may be supposed to be divided ; so that the 
 entire area of the polygon will be 
 
 r 2 
 
 - ... +tan0 n ). 
 
74 MAXIMA AND MINIMA. 
 
 But tan0 n = tan{2ir(0 1 + 6 a + . . + 0^)} = 
 
 where 1 =0 1 + a + + 0-i. 
 
 /. tt=tan0 1 + ta]i0 a +tan03 + . . tan(27r 0J, a min. 
 Now, differentiating with respect to 1? considering the 
 others constant, and remembering that 0j is contained in 1? 
 the assumed sum of the series, we have 
 
 And similarly, any one of the angles is equal to the angle 
 immediately preceding ; hence all the angles are equal, and 
 the polygon is consequently equilateral. 
 
 (8.) Of all triangular pyramids of a given base and alti 
 tude, to find that which has the least surface. 
 
 Let a, I, c be the sides of the base, h the 
 altitude of the pyramid, 0, 0, ^/, the inclina 
 tion of the faces to the base. 
 
 Then, if p be a perpendicular from the ver 
 tex on the side a. sin0=-j .*. = =Acosec0, 
 
 p sm0 
 
 area of face =-ap=-ah cosec0, 
 . . area of the three faces = -^ah cosec0 -f -bh cosec^ -f-dicoseci//, 
 
 u=-~ h(a cosec0-f b cosec^ + c coseci//) (1). 
 
 Also, the base of the pyramid may be divided into three 
 triangles whose altitudes are readily determined ; 
 
 Y f\ = tan 0, . . -7- = cot 0, . . altitude aO=k tan 0, 
 au h, 
 
 area 
 
TWO OR MORE VARIABLES. 75 
 
 .*. area base = ^ ah cot -f ^ bh cot </> + *> ^ cot ^ 
 and putting this area =m 2 , we have 
 
 m 2 = -h(acot6 + bcot(j) + ccotty ..... (2). 
 
 J 
 
 From (1), -= cosec0 cot0 ccosec^/ cot J/ ^ = 0, 
 -^=-1 5cosec0cot</> c cosec i cot t// 1=0, 
 
 :. a cosec 6 cot 0=c cosec v// cot \L > 
 
 du 
 
 d^ 
 I cosec (f> cot $= c cosec ^ cot ^/ > 
 
 d^ . d^ di> 
 
 a cosec 6 cot - = c cosec u/ cot u/ -r- > 
 
 \ . 
 
 I cosec cot fyc cosec if cot ^ > 
 
 /. cosec cot - = ^ cosec rf> cot 0-7- . 
 
 d) W(7 
 
 From (2), =a cot d + b cot <p + c cot \^>, 
 
 2m 2 
 ccotv/;= r -- acotS bcot<j>, 
 
 -c (1 + cot 2 ;/,) ^=a (1 + cot 2 0), 
 a 
 
 -c (1 + cot 2 ^) ft= b (1 + cot 2 0), 
 
 dty_ cosec 2 
 
 </0 c cosec 2 ;// 
 
 _ [ Substitute these values in (3). 
 
 dy_ b cosec^ 
 
 d<f> c cosec 2 ^/ 
 
76 MAXIMA AXD MINIMA. 
 
 , n b cosec 2 6 a cosec 2 
 
 a cosecfl cot --- 5-7= ^ cosecri cot 6 --- 5-* 
 
 c cosec-d c cosec 2 -^ 
 
 . . cot 6 cosec ^ = cot <p cosec 0, 
 
 cosfl 1 cos 6 1 
 sin/9 sin< smp sinfl 
 
 Similarly, by finding the partial differential coefficients 
 
 ) -t considering first \f/ and then constant, it may be 
 ad ay/ 
 
 shown that 0=4/. 
 
 Hence 0=^=il/, or the faces are equally inclined to the 
 base. 
 
 Jr 6 (9.) Required the dimensions of an open cylindrical vessel 
 of given capacity, so that the smallest possible quantity of 
 metal shall be used in its construction, the thickness of the 
 side and base being already determined upon. 
 
 Let a be the given thickness, c the given capacity, 
 x = radius of base inside, y= altitude inside. Then 
 Whole volume v = ?r (x + a) 2 - (y -f a) , 
 
 Interior volume c=7r# 2 y, hence the quantity of metal 
 v c=7r(x+a) 2 (y + a) c=a, minimum, 
 .". (x -\- of (y -}- o}=-& minimum, 
 
 C 
 
 c 1 dy c 2 TTX* c 
 
 But y= :.-=. -- > .*. - = s> 
 
 * TT y? dx TT x? x + a ITS? 
 
 \\Tience x = y= \~j- Therefore the altitude must be 
 made equal to the radius of the base. 
 
TWO OR MORE VARIABLES. 77 
 
 (10.) u=a? t 2>axy+y i find the values of x and y which 
 render u a maximum or minimum. 
 
 x=a ) yCL, u=& minimum when a is positive, 
 
 and a maximum when a is negative. 
 
 (11.) u = aa? Ijx^y -f y 1 , find the values of x and y which 
 make u a maximum or minimum. 
 
 (12.) u=aa?y 2 x^y 1 x^y^ ; find the values of x and y 
 which make u a maximum or minimum. 
 a 
 
 a maximum. 
 2 
 
 (13.) u=(l- ^ - -) (l- ^) ; find the values of x 
 
 * 
 
 y> 
 
 and y which render u a maximum or minimum. 
 
 7T 
 
 (14.) u=a cos 2 #-f-& cos 2 y, where y=--r -\-x\ find the values 
 
 4 
 
 of cos# and cosy which make u a maximum or minimum. 
 
 ! + - rf =1 * 
 
 2 ~~ 2 \/ 2 4- 6 2 2 
 
 2 ), a maximum with the upper, and 
 
 a minimum with the lower sign. 
 
 (15.) Divide a given number a into three such parts #, y, 
 
 and #, that -^--\ -\ shall be a maximum or minimum, 
 J o 4 
 
 and determine which it is. 
 
 (16.) Inscribe the greatest triangle within a given circle. 
 
 The triangle is equilateral. 
 
 (17.) A given sphere is to be formed into a solid composed 
 of two equal cones on opposite sides of a common base, in 
 such a manner that its surface may be the least possible : 
 find the dimensions of the solid, and compare its surface 
 with that of the sphere. 
 
 H 2 
 
78 TANGENTS, NORMALS, AND 
 
 (18.) Show that the greatest polygon that can be inscribed 
 in a given circle is a regular polygon. 
 
 a? ?/2 Z 2 
 
 (19.) In a given ellipsoid, whose equation is -f -75 + -5 = 1 , 
 
 a* o^ c^ 
 
 to inscribe the greatest parallelopipedon. 
 
 If ne, y, z be the half-edges of the parallelopipedon, 
 
 a b c 8abc 
 
 x= > y= -=.> z -=.1 u= - 
 v/3 v/3 %/3 3 
 
 (20.) To find a point P within a given triangle, from 
 which, if lines be drawn to the angular points, the sum of 
 their squares shall be a minimum. 
 
 If A, , C be the angles, a, b, c the sides of the triangle ; 
 
 then CP=l(2a 2 + 2b 2 -c 2 )* 
 o 
 
 The point is the centre of gravity of the triangle. 
 (21.) Divide the quadrant of a circle into three parts, 
 such that the sum of the products of the sines of every two 
 shall be a maximum or minimum, and determine which it is. 
 
 CHAPTER XII. 
 
 TANGENTS, NORMALS, AND ASYMPTOTES TO CURVES. 
 
 If y=f (x) be the equation to a curve, 
 
 y y=-j-(x x) is the equation to a tangent. 
 If M=0 (#, y)=c be the equation to the curve, 
 (x x ) + -^ (/ y)=0 is the equation to the tangent. 
 
ASYMPTOTES TO CURVES. 79 
 
 The equations to the normal are 
 
 dx , du , t ^ du , , , . 
 
 y- y = - T y (x - x ^ and TX (/ - y) ~ T y (*j-*>=<>- 
 
 The tangent=y A/ 1 + (y) Normally A/ 1 + (^) > 
 
 Subtangent = y > Subnormal = y -7- 
 
 The portion of the axis of y intercepted between the 
 
 origin and the tangent is yx=.y^. 
 
 dx 
 The portion of the axis of x so intercepted is x y =x . 
 
 Ex. (1.) Draw a tangent and normal to a given point P 
 in the common or conical parabola. 
 y 2 =ax is the equation to the curve, 
 
 , dy . dy 2a 
 
 Subtangent NT=y=-=2x. 
 y dy 2a 
 
 Hence to draw the tangent, let fall the perpendicular PN, 
 take NT=2AN, and join PT PT will be the tangent. 
 
 Subnormal NG =y~=2a. 
 y dx 
 
 Hence to draw the normal, take NGf=2 AS, and join PG ; 
 PG will be the normal. 
 
 (2.) Let y n =a n ~ l x be the equation to a curve ; find the 
 subnormal and subtangent. 
 
 , dy dy a n ~ l 
 
 ^.^ -_ 
 
 dy ya n ~ l a n ~ l 
 Subnormal NG=y -f=- - T = - 9 = 
 
 ^ >V/>1 /Vl/3//l~~l /M/1*W - jj 
 
 x y* . 
 
 ny n 1 wy n 2 ny n 2 nx 
 
80 TANGENTS, NORMALS, AND 
 
 Subtangent NT= y = ^ =. - = nx . 
 dy a n ~ l y n 
 
 x 
 
 If n=2, y 2 =ax, NG=~, NT=2x, and the curve is a 
 2 
 
 parabola. 
 
 (3.) Let u=x 3 3axi/+y 3 =0 be the equation to a curve; 
 determine the subtangent. 
 
 , 
 
 dx 
 
 . / o \ dy o dyay x 2 
 
 " (y ax )-r= a y x 2 , -/=-? - 
 
 dx dx y 2 ax 
 
 :. Subtangent NT^y= 
 
 * 
 
 dy ayx 2 
 
 (4.) If y 2 =a(x+a) be the equation to a parabola, the 
 origin in the focus ; show that the points of intersection of 
 the tangents with perpendiculars from the focus are deter 
 mined by the equations #,= a, y, = -- 
 
 2i 
 
 S the focus, AS=a, SA T =x, AN=x+a, NP=y, 
 . . . (1), eq n . to curve, 
 
 (2), eq n . to tan., 
 
 T/ 
 
 dx A 
 
 ,= -r-x l ..... (3), eq n . to ppdr. from origin, 
 
 .-. by subtraction, y=_ gx,- ^x.+^x, ...... (4). 
 
 ,., dy 2a dx y 
 
ASYMPTOTES TO CURVES. 81 
 
 2 a 
 
 05,= a, 
 
 (5.) The equation x m y n =.a, which includes the common 
 hyperbola, is said to belong to hyperbolas of all orders. Find 
 the subtangent at a given point in the curve. 
 
 __.. t7 J an dx an 
 
 dy y 2n y n+l dij inx m ~ 1 
 
 /// / ( 
 
 dx n a _ n n 
 
 mx m ~ l n n ~ mx m ~ l 
 
 -^ 17-/TT M/t * / W ** * 
 
 .*. Subtan. NT=y~-= - - = .05?"= x. 
 
 (6.) Given two points A and E, find the locus of P when 
 the angle PEA is double of the angle PAJB, and draw an 
 asymptote to the curve traced by P. 
 
 A the origin, AJB=a, AN=x, NP=y, A = d, =26. 
 
 , 
 AN x BN ax 
 
 y = _ 
 
 a x 
 
 :. y 2 =3x 2 2ax, the equation to the curve. 
 
 Whence, if y=0, x= 
 curve will pass through 0. 
 
 2 2 
 
 Whence, if y=0, 0;= a, and taking A0=. Afi, the 
 o o 
 
82 TANGENTS, NORMALS, AND 
 
 The origin may be changed to by putting x^ = ON, and 
 substituting the resulting value of x in the equation to the 
 
 curve ; whence y=. x (3 4- ] , 
 
 \ OG / 
 
 ." y=dz^v3: is the equation to the asymptote. 
 
 Ifa;=0, y= -j ify=0, =: F^> 
 
 ^^ / 
 .". -y-= v 3= tan 60, and the asymptote cuts the axis of 
 
 x at an Z of 60, and at a distance = from the point 0. 
 
 o 
 
 (7.) If j/ 2 = - be the equation to a curve ; find the 
 equation to the asymptote. 
 
 .*. y= (x+a) is the equation to two asymptotes, and v 
 if x=Q, ya, .*. an asymptote cuts the axis of y at the 
 distance a from the origin; and v ify=0, x= a, .*. an 
 asymptote cuts the axis of x at the distance a from the 
 origin. 
 
 dy 
 Again v -^-=1.= tan 45 or tan 135, . . these asymp- 
 
ASYMPTOTES TO CURVES. 83 
 
 totes cut the axes at an angle of 45, and are consequently 
 at right-angles to each other. 
 
 Putting x=a in the equation to the curve, we have 
 
 ) 
 
 .*. there is another asymptote parallel to the axis of y. 
 
 (8.) If y 2 = (x 1) Vx 2 be the equation to a curve ; 
 find the point and angle at which the curve cuts the axis 
 of x, and the values of x and y when the tangent is perpen- 
 cular to that axis. 
 
 Ifx=0, y _2 
 
 Ify=0, (a?- 
 
 |=(*-i) 2 
 
 Hence, if x = 3, = tan 9 = = = 2, and the 
 
 curve cuts the axis of a? at a distance 3 from the origin, and 
 at an angle whose tangent is 2. 
 
 Again, ifa?=2, </^2=0, :. y-2 = 0, y=2, 
 
 dy 3x5 65 1 
 
 -= =00 when x=2. 
 
 dx 2Vx2 
 
 Hence the tangent cuts the axis of x at an angle of 90, 
 or it is perpendicular to that axis when x=2 and y=2. 
 
 (9.) If from any point P in an ellipse a straight line be 
 drawn to the centre making an angle with the normal, 
 and if I be the inclination of the normal to the axis major 
 
 ., 
 show that tan d = 
 
84 
 
 TANGENTS, NORMALS, AND 
 
 Let CA=a, CB=b, CN=x, NP=y, 
 L CPG=d, CGP=l. 
 
 8 C G 
 
 (O O\ 7 O ^ 11* 
 
 a z x z ) = o z ^- 5 eq n . to ellipse. 
 a- a 2 
 
 =~2 C A r = 7p by a property of the ellipse, 
 
 12> t^ 
 
 a 
 
 - = tan ?, also = 
 x a* x 
 
 =t&n(PGN-PCN)= 
 
 I 2 
 tan? -- rtan? 
 
 1+ tan?--- tan? 
 
 (10.) From the centre C of a circle a radius CR is drawn 
 cutting the chord BD in Jtf, ifP is drawn at right-angles to 
 BD and equal to MR ; determine the locus of P, and draw 
 the asymptotes. 
 
 Let BD, CO be the co-ordinate axes, 
 A the origin, 
 CR=a, CA =c, AM=x, MP=y. Then 
 
 , or 
 y=a Vc 2 + x 2 , the equation required. 
 
ASYMPTOTES TO CURVES. 85 
 
 If x=0, y=a-c=CR-CA = CO-CA=AO. 
 
 =0, x= \= *CF 2 - CA 2 
 
 :. xAD or AB. 
 
 If #=00, 2/= oo. Hence the curve passes from 
 through B and D to infinity. 
 
 To determine the direction of the tangents at these three 
 
 points; -^ = tan = ip -7=^= = if x = 0, :. at the 
 
 dx Vc 2 + x 2 
 
 tangent is parallel to the axis of x. 
 dy x AD 
 
 -g.=tfl,nfl:=:f -- = - : 
 
 dx Vc 2 + x 2 VGA 2 +AD 2 
 
 determines the direction of the tangents at D and B. 
 
 Again, putting x, = ON= CO CJ^= a (c + y), we have 
 y a c x, j and putting yi = NP = x } and substituting 
 these values of x and y in the equation to the curve, the 
 origin will be transferred to 0. Thus 
 
 :. 2/ l 2 =2c^, + ^, 2 , which is the equation to the rectangular 
 hyperbola. 
 
 To find the equation to its asymptotes, 
 
8G TANGENTS, NORMALS, AND 
 
 .*. 2/= (#-f-c) is the equation to the two asymptotes; 
 and Y putting y=Q, we have x=c, and 
 putting x=Q, we have y=+c ; also 
 
 {/ty 
 Y -^-=tanO=l ; .". the asymptotes cut the axis OF 
 
 at / s=45 and 315, at the distance c from the origin 0. 
 Take OT=CA, and draw the lines TS, TS t at Zs=4o 
 and 315 respectively, these will be the asymptotes. 
 
 (11.) The normal to the curve whose equation is y *=kax, 
 
 4 
 
 is a tangent to the curve denned by y 2 = (x 2a) 3 . 
 
 dy 2a y 
 
 y z =zkax, -j-= ) :. y, ?/= (#, #), eq n . to normal, 
 
 9/., . 
 
 /, = 0, then 
 
 ^=^4-2=:part cut off from axis of x. 
 *~27a 
 
 , 
 
 Jx y x 2a dj 3 
 
 ^/x 2 , _ x-j-4^ 
 
 /. x y - = x - (x 2) = - = part cut off from 
 wy o o 
 
 axis of x. 
 
 Hence, that the normal and tangent may cut the axis of 
 x at the same point, we must have the equation 
 
 G 
 
 % 
 
 But, the angles they make with the axis of x ought to be 
 the same, and since 
 
 dx y dy 3 y 
 
 -r-=^> and = 
 
 dy 2a dx. 2 x 2a 
 
ASYMPTOTES TO CURVES. 87 
 
 . y_ 3 y y2 4# _9y*_ _4_ _ 
 
 " 2a 2\-2a a 2 a (x-2a) 2 3a ( *~~ 
 
 :. 3=x 2a, x=3#-f2a, the same as before. 
 Hence, the normal and tangent, cutting the axis of x in 
 
 the same point and at the same angle, must be coincident. 
 (12.) In the curve defined by j^ax^ + aP prove that the 
 
 portion of the axis of -y intercepted between the origin and 
 
 the tangent =|.(--^-) 
 
 dy 
 
 3 ( y^ X s ) 2a% 2 ax 2 a x 2 
 ~ " 
 
 a x 2 a x 2 a / x \ 
 
 : ~~2 \/. _!_*./ 
 
 3 {(a + x)x 2 }* 3 (a + x)%x% 3 
 (13.) If y%=ctfx% ; draw a tangent to the curve, and 
 show that the part of the tangent intercepted between the 
 axes = a, and that perpendicular on tangent = Vaxy. 
 
 2. 2 222 V P 
 
 # =d 3 #3", X% =.CC% 7/3" 
 
 ?. -* ^- - ,-i ^_ ^^ 
 
 3 cb a dx~ x* 
 
 x -~- = x s x = 
 
 7 ^ 
 
88 TANGENTS, NORMALS, AND 
 
 Now DT 2 = 
 
 :. DT=a= part of tan. intercepted between the axes. 
 FD AD AD* af , 
 
 iax?i/3= length of perpendicular on tangent. 
 
 (14.) Suppose a rigid rod BP slides along the line Ax in 
 such a manner that its extremity P shall f 
 be constantly in a given curve whose equa 
 tion is y=f(x\ and let BQ be an n th part 
 of BP- } determine the equation to the locus J; 
 of ft 
 
 Let BP=a, AN=x, NP=y, AM=x t) MQy,. Then 
 
 :NP::BQ:BP, or y, : 
 
 :. y , = - / \x, v a 2 n 2 f/ 2 > , the equation required. 
 
 n (_ n ) 
 
 (15.) Determine the subtangent to the curve of which the 
 normal = 2 a 2 (abscissa) 3 . 
 
 Let x be its abscissa, y its ordinate. Then 
 dy dy_ ( 
 
 Normal PG=y -j-t .". y-y-=2, an equation 
 
 dx dLx 
 
 ently derivable by differentiation from * = -> 
 
 2> 2i 
 
 :. y=ax 2 is the equation to the curve. 
 
ASYMPTOTES TO CURVES. 89 
 
 dy 2a 2 x* 2aW dx 1 
 
 
 
 , .. 
 
 ax y ax 2 dy 2 ax 
 
 dx ax 1 x 
 
 Now 
 
 The equation to, the curve may be put into the form 
 x 2 =-y, therefore the curve is a parabola, whose parameter 
 
 is > and whose line of abscissae is perpendicular to the hori 
 
 zontal axis. 
 
 / - -*\ 
 (16.) The equation to the catenary is 2y=c \e c +e c ) ; 
 
 find the length of the normal. 
 dy c ( 2 
 
 dj? 4 
 
 rdy^ 
 1 + - ;= 
 
 ,.*/: 
 
 /. normal P= 
 
 (17.) If y n -(a + lx)y n -i + (c + ex+fx 2 )y n - 2 -&ie. = Q be 
 the equation to a curve of w dimensions, prove that, if each 
 ordinate be divided by the corresponding subtangent, the 
 sum of the quotients will be a constant quantity. 
 
 Let rj, r a , r a , .... r n be the values of / which satisfy 
 the given equation, and 
 
 s i) s -2> S z> s n the subtangents corresponding to these 
 values of y ; then, by the theory of equations, 
 
 i 2 
 
90 TANGENTS, NORMALS, AND 
 
 . d^ dr a dr 3 dr *_i . 
 
 ~7 I 7" i 7~ n ; } 
 
 ax dx ax ax 
 
 and, taking the differential expression for the subtangents, 
 
 r n dx 
 
 s. 2 = 
 
 . . . n 
 ar. 2 dr 
 
 8 1 dx s.j dx s n dx 
 
 Hence -\ --\ - . . . + =b. 
 
 (18.) If y* x 4 +2bx 2 y=Q be the equation to a curve; 
 find the equation to the asymptote. 
 Assume y=.xz, then 
 
 x=- - -) y = ^\ - 1 which both become infinite 
 1 z 1 z 
 
 when z 4 = 1 or z= 1 . 
 
 AT) 
 ~~ X ~ 
 
 =. - 3 - -> which, when z=l, and consequently 
 
 , An lx b b 
 
 # = oo. becomes -4 Z>= 7= 7= 
 
 2*-ffi 2 2 
 
 # 
 
 Hence y=x ) y=x are the equations to 
 
 A J 
 
 two asymptotes. 
 
ASYMPTOTES TO CURVES. 91 
 
 (19.) Investigate an expression for the subtangent : and 
 in the parabola of the n th order, whose equation is y=ax n , 
 find the subtangent and subnormal. 
 
 Subtangent = x, subnormals na 2 a 2n ~^. 
 
 b 2 
 (20.) The equation to the ellipse being y 2 = 2 (2axx 2 ) j 
 
 find the subtangent and subnormal. 
 
 2 ax x 2 I 2 . . 
 
 bubtangent = > subnormal^ - (a x). 
 
 ax a 2 
 
 (21.) Prove that equals the tangent of the angle at 
 
 CtOG 
 
 which a curve, referred to rectangular co-ordinates, is inclined 
 to the axis. 
 
 (22.) y 2 =a 2 x 2 being the equation to the circle, the 
 origin at the centre, show that the curve cuts the axis of x 
 
 O 
 
 at an angle of 90. 
 
 (23.) y 2 =%axx 2 being the equation to the circle, the 
 origin in the circumference, find the subtangent and normal. 
 
 Subtangent= > normal=. 
 
 a x 
 
 (24.) If an ordinate NP in an ellipse be produced until it 
 meets the tangent, drawn from the extremity of the latus 
 rectum, in T ; prove that the distance of P from the focus 
 is equal to the distance of T from the axis of abscissae. 
 
 (25.) In the ellipse, if it be assumed that o? = #cos; 
 prove that the equation to the tangent will be 
 Ice cos 1 4- ay sin t= db. 
 
 (26.) Find the locus of the intersection of pairs of tangents 
 to an ellipse, the tangents always intersecting each other at 
 right angles. x 2 + y 2 = a 2 -f I 2 . 
 
93 TANGENTS, NORMALS, AND 
 
 4$ 
 
 (27.) y 2 =-^ being the equation to the cissoid of 
 
 Diocles, find the equation to the tangent, and show that 
 there is an asymptote which cuts the diameter at its extre 
 mity at right-angles. 
 
 Equation to tan. y t =. -j ^ r {(3a <c) x { ax] 
 
 (28.) Prove that half the minor axis of an ellipse is a mean 
 proportional between the normal and the perpendicular from 
 the centre upon the tangent. 
 
 (29.) In the logarithmic curve, whose equation is y=a x , 
 show that the subtangent is equal to the modulus of the 
 system whose base is a. 
 
 (30.) Prove that the curve whose subnormal is constant 
 is a parabola. 
 
 I 2 
 (31.) In the hyperbola, whose equation is y 2 = ^ (2ax + tf 2 ), 
 
 show that ?/=+ (# + #) is the equation to two asymptotes 
 
 passing through the centre and equally inclined to the axis 
 of x. 
 
 (32.) Draw the rectilinear asymptotes of the curve defined 
 ^7 y 4 + x 3 y= A 2 , and determine the form of the curve at 
 the origin. 
 
 (33.) Let or 3 y 3 + ^ 2 =0 be the equation to a curve; 
 
 show that the equation to the asymptote is y=#-f-. 
 
 o 
 
 (34.) If ay*=la?c 2 xy be the equation to a curve ; show 
 
 7 i 2 
 
 at y= (-) ( x -- 1 j) i s ^ ne equation to the asymptote. 
 \dl \ 3 as //k/ 
 
 (35.) In the common parabola, whose equation is y^^ 
 
ASYMPTOTES TO CURVES. 93 
 
 find that point at which the angle, made by a straight line 
 
 from the vertex with the curve, is a maximum. 
 
 x=a. 
 
 (36.) A rectangular hyperbola, and a circle whose radius 
 is 2 a, have the same centre ; find the angle of intersection 
 
 of the two curves. v / 15 
 
 Angle = tan" 1 j 
 
 (37.) Find that point in an ellipse at which the angle 
 contained between the normal and the line drawn to the 
 centre is a maximum. 
 
 (38.) Determine the angle at which the curve, called the 
 lemniscata of Bernouilli, whose equation is (y 2 + tf 2 ) 2 
 = 2 a 2 (3? y 2 ), cuts the axis of x. 
 
 (39.) If A be the vertex, P and Q corresponding points in 
 the cycloid and its generating circle, prove that the tangent 
 at P is parallel to the chord AQ. 
 
 (40.) The centre of an ellipse is the vertex of a parabola, 
 the axis of the parabola intersects the axis of the ellipse at 
 an angle of 90, and the curves also intersect each other at 
 right angles ; show that major axis : minor axis :: v2 : 1. 
 
 (41.) If y 1 =.mx + nx 2 , show that an asymptote cuts the 
 
 axes at points indicated by x= -- and y = f 
 
 2n 2n^ 
 
 (42.) Show that the locus of the intersection of tangents 
 to the rectangular hyperbola and perpendiculars upon them 
 
 from the centre is the lemniscata. 
 
 (x + a} 5 
 
 j - TO 
 
 determine the distance of its minimum ordinate from the 
 
 (43.) Draw the asymptotes of the curve y 2= j - TO 
 
 (44.) Find that tangent to a given curve which cuts off 
 from the co-ordinate axes the greatest area. 
 
94 TANGENTS, NORMALS, ETC. 
 
 (45.) Draw a tangent to the curve, whose equation is 
 
 m-l 
 
 y = ax m , and show that the tangent always cuts from the 
 axis of y a portion equal to an m th part of the ordinate at the 
 point of contact. 
 
 (46.) If y*+3? 3# 2 =0, show that y= x+\ is the 
 equation to the asymptote, and that the maximum ordinate 
 is at the point indicated by #=2. 
 
 (47.) If C be the centre of an ellipse, and NP any ordinate, 
 and if in NP a point Q be so taken that its distance from 
 shall be equal to NP ; show that the locus of Q is an ellipse 
 whose major axis is the minor axis of the given ellipse. 
 
 (48.) Draw a tangent to the curve whose equation is 
 
 a? 
 
 y=, - 7,5 and determine whether the curve has an asymp- 
 
 a 2 -f x 1 
 
 tote. 
 
 (49.) ABD is a semicircle, centre C and diameter A D ; 
 EF is a chord parallel to AD, CQR a radius cutting EF in 
 Q; QR is bisected in P. Find the locus of P. 
 
 (50.) Show that the curve, whose equation is 
 axy=Q, has a rectilinear asymptote at the distance b from 
 the origin, and also a parabolic asymptote, whose equation 
 
 Q ~\ 2 
 
 is ay b 2 = (a; -- b\ , the latus rectum of the parabola 
 4 \ 2i I 
 
 being a, and its axis parallel to the axis of y. 
 
 (51.) BAG is a triangle, right-angled at A a straight rod 
 moves through the fixed point C, while one end slides down 
 the line BA : show that the curve described by the other 
 end is a conchoid whose equation is oc 2 t/ 2 =^(x b) 2 (a 2 # 2 ), 
 and determine its subtangent. 
 
POLAR CO-ORDINATES. 
 
 95 
 
 CHAPTER XIII. 
 
 POLAR CO-ORDINATES. SPIRALS. 
 
 Ifr=/(6)), or/?=/(r), smdu=-> then 
 
 Tangent of angle (0) contained by radius vector (r) and a 
 
 tangent to the curve, is tan SPY=r = u 
 
 dr du 
 
 Perpendicular on tangent, 
 
 SY= P = 
 
 Subtangent ST=r 2 
 
 I 
 
 du* 
 
 IA= area ANP, -j-= 
 
 Ex. (1.) Find the polar equation to the common parabola. 
 SP=r, 
 
 :.r + rco$d=2a, 
 
 2a 
 
 1-fcosO 
 
 (2.) The equation to the spiral of Archimedes is r=a9 ; 
 find the angle between the radius vector and tangent, and 
 the subtangent. 
 
 =a, 
 dt) 
 
 dr 
 
 Subtangent ST=r 2 =- 
 dr a 
 
96 POLAR CO-ORDINATES. 
 
 (3.) If r=a (1 + cos0), find the equation between p and r. 
 
 1 1 du 
 
 * 
 
 u~ a + aco*6 dt> 
 
 \ 2 ft 2 sin 2 a 2 l--cos 
 
 1 1 2a 
 
 But cos0= -- #, rt 2 cos 2 0= --- \-a 2 , 
 u u* u 
 
 2 
 
 
 U U A U 
 
 (4.) The tangents at the vertex and extremity of the 
 latus rectum of a conic section intersect ; prove that the 
 distance of the point of intersection from the vertex is equal 
 to the distance of the focus from the vertex. 
 
 Let A be the vertex, S the focus, and T the point of in 
 tersection. 
 
 The equation y= ^2 ax~+. 3? will, by using 
 (i 
 
 the negative sign, comprehend all the conic 
 sections excepting the hyperbola ; and, by using 
 the positive sign, it is the equation to that 
 curve. 
 
 Alsoy, y=i~-(x { x) is the equation to the tangent. 
 
 Differentiating the assumed equation, = 
 
 dx a 
 
SPIRALS. 97 
 
 and substituting the values of y and ~ in the equation to 
 the tangent, we have 
 b 
 
 (x l x). 
 a 
 
 But at the origin #, = (), and x=AS=m suppose. Then 
 
 /- - 5 
 \ v 2am-\-m* 
 a a 
 
 bin a+ 
 
 _ bm 
 
 Now a 2 = 2 -J- (a+m) 2 , by a property of the curve, 
 
 /. 2am+m 2 =b 2 , :. TA=y>=m=SA. 
 
 (5.) In the ellipse, if p be the perpendicular from the 
 centre on the tangent, and r be the distance of the point in 
 
 the curve from the centre, prove that p 2 = 
 
 Perp r CP=r, L PGN= Q, then x = r cos d, y = r sin ; 
 equation to the ellipse. 
 
 ~ 
 
 
 _ 2 
 
 H To - ^*" I o H 
 
 o o 
 
 V a 2 b 2 
 
 a 2 b 2 
 
 b 2 cos 2 H- a 2 sin 2 a 2 (1 e 2 ) cos 2 d + a 2 sin 2 6 
 
 b 2 
 
 where 1 e 2 =-z 
 a 2 
 
 a 2 b 2 b 2 
 
 (! 
 
 du 1 .11 ^ 2 cos0sin0 
 
 dO b 2 u b Vl-e*Go$ 2 i) 
 
 K 
 
98 POLAR CO-ORDINATES, 
 
 2 / du \ 2 1 ~~ ^ 2cos2 ^ C 
 f = ~~ 
 
 2 du \ 2 1 ~~ ^ 2cos2 ^ Cos2 # ( 1 cos 2 ,9) 
 ~~ 
 
 l-2e 2 cos 2 0-M 4 cos 2 
 
 But = 
 ?- 
 
 i 4 i 2 
 
 >2_ 
 
 (6.) In the ellipse, if A t be the origin, the equation is 
 j/2 = _l(2a# tf 2 ) : let S be the pole, A l SP=d, and 
 
 SP=r ; show that the equation referred to polar co-ordi- 
 
 a (1 -e 2 ) 
 nates is r= - 
 
 (7.) The equation to a curve being y= 
 determine the polar equation, and show that an asymptote 
 cuts the axis of abscissae at an angle of 45, and at a dis 
 tance = from the origin of co-ordinates. 
 
 ui 
 
 (8.) In the hyperbola, if $ be the pole, the polar equa- 
 
SPIRALS. 99 
 
 / O "I \ 
 
 tion will be r= - > if the centre be the pole, the 
 
 polar equation will be r= - - 
 
 (9.) Show that the polar equation to the lemniscata of 
 
 g8 
 
 Bernouilli is r 2 =2a 2 cos%0, and that p = 7^r 
 
 2 a* 
 
 (10.) Show that the polar equation to the conchoid of 
 
 Nicomedes is r=#H -- the equation between rectangular 
 cosfl 
 
 co-ordinates being x 2 y 2 =(a + x) 2 (b 2 x 2 ). 
 
 (11.) Show that the equation r = - represents two 
 
 zfcir 
 
 polar curves, one having an exterior and the other an interior 
 asymptotic circle, and exhibit the general form of the two 
 spirals. 
 
 (12.) The polar equation to the cissoid of Diocles is 
 r=2a tan 6 sin 6 . Prove this. 
 
 a 2 
 
 (13.) The equation to the lituus is r 2 = ; show that the 
 
 
 
 subtangent =2av / 0. 
 
 (14.) In the cardioid r=a (1 cos9), and if r, be a radius 
 in the direction of r produced backwards, r,=a(l-Hx>B0) : 
 show that 20 = 0. 
 
 (15.) If the polar equation to a hyperbola, referred to its 
 
 focus, be r=^ -- - j show that there are two asymptotes 
 
 1+0COS0 
 
 intersecting the axis of x at a distance ae from the origin, at 
 
 angles whose tangents areH and -- respectively. 
 
 (i ci 
 
 (16.) If 0= -- be the equation to a spiral; show 
 
100 SINGULAR POINTS, 
 
 that a circle whose radius is 2 a is an asymptote to the 
 
 spiral. 
 
 a n 
 (17.) If 0= and na n =-l n ; show that the equation 
 
 between the radius vector and perpendicular on tangent is 
 
 l n r 
 P= 
 
 CHAPTER XIV. 
 
 SINGULAR POINTS. TRACING OF CURVES. 
 
 A curve is convex or concave to the axis according as 
 
 d 2 y 
 y and have the same or opposite signs. 
 
 To determine whether there be a point of contrary flexure, 
 
 d 2 y 
 
 we put - = or oo ; and if a be one of the values of x so 
 dx 
 
 found, we substitute successively a + h and a h for x in 
 ^ > then if - have opposite signs, there will be a point of 
 contrary flexure denoted by x=a. 
 
 At a point of contrary flexure in polar curves -~z=0. 
 
 If any values of x and y make = > this circumstance 
 
 generally indicates a multiple point. 
 
 / d 2 u \ 2 /d 2 u\ /d 2 u\ 
 
 For a true double point - TT, -7-7, I > . 
 
 \dxdyi \dx 2 ) \dy 1 ) 
 
 i d 2 u \ 2 /d 2 u\ f d 2 u\ 
 
 For a point of osculation I - 7- ) I -7-,, -7-7; }=(). 
 
 \dxdyi \dx 1 l \di/ 2 / 
 
 i d 2 u \ 2 /d 2 u\ ,d 2 u\ 
 
 For a coniugate point - j- -;, -7-7, < 0. 
 
 J ^ \dxdyl \dx 2 ) \dy 2 ) 
 
TRACING OF CURVES. 101 
 
 At a cusp, if x=a, ~ has but one value; and, substituting 
 ctoc 
 
 d*y 
 
 successively a-\-h and a h for x, - has two values. 
 
 ctx 
 
 For the ceratoid or cusp of the first species, the values of 
 
 <* 2 yi 
 
 have opposite signs. 
 clx 
 
 For the ramphoid or cusp of the second species, the values 
 
 d 2 y 
 
 of - have the same sign. 
 dor 
 
 Ex. (1.) If the equation to a curve be y= v zP + cx 4 ; 
 
 ci 
 
 show that the origin is a point of osculation, ascertain if 
 there be any maximum ordinate, and determine the general 
 form of the curve. 
 
 It is obvious that, by giving x successive positive values 
 from to oo , y will have successive positive and negative 
 values from to oo , consequently there are two similar 
 branches extending from the origin to infinity, one branch 
 on each side of the axis of x to the right of the axis of y. 
 
 dy 1 5.r 4 + 4ttr s x 5x + c 
 Now = - - -- = -- =0 when #=0, 
 
 dx a SvxP+cx 4 2a Vx + c 
 
 and Y when #=(), y also =0, and ~- has 
 
 dx 
 
 two values, one positive and the other ne 
 
 gative, each =0, therefore the axis of x is 
 
 a common tangent to the two infinite branches at the origin j 
 
 hence the origin is a point of osculation. 
 
 Again Y y= vx + c; when# = c, y = 0, and while 
 
 x takes successive negative values from to c, y will take 
 successive positive and negative values from to again, 
 
 K 2 
 
102 SINGULAR POINTS, 
 
 and therefore to the left of the axis of y there is a loop or 
 nodus. 
 
 dy x 5# + 4c 4 
 
 And v = --- . = 0, ..r-f4<?==0, and#= -- c 
 
 dx 2a V x + c 5 
 
 determines the position of the maximum double ordinate ; 
 
 and v -p= tan 0=oo when#= c, the tangent at this point 
 dx 
 
 intersects the axis of x at right-angles. 
 
 Take AE c, and draw the tangent TBtLAB, take 
 
 AN c, and draw the double ordinate PNp = (- 
 
 4 
 
 which is the value of 2y corresponding to #= -c; the 
 
 o 
 
 loop will pass through A, P, J3 } p. 
 
 V^J 
 (2.) Trace the curve, whose equation is y =/=( #) ; 
 
 va 
 
 and show that there is an oval between #=0 and x=a ; de 
 termine the position of the maximum double ordinate, and 
 exhibit the form of the exterior branch. 
 
 v^ 
 Firstly, y=-(a-x) = 
 
 V a 
 
 Let a?=0, /. y=0, 1 Take yl.g^a. "A 
 
 #<#, yis:, Then, v while 
 
 x=a, y=0, > ^ increases 
 
 x > a, y is impossible. from to a, y has positive 
 
 Putting # for#, y is impossible. J and negative values from 
 
 to again, . . there is 
 a maximum ordinate somewhere between A and J5, and AE 
 
 is the axis of an oval. 
 
 3 ^ 
 
 //y /- 1 2* S^ 3 /- 
 
 Now = Va -- 7= -- 7== - -- 7= v#=0, 
 dx 
 
x<a, ys, 
 x=a, y=2a, 
 y is db, 
 
 TRACING OF CURVES. 103 
 
 . . 7== T=-> 3x=a, .". #= denotes the point where 
 v# v# 3 
 
 the maximum double ordinate cuts the axis of x. 
 
 Secondly, y= (a + x). 
 
 va 
 
 = 0, .*. y=0, "] Draw EP^a. Then, 
 
 Y while # increases from 
 to infinity, y has posi- 
 tive and negative values 
 from to infinity ; there 
 Putting x for #, y is impossible. J is a branch above and 
 
 below the axis of x exte 
 rior to the oval. 
 
 "No curve exists to the left of the origin. 
 (3.) y 2 (a 2 + x 2 )=x 2 (a 2 x 2 ) is the equation to a curve; 
 trace it, determine the angles at which it cuts the axis of x, 
 and find its maximum ordinate. 
 
 y=x - v feS- 
 
 If x=0 } theny=0 Put x for x, then 
 
 x<a, y is possible if x=Q, y=0 
 
 x=a, y=0 x<a, y is possible ip 
 
 #>#, y is impossible. x=-a y y=0 
 
 xXi-j y is impossible. 
 Take AJB=a, Ab=a, in the axis of a?, and the curve 
 
 will pass through the points A, B, b. 
 
 And Y when x > a, y is impossible, the curve cannot 
 extend beyond , I. 
 
104 
 
 SINGULAR POINTS, 
 
 dx 
 
 and putting # = and db in this expression, we have 
 
 dx (a 2 )* (a 2 )* a 3 -a 
 
 -=oo =tan 90. 
 
 = = 1 = tan 45 or tan 135. 
 
 .". the two tangents at the point A are inclined to the 
 axis of x at / s=45 and 135 respectively, and the tangents 
 at B and b are J_ to the axis of x : . . the point A is a 
 double point. 
 
 VcP3? 
 ~2 - 5* 
 
 a max. 
 
 Hence the greatest ordinate cuts the axis of x at points 
 
 denoted by # = #v v/2 1 and a\/ V^2 1, and the 
 length of this ordinate may be ascertained by substituting 
 these values of x in the equation to the curve. Thus 
 
 ; - v / 2 -a 
 1. A/ 
 V 
 
TRACING OF CURVES. 105 
 
 (4.) If y= .f X , show that there are points of contrary 
 
 flexure when #=0 and \/3, that the curve cuts the axis 
 of x at an angle of 45, that the axis of x is an asymptote to 
 the two infinite branches, and that there are maximum or- 
 dinates when x=+a and a. 
 
 a?x 
 Let #=0, . . y=0 Put x for x, then y= 5- 
 
 x<a, y is + Let #=0, . . y=0 
 
 a 
 
 x=a, y~^ x<a, y is 
 
 a 
 x>a, y is + x=a, y=- 
 
 ^=00, y=0. x>a, y is 
 
 Take AB=a, Ab=a, and draw 
 the ordinates .Z?$, 5y, equal to -and 
 
 respectively, the curve will pass 
 through the points A, Q, <?,, its right- 
 hand branch being above the axis of #, and its left-hand 
 branch below it, the two branches meeting that axis again 
 only at an infinite distance from the origin A . . . the axis 
 of x is an asymptote to the two infinite branches. 
 
 _ 
 
 " W ~ 
 
 dx~ (a2 + x?) 2 " (a 
 
 2 a*x (x 2 -3 a 2 ) /on 
 
 = / o ; 2 \s => tf*=as/3 or 0. 
 (a 2 -|- or 2 ) 3 
 
 Substituting aSSJi, a \/~3 + h respectively for x, we have 
 
106 SINGULAR POINTS, 
 
 which is negative, since k< a \/3 ; 
 
 which is positive. 
 
 Hence #=& v3 indicates a point of contrary flexure ; and, 
 substituting this value of x in the given equation, we have 
 
 V=^- Take AN=a*/3, and draw NP=^ -, when 
 4 4 
 
 P will be a point of contrary flexure; 
 
 Also substituting k, 0-f A respectively for #, 
 
 d 2 y _2 a 2 h (k 2 - 3 a 2 ) 
 
 one positive, the other negative. :. the origin .4 is also a 
 point of contrary flexure. 
 
 d*y 
 
 Hence also, y being positive and ^ to the left of NP 
 
 Ct(JC 
 
 negative, the curve from A to P is concave to the axis of #, 
 and consequently beyond P it is convex. 
 
 Again / as x increases y at first increases and afterwards 
 decreases, having various finite values between its primary 
 value and its ultimate value 0, there will be a maximum 
 ordinate somewhere on each side of the origin. 
 . dy a-* 2 
 
 But when #=0, y=-^- Draw BQ=~, it will be a 
 
 2i Ji 
 
 maximum ordinate. 
 
TRACING OF CURVES. 107 
 
 By substituting for x in ~- we have 
 
 4 
 
 = =1= tan 45. .*. the curve cuts the axis of x at 
 a 4 
 
 the origin A at an Z of 45. 
 
 / d 2 I ffiZ 
 (5.) \iy=x* A / 2 - r, ; show that the branches of the 
 
 curve pass through the origin, and are contained between 
 two asymptotes perpendicular to the axis of x. 
 Let #=0, .". ?/=0 Put x for x, then 
 
 y is possible if #=0, y=0 
 
 y^ 00 #<#, y is if 
 
 y is impossible. x=a, y= GO 
 
 a? >a, y is impossible. 
 
 Take ^j3=a, ^6= a; then, since at the 
 origin J. the ordinate is 0, and then as x in 
 creases the ordinates increase until x=.a, when 
 an infinite ordinate passes through and, 
 since the values of y are both positive and ne 
 gative, a branch extends on each side of the axis of x. 
 
 Also, since when x is negative, the ordinates take values 
 exactly corresponding to those when x is positive, the curve 
 has similar branches to the left of the origin. 
 
 , , dy 
 
 Again SB -- - - j- : and, putting x(j and 
 dx (a 2 x 2 )? (a 2 -f- x 2 ) 1 * 
 
 a in this expression, we have tan0=-^-=;l and GO. 
 
 . . tan = 1 = tan 45, tan = 1 = tan 135, tan = oo 
 =taii90. 
 
 Hence a tangent to the curve cuts the axis of x in the 
 origin A at an angle of 45, another through the same point 
 
108 SINGULAR POINTS, 
 
 at an angle of 135 : and at B a tangent to the curve is J_ 
 the axis of #, and is coincident with the infinite ordinate. 
 This tangent is consequently an asymptote, the branches of 
 the curve do not extend beyond it, and they are convex to 
 the axis of x. 
 
 (6.) If (y Vf=(xaf] show that there is a ceratoid 
 cusp when x=a, and that the tangent at that point is pa 
 rallel to the axis of x. 
 
 If x=a, y = l. Take AB=a, P=l, then P is the point. 
 
 Now y i= (# )$, /. -/-=-(x a)? = 
 
 dx A 
 
 when xa ; . . tan 0=0, and the tangent to the curve at 
 the point denoted by x=a is || to the axis of x. 
 
 Again - = +: (a a)^=Q when x=a ; 
 and, putting a + h, ak successively for x, 
 
 =:t-v which has two values, one +, another . 
 ax* 4 
 
 d*y 15 / 7 . . . 
 
 -- = +--V h, which is imaginary : 
 
 dx 2 4 
 
 dn d u 
 and since if x= a, -j-=0, ~T~2 = > an( * ^ x^=a h, they are 
 
 both impossible . . the curve cannot extend y 
 
 d 2 )/ 
 to the left of P : also v if x=a + h, - has 
 
 two values, one positive and the other negative ; /, at the 
 point P there is a cusp of the first species. 
 
 aS 2 . 
 (7.) Show that the curve, whose equation is r=- 2 _ , > has 
 
 a point of inflection when r= > and rectilinear and circular 
 asymptotes. 
 
TRACING OF CURVES. 109 
 
 r0 2 -r=ad 2 , (r-a)6 2 =r, :.6= A /_!L- . 
 
 V r a 
 
 r a r 
 
 de_ (r-o) 2 a_ 
 
 ^" 2 A /~ " ~2r*(7-a)f 
 
 \ / 
 
 V ~a 
 
 x / - ^ - r 
 - r 7 v4 (r 6) 3 + a 2 r 
 
 Hence there is a point of contrary flexure, when r=^- a. 
 
 2 
 
 1 O 2 1 
 Again -= - j- -^ e ^ r become infinitely great, then 
 
 != =0, .-. e 2 -i=o, e=i. 
 
 r GO 
 
 and, when r becomes infinitely great, 
 rill 
 
 r a a a 10 
 
 /. Subtangent ST=r 2 =+ : 
 
 UT L. 
 
 L 
 
110 SINGULAK POINTS, 
 
 and, since ST remains finite while SP is infinite, a tangent 
 may be drawn which will touch the curve at a point infi 
 nitely distant from the origin ; this tangent is therefore a 
 rectilinear asymptote : and Y and ST have each two 
 values, /. there are two rectilinear asymptotes. 
 
 1 2 -l 6) 2 -l 1 
 
 Again, let r=a, :. -= r , l^-^!--, 
 
 /. 0=oo when r=a. 
 
 Also 6= A / which is impossible when r<a. 
 
 V r a 
 
 Hence / r=a makes 6 infinite, and r<a makes 6 im 
 possible, there is an asymptotic Q> radius =a, within the 
 curve. 
 
 In the logarithmic and many other spirals the curve makes 
 an infinite number of revolutions about the pole before 
 reaching it ; hence the pole may, in such instances, be con 
 sidered as an indefinitely small asymptotic circle, that is, an 
 asymptotic circle whose radius =0. 
 
 The equation to the logarithmic spiral is r=a e , or r=ae m6 , 
 
 e 
 or r=ce u ; r increasing in a geometric ratio, while 6 increases 
 
 in an arithmetic ratio ; the radii including equal angles are 
 proportional. Its evolute and involute are similar to the 
 original spiral. 
 
 (8.) Trace the curve whose equation is r=a (2 cos + 1). 
 Let 0=0, .-. r-a (2 + 1)=3, 
 
 0=30, r=a 
 
 0=60, r=a 
 
 0=90, r=a(0 + l)=a, 
 
TRACING OF CURVES. Ill 
 
 Let 0=120, ..cos0=-cos60=--, r=a( 1 + 1)=0, 
 
 J 
 
 /o 
 
 0=150, cos0=-cos30=- > r=( 
 
 which is < a, 
 
 = 180, cos0=-l, 
 
 30= -- jp> r=a 
 
 which is < a, 
 
 0=210, cos0=-cos30= -- j> r=a( v/3-fl), 
 
 0=240, cos0=-eos60=--> r=0, 
 
 2 
 
 0=270, cos0=0, r 
 
 0=300, cos0=cos60, *=(! + l)=2a, 
 
 0=330, cos0=cos30, r=a(\/3 + l), 
 
 which is 
 0=360, cos0=l, r=a(2 + I)=3a. 
 
 Divide the Q ce of a O into 12 equal parts, and draw 
 radii through the points of division. Take AB =3 a, AP, 
 Ap each =a(S3 + 1), ^t<7, ^1A" each =2a, ^Z>, ^^ each 
 
 Take^.E", ^16 ?/ each=6c( \3 + l), and^^^-a. These 
 three, being negative values of r, must be measured in an 
 exactly opposite direction, as AE, AF, AG. 
 
 The curve, which is the trisectrix, will pass through the 
 points B, P, C, D,A,H,K,p } and the interior oval will pass 
 through A, E, F, G. 
 
 Taking r=a(2cos0 1), a precisely similar curve is pro 
 duced, but turned the contrary way. 
 
 Taking for 0, the same curve is produced, 
 v 2cos(-0)=2cos0. 
 
112 SINGULAR POINTS, 
 
 (9.) Show that the curve, whose equation is (j/ 2 4-^r 2 ) 3 
 = 4a 2 .r 2 y 2 , has a quadruple point at the origin, and that 
 there are four loops or ovals j namely, one in each 
 quadrant. 
 
 Let the equation be transformed into one under polar co 
 ordinates, putting #=r cos 0, y=rsin0. 
 
 (r 2 sin 2 + r 2 cos 2 0) 3 = 4 a Vsin 2 r 2 cos 2 0, r 6 = 4 2 r 4 sin 2 cos 2 0, 
 r 2 =4a 2 sin 2 0cos 2 0, r=2asin0cos0, :. 
 
 1st quad. If 0=0, r=0, By put- 
 
 2 for the 
 
 v3 curve is 
 
 0=30, r=asm6Q=-^-a, . _ 
 
 2 reproduced. 
 
 0=45, r = a sin 9 = a, Take the several 
 
 -v/3 values of r at the 
 0=60, r=asin!20= a, ,. 
 
 2 corresponding 
 
 0=75, r=asin!50=-> 
 
 2 In the second 
 
 = 90, r = a sin 1 8 = 0, and fourth quad- 
 
 . rants, the values 
 
 2nd quad. = 105, r=asm210= -> 
 
 2 of r, being nega- 
 
 a tive, must be 
 3rd quad. 0=195, r=sm390=^ 
 
 measured in op- 
 
 **A a posite directions. 
 4th quad. 0=285, r=sm570= - * 
 
 * Hence, there 
 
 will be an oval whose axis =a in each quadrant : and the 
 origin is a quadruple point. 
 
 (10.) Ifr=atan0, show that the asymptotic subtangent 
 is a, and that the curve is included between vertical asymp 
 totes. 
 
TRACING OF CURVES. 113 
 
 Let 0=0, :.rQ, Let 0=- 
 
 3;r 
 
 0=45, r=a, 0=-^-, r=co 
 
 r=-a, 
 
 0=135, r=-a, 6=2*, r=0. 
 
 6=*, r=0. 
 
 Take therefore SJ3=a at an angle of 45 with the axis 
 of x, the curve will pass from the origin $ through B to 
 infinity. 
 
 And Y those lines are said to be || which coincide only 
 at an infinite distance, and Y the asymptote will ultimately 
 coincide with the curve and consequently with SP when 
 both are infinite, .*. the asymptote must be drawn || SP. 
 
 There are similar branches in all the four quadrants. 
 
 dr dd 1 
 
 Now -=#(l-ftan 2 0), = > 
 
 d dT 
 
 n dd a 2 tan 2 tan 2 _ ^__ 
 
 oo J 
 
 /. ST=r 2 =a, the asymptotic subtangent. 
 
 Take OT=, and draw ^T, \\SPj TP, produced is the 
 asymptote. Hence, this curve is included between vertical 
 asymptotes. 
 
 (11.) #=a(l cos0), y=aO are equations to the curve 
 called the companion to the cycloid ; find the points of 
 contrary flexure. 
 
 Let BDQ be the generating circle, centre 0, vertex D, 
 radius -a, DM=x, MP=y, L DOQ=Q. 
 
 L 2 
 
114 SINGULAR POINTS, 
 
 Let 0=0, /.#=0, y=0, A 
 
 -> a - 
 
 7T 
 
 0=- + a, . . cos0= sma. ^_ 
 
 #=a(l + sina),"! . 
 
 I which increase 
 
 y = a (- -f a V f as a increases. 
 Let a=-j 0=?r, .*.cos0= 1, cos0=l, 
 
 Putting for 0, a similar curve is produced on the 
 other side of the axis of x. 
 
 Now -^=- 
 
 Jd7 
 
 = 0. if x=a. 
 
 Substituting a + h, ah respectively for x in this expres 
 sion, we have 
 
 . 
 which is positive, 
 
 -a{a-(g-h)} 
 
 - . 
 
 = -- j which is negative : 
 
 .*. there is a point of contrary flexure when x=-a, y=a. 
 
 DO a. Take OJK = ~ a, Or=-?-a, each = arc Dn, 
 
 JBA=7ra=&rc DQB ; the curve will pass through D, JK, A, 
 and JK, r will be the points of contrary flexure. 
 
TRACING OF CURVES. 
 
 115 
 
 (12.) Show that the curve y 4 + 2 axy 2 aa? = has a triple 
 point at the origin, and determine the position of the 
 
 tangents. 
 
 dy 
 x 2yp-\-f)-^a^=Q, where ^=; 
 
 (4 yS _|_ 4 a #y)j9 = 3 ax 2 2 ay 2 , 
 
 Sax 2 2 ay 2 
 
 =__- =^ ^ ^=0 and y= 
 
 
 
 .*. there may be a multiple point. 
 Differentiating numerator and denominator, 
 
 
 
 .. 
 
 7: if ^r= 
 
 
 
 Differentiating as before, 
 Qa 
 
 . _ 
 
 ~ 
 
 if ^0 and y=0. 
 
 Sap 4p 
 
 ..-----, , 
 
 .*. the origin is a triple point; and / tanfi^r- 1 -^:-! -- 
 
 . dx V2 
 
 and = -- j=. and also =00, .". the tangents cut the axis 
 
 v2 
 
 at / s=tan -1 ( -~\ and tan" 1 ( -- =L\, and at right-angles. 
 
 * These repeated differentiations are sometimes tedious : they may, 
 however, in such cases as this, be simplified by considering p constant, 
 as no error will arise from that assumption. Thus, instead of this 
 equation, we should have had, by considering p in the previous one 
 
 ,, 
 
 constant, p=^-, ; -. - ; : > whence = -^ as above. 
 >r * 
 
116 SINGULAR POINTS, 
 
 (13.) In the diameter AB of a circle take a point C t - draw 
 a chord AP and an ordinate PN, and CQ parallel to AP, 
 meeting PN in Q : trace the curve which is the locus of Q. 
 AB=a, AC=b, 
 
 NP= vax x 2 , equation to Q, 
 
 CN : AN :: NQ : NP, or 
 x b : x \\ y \ 
 
 the equation to the curve which is the locus of Q, 
 
 Let y:=rO, .". x=l> and =a , let#>a, y is impossible. 
 y has finite values positive and negative when x > b and < a. 
 
 Hence the curve will pass through C, Q, B, and form an 
 oval. 
 
 By the question no part of the curve can be to the left of C. 
 
 (14.) A rod PQ passes through a fixed point A ; find the 
 equation to the curve described by P when Q moves in the 
 circumference of a circle of given radius, and trace the curve. 
 
 PQ=E= length of rod, diameter 
 of Q BQ=.a, AB=b, qp position of 
 rod when Q has moved along the 
 
 Euc. iii. 35. 
 
 Let A q= r, 
 
 in 2 
 
 r 2 (b + c) cosO-r=bc, 
 
 And Y Ap = qp Aq=J% r, by giving successive values 
 to 0, and taking the corresponding values of r, the curve, 
 
TRACING OF CURVES. 
 
 117 
 
 which is the locus of P, will be traced. If ED be the posi 
 tion of the rod when Q has described a ^O> PJ)=Q. 
 Hence the curve is an oval, whose axis PD=a. 
 
 (15.) The equation to the spiral of Archimedes is r=ad j 
 trace the curve, and show that the origin is a point of con 
 trary flexure. 
 Let 0=0, /. r=0, 
 
 3-1416_ /.^ 8 ^v 
 
 ( ^ " } 
 =a (1-5708), \ 
 
 r=a- 
 
 4 
 3-1416 
 
 r=a(3-UlQ), 
 
 Take the angles, and draw the corresponding lines for the 
 values of r, and the curve may be traced. 
 
 Put for 0, and the values of r, being negative, must 
 be measured in a directly contrary direction. 
 
 ^-1 
 dr a 
 
118 SINGULAR POINTS, 
 
 and ~ changes sign immediately before and after the 
 
 origin. 
 
 .". the origin is a point of contrary flexure. 
 
 In the figure, if r commences its revolution above the 
 axis of x in the first quadrant, the branch of the spiral 
 ABCDEF will be generated. If negative values be given 
 to 6, and r be measured in a directly opposite direction, the 
 branch represented by the dotted line will be traced ; and 
 we shall have the double spiral. If r commences its revolu 
 tion upwards in the second quadrant, two branches will be 
 generated, similar to the others, but turned in a contrary 
 direction, and intersecting them in the horizontal and ver 
 tical axes. 
 
 This spiral was invented by Conon : but Archimedes dis 
 covered its principal properties. 
 
 If a fly were to move uniformly from the nave of a wheel 
 along one of the spokes whilst the wheel revolved uniformly 
 about a fixed axis, the fly would describe this spiral. 
 
 Teeth of this form are applied in the construction of 
 engines in which uniform motion in a given direction is 
 required. 
 
 (16.) Two points start from the opposite extremities of 
 the diameter of a circle, and move with uniform velocity in 
 the same direction round the circumference, their velocities 
 are in the ratio of 2 : 1. Determine the locus of the bisec 
 tion of the chords which join the positions of the two points, 
 and find the polar subtangent of the curve. 
 
 Let the diameter AB = 2 #, and A be the 
 position of the point which moves with a 
 velocity equal to double that of the point 
 at B. Now when this latter point has made 
 
TRACING OF CURVES. 
 
 119 
 
 half a revolution, the former will have made a complete 
 revolution, and consequently the two points will coincide 
 at A. Again, the motions continuing, if we take any arc 
 A C, and bisect it in D, C will be a position of the point 
 which started from A, and D the corresponding position 
 of the point which started from B. Draw the chord CD, 
 bisect it in P, and join OP, 00, OD. 
 
 Let be the pole, OP the radius vector =r, ^AOP=6, 
 
 then POD=-> 777; cos POD, or = cos-0, the equation 
 o O-tJ a o 
 
 to the locus of P. 
 
 To find the polar subtangent, 
 
 1 r 
 cos-0= ) 
 o a 
 
 . 1 dd 1 
 sm-0 =-> 
 6 dr a 
 
 dr . 1 
 
 asm-0 
 o 
 
 a A/l-cos 2 - 
 
 -0 
 
 flf - ; 
 
 To trace the curve, 
 Put = 0, then cos = 1, 
 
 = 45, 
 
 0=90, 
 
 0=135, 
 
 0=180, 
 
 ==the polar subtangent. 
 
 00830= 
 
 
 
 r=aco$ 6. 
 o 
 
 r=a, 
 
 ^r-j r=a 
 
 AX l a 
 
 COS 40 = -7= J r = 7- 
 
 ^2 J\ 
 
 
120 SINGULAR POINTS, 
 
 */3_1 
 
 Let 0=225, then 
 
 2 \/2 2 A/2 
 
 0=270, cos90 = 0, r=0, 
 
 \/3-l \/3- 
 
 
 
 
 2 A/2 
 
 0=360, 
 
 cos!20= 
 
 1 
 
 0=405, 
 
 cos!35 = 
 
 7= 
 
 0=450, 
 
 cos 150 = 
 
 A/3 
 
 ~~2" 
 
 ._ 
 
 2v/2 
 
 00,165=- ^ + 1 
 
 2v/2 2v/2 
 
 0=540, cos!80 = 1, r= a. 
 
 The negative values of r, which are 
 measured in an opposite direction, are 
 distinguished in the figure by dotted 
 lines. 
 
 By giving negative values to the 
 same curve would be produced, but 
 turned in a contrary direction. 
 
 (17.) If a 2 y=3lx 2 of 1 ; show that there is a point of 
 
 2b 3 
 contrary flexure when x=. b, and j/= - 
 
 (18.) If i/=2a A/ 5 show that there are two 
 
 points of inflexion when # = > y=. + 
 
 (19.) If ax% (x a)y%= be the equation to a curve; 
 show that there is a point of contrary flexure when #= 2a. 
 
TRACING OF CURVES. -121 
 
 (20.) If y=ax + bx 2 ca? ; show that there is a point of 
 
 inflexion when #= > and y=7^ n (9< 
 oc 27c 2 
 
 (21.) If y = c + (x a) 2 (x I)?; show that there is a 
 double point when x=a, and y=-c. 
 
 x 2 
 (22.) If y=-g (a 2 x 2 } ; show that there are points of 
 
 . - . , a 5a 
 
 inflexion when x = db 7= > y = 
 
 be three equations having no mutual relation, and x becomes 
 
 infinitely great in each ; prove that in (1) y = oo , and -y-= oo , 
 
 dx 
 
 in (2) y=oo , and =1, and in (3) y=0, and =0. 
 
 (24.) If y 2 (^ a 2 )=^ 4 ; show that the equations to the 
 asymptotes are y=+x, y=x } and that the curve lies 
 above the asymptote : also show that the curve has two 
 branches touching the axis of x at the origin, both being in 
 a plane perpendicular to the plane of the paper, between 
 two asymptotes which cut the axis of x at right- angles 
 when x=+a, x=a; show that beyond these asymptotes 
 the curve is in the plane of reference, and approaches nearest 
 to the axis of x when ^=^^2, again receding towards the 
 asymptotes whose equations are y=db# } and intersecting 
 them at oo in a point of inflexion. 
 
 (25.) If y^ + tf 3 2a.r 2 =0 ; show that the equation to the 
 
 asymptote is y=- &-{-) that at the origin there is a cusp 
 o 
 
 of the first species, the two branches being above the axis 
 of x and concave to it, that the curve cuts the axis of x at 
 
 M 
 
122 SINGULAR POINTS, 
 
 right-angles at a point denoted by x=2a, where there is a 
 point of inflexion, beyond which it approaches the asymp- 
 
 2a 
 
 tote whose equation is y= x + ^ ; show also that there is a 
 
 o 
 
 maximum ordinate whose length is - v^J, when #= 
 
 o o 
 
 (26.) If r = =. ) show that p = and that 
 
 VO v / ! 
 
 there is a point of inflection when r=-av 2, the curve being 
 concave towards the pole when r is less than a v/2, and con 
 vex towards it when r is greater than a \/2. 
 
 (27.) y=a + a*(xa)%; determine the nature and posi 
 tion of the cusp. 
 
 yA 
 
 (28.) 3/ 2 =-o - 3 being the equation to a curve referred 
 o/ x 
 
 to rectangular co-ordinates ; show that the equation between 
 polar co-ordinates is r=:&tan0, and that the equation be 
 tween the radius vector and the perpendicular from the pole 
 
 ar* 
 
 upon the tangent is = / . n ^^^ > show also how the 
 
 1 4 
 
 branches of the curve are situated with regard to the plane 
 of reference. 
 
 (29.) If 0= - ) show that a line drawn parallel to the 
 r a 
 
 prime radius or axis, at the distance a above it, is an 
 asymptote to the curve, that, when is +, the curve has 
 an interior asymptotic circle, and when 6 is , it has an 
 exterior asymptotic circle. Trace the curve, and show that 
 the rectilinear asymptote is a tangent to the asymptotic 
 circle. 
 
 (30.) The equation to the Cardioid is r=a (1 -fcosO) ; 
 trace the curve. 
 
TRACING OF CURVES. 123 
 
 (31.) If r=a > trace the curve, and show that 
 
 6 smfc) 
 
 there is an asymptotic circle, radius =a, and that the curve, 
 coming from infinity, continually approaches the convex 
 circumference of the asymptotic circle on one side of the 
 diameter, and the concave circumference on the other side 
 of the diameter. 
 
 ?/ / X" -\- n 
 
 (32.) The equation to a curve being = A/ -, 
 
 3 x V x 2 a 2 
 
 show that it has asymptotes, at right-angles to the axis of x, 
 at points denoted by x= + a, x= a, and other asymptotes 
 cutting the axis of x at 45, and 135, respectively; that 
 
 there are minimum ordinates when x= a\f v/2 + 1. De 
 termine the value of these ordinates, and show the position 
 and direction of the branches of this curve. 
 
 (33.) y=a: (ax a 2 )? ; determine the nature and posi 
 tion of the singular point. 
 
 (34.) aPy 2 + a 2 y 2 a 4 =0 is the equation to a curve j show 
 that its asymptote coincides with the axis of x, and that 
 there are points of inflexion above that axis at distances 
 
 /2 /2 
 
 equal to + a \J > and a A/ from it, and at dis 
 tances equal -\ -/= and -=. from the origin of co-ordi- 
 
 v 2 v 2 
 
 nates. 
 
 (35.) If x 3 ?/ 3 =a 3 ; show that the curve cuts the axis 
 of x at right-angles, at the distance a from the origin, that 
 at each of these points there is an inflexion, the part of the 
 curve between them being concave to the axis, the part to 
 the left of the origin being convex, and the part to the right 
 of the point denoted by x=a, concave. 
 
124 SINGULAR POINTS, 
 
 (36.) If (a? a) 5 = (y #) 2 ; show that the common tangent 
 to the two branches of the curve is inclined to the axis of x 
 at an angle of 45, that the curve cannot extend to the left 
 of the point denoted by x-=a, and that, at the distance a 
 above that point, there is a cusp of the first species. 
 
 3 / C 4_ aa a 
 (37.) If y= A / r- be the equation to a curve ; 
 
 c$ 
 
 show that there is a point of inflexion at the distance - 
 
 b* 
 
 above the origin, and another in the axis of x, at the dis 
 tance - from the origin. 
 a* 
 
 (38.) 3/=csin- is the equation to the curve of sines; 
 
 show that, at all the intersections of this curve with the 
 axis of x, there are points of contrary flexure. 
 
 (39.) ^ 2 =a 2 + a? \/2a 2 a?* being the equation to a curve ; 
 show that its branches intersect the axis of x at angles 
 
 =tasi- l . 7= and tan-^v/2, that there are four double 
 
 points in the axes of co-ordinates, at the distance a from the 
 origin, and that the branches form two intersecting ovals. 
 
 (40.) If r 2 =a 2 sin20 ; show that there is an oval in each 
 of the first and third quadrants, and that no curve exists in 
 either the second or fourth quadrants. 
 
 (41.) If the equation to a curve be a?+f 2 A/<M?y=0 ; 
 show that the axes are tangents, that /? = and oo , and 
 that the origin is a double point. 
 
 (42.) If tan3e= - and tan0= define a curve; 
 v x ax 
 
TRACING OF CURVES. 125 
 
 show that it has a maximum ordinate at the point denoted 
 rs 
 
 by x=a ( 1 ) , and trace the curve. 
 
 \ 2i I 
 
 (43.) Trace the curve, whose equation is Zay^ + Sa 2 ?/ 2 
 + 2a 2 x 2 =a 4 + x 4 , and determine the different angles at which 
 it cuts the axis of x. 
 
 (44.) Transform the equation (ax)y 2 =x* from rectan 
 gular to polar co-ordinates, and trace the curve. 
 
 (45.) Trace the curve, whose equation is f/ 3 by 2 ax 2 
 =0, and determine whether it has a point of contrary 
 flexure. 
 
 (46.) Prove that, in the logarithmic spiral, the equation 
 to which is r=ae m6 , the tangent constantly makes the same 
 angle with the radius vector. 
 
 *2 O 
 
 (47.) Trace the curve, whose equation is = ? and 
 
 x 1 a x 
 
 ascertain the angles at which it cuts the axis of x. 
 
 (48.) If the hour and minute hands of a watch were of 
 equal length, and an elastic thread, so extensible as not to 
 impede their motions, were attached to the extremity of 
 each index, the thread representing a straight line of va 
 riable length, from to the diameter of the dial-plate ; 
 determine the polar equation to the curve which would be 
 described by the middle point of the thread, and trace that 
 curve. 
 
 (49.) If perpendiculars be drawn to the diameter of a 
 circle, and from each of them a part be taken, measured 
 from the diameter, equal to half the sine of twice the arc 
 which it cuts off, the arc being measured from the same ex 
 tremity of the diameter; show that the equation to the curve 
 
 M 2 
 
126 SINGULAR POINTS, 
 
 passing through the points thus determined is a lemniscata, 
 whose equation is y= */a 2 x 2 , and trace the curve. 
 
 a 
 
 x ( x c \s 
 
 (50.) If y= + l } there is an isolated point, de- 
 a? 
 
 termine its position, and exhibit the form of the curve. 
 
 (51.) Ino^loga? 3,- 2 y-f ^^O, show that the origin is a 
 point d arret ; and in y-\-yer x #=0 a point saillant, the 
 branch corresponding to the negative values of x starting at 
 an angle whose tangent is 225. 
 
 (52.) Transform (x 2 +y 2 ) 3 =a 2 y 4 to an equation between 
 polar co-ordinates, show that the pole is a quadruple point, 
 and exhibit the form of the curve. 
 
 (53.) Show that the curve, the equation to which is 
 ay 2 =(xa) 2 (xb), has a singular point when x=a, a con 
 jugate point if I is greater than a, and a double point if a 
 is greater than b. 
 
 (54.) ACB is a semicircle whose diameter is AB; draw an 
 ordinate NO and a chord A G, then NP being taken in the 
 ordinate, always equal to the difference between the chord 
 and the corresponding abscissa, show that the locus of P is a 
 parabola, and that there is a maximum ordinate when the 
 abscissa and corresponding ordinate are equal. 
 
 (55.) Show that the curve, whose equation is y= 
 
 has three points of inflexion ; and that, when x=- vab, the 
 tangent is parallel to the axis of x. 
 
 (56.) If r=a6 n ; show that there are points of contrary 
 
 n 
 
 flexure when r=0, and r=a ( w 2 n) 2 ; and that this equa 
 tion comprehends those of the spiral of Archimedes, the 
 
TRACING OF CURVES. 127 
 
 lituus, the hyperbolic or reciprocal spiral, and an infinite 
 number of spirals. 
 
 (57.) Show how the trisectrix, the equation to which is 
 r=a (2cosO 1), may be used to trisect an arc or angle; and 
 explain the difference between the generation of this curve 
 and that of the cardioid. 
 
 (58.) Prove that the angle at which the logarithmic or 
 equiangular spiral, whose equation is r=a e , cuts the radius, 
 is constant, and that the radii which include equal angles 
 are proportional. 
 
 (59.) If x = a (d esind), and y=.a(\ ecosQ) define 
 the trochoid ; show that, at a point of contrary flexure, 
 
 (60.) A circle, which continues constantly in the same 
 plane, rolls, like a carriage wheel, along a fixed horizontal 
 line ; the curve described by a point in the circumference is 
 
 the cycloid. Find the equations -^ = ( - - ) 2 , and 
 
 ay \2a-yl 
 
 dy _i^ax\\ 
 dx \ x I 
 
 (61.) Ascertain the loci of the transcendental equations 
 
 (1) / #2_|_ cos , rv /_ ^ 
 
 (2) y = 3? vfl a sec 2 #. 
 
 (62.) Show that, in curves referred to polar co-ordinates, 
 
 ds r 2 
 
 s being the length of the spiral, = Investigate the 
 
 do p 
 
 r n+2 
 
 equation between r and 6 when 2 = - > and between 
 
 n n 
 
 p and r when r= 
 
 (63.) If a, and b, be two conjugate diameters of an ellipse, 
 
128 SINGULAR POINTS. 
 
 y the angle they make with each other, and ^ h ,^ =^ 
 * a u** IT 
 
 the polar equation to the ellipse referred to the centre ; 
 prove that a, 2 -\-b l 2 =a 2 -{-b 2 , and a t b, = ab cosec^. 
 
 (64.) Trace the curve, whose equation is ay 2 =x? bx?, 
 and determine the number and nature of its singular points. 
 
 (65.) Let JBACbe a parabola, A the vertex, and EG the 
 latus rectum ; in BC take M and N equidistant from B and C, 
 draw MD and NE perpendicular to 0, to meet the curve 
 in D and E, draw CD cutting NE in P. Determine the 
 equation to the locus of P, and trace the curve. 
 
 (66.) A straight line DAE, at right-angles to the dia 
 meter ACE of a circle, moves, parallel to DAE, along the 
 diameter, whilst a line which at first lies on the radius CA, 
 revolves with a uniform angular motion about C, intersecting 
 the other moving line in P } show that the equation to the 
 
 T^tXj 
 
 curve traced out by P is y=.(a tf)tan ; that the curve, 
 
 Zci 
 
 which is the quadratrix of Dinostratus, has an infinite 
 number of infinite branches intersecting the axis of #, and 
 that the moving parallel is an asymptote to two infinite 
 branches. Show also that, if this curve could be geometri 
 cally described, the ratio of the diameter of a circle to its 
 circumference would be determined. 
 
 (67.) A globe, whose radius is a b, vibrates in a hollow 
 hemisphere, whose radius is a, in such a manner that a great 
 circle of the globe coincides with a great circle of the hemi 
 sphere ; determine the curve traced out by the highest point 
 on the globe in one revolution, and exhibit the polar equa 
 tion. 
 
129 
 
 CHAPTEE XY. 
 
 CURVATURE OF CURVED LINES. RADIUS OF CURVATURE. 
 EVOLUTES. 
 
 Rectangular Co-ordinates. 
 
 If the equation to the osculating circle, or circle of curva 
 ture, be R 2 =(x a) 2 + (y /3) 2 , and if p be put for -j-> and 
 
 q for -> R being considered positive when the curve is 
 
 concave to the axis of x, and negative when the curve is 
 convex j then 
 
 a and /3, being the co-ordinates of the centre of the radius of 
 curvature, are the co-ordinates of the evolute of the curve. 
 If u=0 be the equation to the curve, 
 \ 2 d 2 u du du 
 ~ 
 
 dxdy d^dy\dx df 
 
 dy 
 
 The middle term of the numerator in this expression 
 vanishes when the value of u is the sum of two parts, one 
 involving x and the other y. 
 
 The distance from a point in the curve to the intersection 
 of two consecutive normals is the radius of curvature at that 
 point. 
 
 The normal to the curve is the tangent to the evolute. 
 
130 RADIUS OF CURVATURE, 
 
 Polar Co-ordinates. 
 
 If R be the radius of curvature as before, r the radius 
 vector, 6 the angle traced out by r, and p the perpendicular 
 upon the tangent, 
 
 dr 
 
 (fjM" \ 
 r 2 + fM 
 
 dr 2 d 2 r 
 
 dr 
 The semi-chord =p - = 
 
 dr* d*r 
 
 - * _ 
 
 dO 2 dti 2 
 
 To find the equation to the evolute to a spiral ; r and p 
 being taken as co-ordinates of the involute, r, and p, as cor 
 responding co-ordinates of the evolute, we must eliminate 
 R, r and p from the four equations 
 
 p=f(r), P ,=(*-rf, R=r^, r?=*+lP-2Sp. 
 
 Ex. (1.) To determine the radius of curvature at any 
 point in the common parabola. 
 
 y 2 =4m^, the equation to the curve, 
 
 dy dy 2m 
 
 2y -=4m, .. p=-= > 
 
 y dx dx y 
 
 _d^/_ 2m dy_ _ 2m 2m_ 4m 2 
 dx 2 y 2 dx y 2 y i/ 3 
 
 4m 2 
 
 <7 f 4m 2 w * 
 
 Since this expression for the radius of curvature diminishes 
 as x diminishes, R is least when #=0, and then R=2m 
 
EVOLUTES. 131 
 
 = half the latus rectum ; hence in the parabola the point of 
 greatest curvature is the vertex. 
 
 (2.) The equation to the rectangular hyperbola, referred 
 to its asymptotes is %?/=m 2 \ find the radius of curvature. 
 dy dy m 2 in 2 
 
 v^+y=v> x ^r -- :.p -- 2 
 
 dx dx x x* 
 
 . _ 
 
 q z* 2m 2 ~ 2m 2 
 
 (3.) If the equation to a circle be x 2 a(x 
 find the radius of curvature. 
 
 2 2 2* 
 
 (4.) Find the radius of curvature to the hyperbola, and 
 determine the equation to its evolute. 
 
132 RADIUS OF CURVATURE, 
 
 I 2 
 
 2/ 2= 2 (a 2 a 2 ), the equation to the curve, 
 a 
 
 dy 
 
 a 2 * 2 -a 4 
 
 _72 
 
 ~ X 
 
 Hence 7?=_ C 1 +P 2 )% = (a 2 x 2 + b 2 z 2 -a*)% (x 2 -a 2 )% 
 
 a 3 ^ 2 a 2 % ab 
 
 _(a^ 2 .T 2 -a^_ {a 2 ( e 2 x 2 -a 2 ) }f _ 
 a 4 b a 4 b 
 
 ( e 2^2 _ ^2\f 
 
 = ^ - ; = radius of curvature. 
 ab 
 
 To find the equation to the evolute, 
 
 _ _ 
 
 . 22 
 _ y (e 2 x 2 a 2 ) 
 
EVOLUTES. 133 
 
 2 * 2 a 2 I 2 } 
 
 ____ 
 
 a? 
 
 =^ -- sr ." a=-^ 
 a 2 a 2 
 
 y* - 2 _ 
 ~~ 
 
 \ / \ / 
 
 the equation to the evolute. 
 (5.) Show that, in the catenary, the radius is equal but 
 opposite to the normal. 
 
 y=-(e a + e a ) } the equation to the curve, 
 
 X _X_ X _X_ 
 
 dy _a ie a e a \_e a e 
 
 dx 2i \ a a 1 a 
 
 2x 2x 2x %x x x 
 
 -+2 + i"^ 
 
 = y 3 / fl 2 \ y 2 
 
 a 3 \ 2/ 
 
 N , 
 
134 RADIUS OF CURVATURE, 
 
 7 9 
 
 l+-^ r =y 
 dke 2 
 
 O 
 
 Hence the radius of curvature is equal but opposite to the 
 normal. 
 
 (6.) Determine the radius of curvature and the evolute 
 of the cycloid. 
 
 , CD=2a. 
 
 y x-\- v 2 ay ?/ 2 
 
 =:versm -- = - ., the equation to the curve. 
 
 a a 
 
 q 
 
 Now CF*= 
 
 :. CF= v/2y, .*. 7?=2C^=radius of curvature. 
 
 To find the equation to the evolute, 
 
 
 Substituting these values of a and ft for the co-ordinates 
 in the equation to the curve, we have 
 
 ft ~ 
 
 -=versm -- - -- ..... (1) 
 a a 
 
EVOLUTES. 135 
 
 Taking CA t =CD, and A { B>, parallel to AB, as the axis of 
 the abscissae, and substituting /3, 2 a for /j, and TT a, for a, 
 TTO, being equal to AC; the origin will be transferred to A , 
 and equation (1) will become 
 
 A / a, + v/2a/3,-A 2 \ 
 
 2 =versin (TT 
 \ 
 
 =versm 
 
 z 
 
 And v A,N^ t =(x,,, and N l P l =i3 l , this equation to the evo- 
 lute is the equation to another cycloid originating at A tt and 
 whose generating circle is equal to that of the given cycloid, 
 but moves in an opposite direction. 
 
 (7.) Show that, in the common parabola, the chord of 
 curvature through the focus is equal to four times the focal 
 distance ; and find the length of the evolute in terms of 
 the focal distance and the distance between the focus and 
 vertex. 
 
 Let the focal distance SP=r, the per 
 pendicular from the focus upon the 
 tangent, SY=p, and J)JS=2SA=2a=c. 
 
 Then, by a property of the parabola, SY 2 =SP-SA, 
 
 ,y_cr f dp _c dr 4/> 
 
 2 dr 2 dp c 
 
 Chord =2#- = = = 4r=4/S7 > . 
 dp c c 2 
 
 Again, y 2 =4aa?, the equation to the curve, 
 
 dy 2a 4a 2 
 
 . . = > /. p z =, 
 
 ax y y 
 
136 KADIUS OF CURVATURE, 
 
 _(Py__ _2a dy__ 2a 2a_ a 2 
 
 dx 2 y 2 dx y* y~ y 3 
 
 q 
 
 a? 
 
 f\ (V T>2. 
 
 Hence, length of evolute s=Rc= -- 2SA 
 
 4* 
 
 The form of the evolute,* which is a semi-cubical parabola, 
 is represented in the figure, by the lines ev, ev t . 
 
 (8.) Find the value of the radius vector in the spiral of 
 Archimedes, when the radius of curvature equals the chord 
 of curvature. 
 
 r=a9, the equation to the curve, 
 
 dr , dr rV r 2 p 2 
 -=a. But-=- E. 
 
 But 
 
 Hence 
 
EVOLUTES. 137 
 
 -u , o 
 Now, chord = 2 P - = 
 
 r 4 
 
 
 And, comparing this value of the chord with the value 
 of the radius of curvature, already determined, it appears 
 that radius = chord if (r 2 + a 2 )* = 2 r (r 2 + a 2 ), or 
 
 2 , 
 
 = a 
 
 (9.) To find the radius of curvature in the semi-cubical 
 parabola. 
 
 2^ 
 3/ 2 = > the equation to the curve, 
 
 Od 
 
 dv 2^ 2 dy a? 
 
 2y~= , :,p-j-, 
 
 dd a dx ay 
 
 I /yj I _ _ _ _ -^ . ._ __ 
 
 A I r - A I 9 O - 9 9 """ O 9 9 
 
 2 y 2 a 2 y 2 3 a 2 y 2 
 
 x 1 
 
 ^axy ax 1 -- 
 
 _ a ~y _ 2 axy lap p _ ay 
 
 == 
 
 3 
 
 ^v^ : 
 
 JNOW Jft= 
 
 q 
 
138 
 
 RADIUS OF CURVATURE, 
 
 (10.) Find the radius of curvature and chord of curvature 
 in the cardioid. 
 
 r=a (I + cos0), the equation to the curve, 
 
 1 - dd p 
 
 - ~ 
 
 dr . . dti 
 
 = asind, = 
 dti dr 
 
 -p , 
 
 .But = 
 
 1 
 
 1 
 
 But v cos0=r a, 
 
 =r 2 2ar + a 2 , 
 
 .dr 
 Jp 
 
 . _ 
 
 Hence 
 
 =T = =.=- Vzar. 
 dp 3V2ar 3 
 
 (11.) If R and R, respectively represent the radii of cur 
 vature of an ellipse at the extremities of two conjugate dia- 
 
 q / / 
 
 22 / 4 /0* 
 
 meters : show that R* -f R?= A / TT+ A / -* 
 
 V b 2 V a 2 
 
 Let -P^, Qq be two diameters, then if the 
 tangent at Q be parallel to Pp, or if the tan 
 gent at P be parallel to Qq, they will be con 
 jugate diameters. 
 
 = 0, W=r,, 
 
 1 r//> 
 
 ~ 
 
 (1) 2 Iog^= 
 2 dp_ 
 
 pdr~d 2 + I 2 i 3 
 
 r 
 
EVOLUTES. 139 
 
 p 
 
 ab 
 
 dr ( 
 
 (aft)* (aft)* 
 
 Hence 
 
 But since, in an ellipse, the sum of the squares of any two 
 
 conjugate diameters is equal to the sum of the squares of the 
 
 major and minor axes, therefore (2a) 2 + (2b) 2 =(2r) 2 
 
 or a 2 -+- ft 2 = r 2 + r 2 , 
 
 The form of the evolute of an ellipse is represented in the 
 figure. 
 
 (12.) Find the equation to the evolute of the logarithmic 
 curve. 
 
 X 
 
 y=ae a , the equation to the curve, 
 
 dy - 1 - y d^y 1 dy 1 y y 
 
 p=-f-=ae u -=e a =-> q=-^=~ -f-=-.?-=*-, 
 
 ax a a . dx* a dx a a a* 
 
 Now f - 3= - 
 
 y l q a 2 y 
 
 16 
 
140 RADIUS OF CURVATURE, 
 
 __ da dn y da, 
 
 But = -^= L, y=<*> ; 
 
 dj3 dx a dfi 
 
 . . a = - - ~~ 7 is the equation to the evolute. 
 ap 4 
 
 (13.) If D be the point of intersection of the directrix and 
 axis of the common parabola, and PN, QM ^- 
 
 be ordinates of corresponding points in the pa- /T\ 
 
 rabola and its evolute; show that DM=3DN. " 
 
 I 
 The evolute of the common parabola is the l 
 
 semicubical parabola. 
 
 The normal to the curve is the tangent to the evolute. 
 the equation to the common parabola, 
 
 4 
 /3 2 = -(a 2 a) 3 , ..... semicubical parabola, 
 
 , y=- -T- (x, x), equation to the normal, 
 
 Let y=0, then x t =x-\- ( 2,a, the part cut off from the axis 
 of x by the normal to the curve. 
 
 Again 21og / 3=log^+31og( a -2a), 2 =3>~JL_, 
 
 .-. a _^3 = ___, the part cut off from the axis of x by 
 the tangent to the evolute. 
 
 Hence x + 2a= - 3# + 6a=a-j-4a, 
 
 o 
 
 ;. 3^=a 2 a. 
 
EVOLUTES. 
 
 x=DN-a, 
 
 141 
 
 But DN=a + x, 
 
 DM=a + a, a=Ma, a-2a=J)M3a. 
 :. DM- 3a= 3DN- 3a, :. DM= 3DN. 
 
 (14.) In an ellipse, e being the eccentricity, determine 
 the radius of curvature in terms of the angle made by the 
 normal with the major axis. 
 
 Normal PG 
 
 =y\/ ] 
 
 dx 2 
 
 Now y=- "/a 2 x 2 , the equation to the ellipse, 
 
 _dy b x 
 
 .". p -j-= -, > 
 
 dx a \f a 2 x 2 
 
 _a 4 - (a 2 -b 2 ) x 2 _ a 2 - e 2 x 
 +P +a 2 a 2 ^~ d 2 a 2 -x 2 ~ a 2 -x 2 
 
 and q= 
 
 ba 
 
 Hence -R= 
 
 Now sin 2 ri>= 
 
 q la 
 
 a 2 x 2 
 
 1 -f p 2 a 2 e 2 x 2 
 
 a 2 (l-e 2 ) 
 
 x __ L. 
 
 - . 
 
 1 e^sin 
 
 n 
 
 I 
 and, substituting this value of a 2 e 2 x 2 in equation (1), 
 
142 RADIUS OF CURVATURE, 
 
 7?= = 
 
 (1-<? 2 )*.(1 
 
 (15.) An inextensible cord AB is attached to a stone 
 at B, and a person holding the other extremity of the cord, 
 moves with it at right-angles to AB uniformly along the 
 straight line AC ; it is required to determine the equation 
 to the curve described by the stone, and to find its evolute. 
 
 Let the person be supposed to move in the direction AC 
 until he arrives at any point T, while the 
 stone moves along the curve BP ; the cord 
 will then be in the position PT, and since up 
 to this moment the stone has never been so 
 near to the line AC as it now is, the line PT produced 
 would not cut the curve JSP hence PT, or the cord in any 
 position, is a tangent to the curve. 
 
 Let AN=x, NPy, AB=a ; then 
 
 =y> and NT*=PT*-NP*, 
 
 ?/ 2 I } = a 2 y 2 , . . y = db v/a 2 y 2 , the equation 
 
 required. 
 
 Hence the curve is the tractory, and AC is its directrix. 
 
 The equation may be readily reduced to y 
 
 a form in which it is frequently given. 
 
 /dx\ 2 1 a 2 
 
 To determine the evolute ; (-)==: 1 
 
 \dyl p 2 y 1 
 
EVOLUTES. 143 
 
 dy y 
 
 Agajn = = 
 
 
 OLX v a 2 y 2 , 
 
 dfi d8 dx a 2 
 
 Hence = = 
 
 da dx da 
 
 da 
 
 dx d 2 y 2 
 
 a* 
 
 J 
 
 the equation to the evolute. Hence the evolute to the 
 tractrix is the catenary. 
 
 (16.) The equation to a circle being y^=(a 2 x 2 )^ ; prove 
 that the radius of curvature equals a . 
 
 /yi2 aj2 
 
 (17.) g +^2 = 1 being the equation to the ellipse; show 
 
 that the radius of curvature is - - ; > where the eccen- 
 
 ab 
 
 tricity e= -- 
 
 x 3 
 (18.) In the cubical parabola, whose equation is y= 
 
 (a 4 -f x 4 )? 
 show that the radius of curvature is a , 
 
 (19.) Prove that in the circle, parabola, ellipse, and hy 
 perbola, or in any plane curve whose equation is of the 
 second degree, the radius of curvature varies as the cube of 
 the normal. 
 
144 RADIUS OP CURVATURE, 
 
 (20.) The equation to the rectangular hyperbola is 
 y 1 x 2 + a 2 = ; show that the radius of curvature is 
 
 and that the equation to its evolute is 
 
 (21.) Determine the radius of curvature to the curve * 
 called the tractrix, the equation being y=\/c 2 y 2 . 
 
 / 
 
 y 
 
 (22.) The polar equation to the lemniscata of Bernouilli 
 
 a 2 
 is r 2 =o 2 cos20 ; show that the radius of curvature is 
 
 O T 
 
 (23.) Prove that the length of the arc of the evolute in 
 tercepted between two radii of curvature is equal to the 
 difference between the lengths of those radii. 
 
 (24.) Show that in the common parabola, whose equation 
 is y 2 =4#x, the radius ot curvature is greatest at the vertex, 
 that the radius of curvature at that point is half the latus 
 rectum, and determine the equation to the evolute. 
 
 (25.) If N be the normal and R the radius of curvature 
 to a point in the ellipse ; prove that 
 
 (26.) r=f=. being the equation to the lituus ; show that 
 the radius of curvature is 
 
 (27.) If r=f(d), find an expression for the radius of cur 
 vature, that is, prove that 
 
EVOLUTES. 
 
 145 
 
 (28.) The equation to the logarithmic or equiangular spiral, 
 referred to p and r, is p=mr j show that the radius of cur 
 vature is ~> and that to this spiral the evolute is a similar 
 m* 
 
 spiral. 
 
 at2 yi& 
 
 (29.) -f ^~2 = l being the equation to the ellipse ; show 
 
 that the equation to its evolute is (aa)%+(l>l3)%=(a 2 1 2 )%, 
 and exhibit its form and position with respect to the centre 
 of the ellipse. 
 
 (30.) In the hyperbola, the focus being considered as the 
 pole, the length of the perpendicular on the tangent is 
 
 \ 
 
 - j show that the chord of curvature through the 
 
 . 
 focus is 
 
 a 
 
 (31.) The equation between p and r in the epicycloid 
 is (c 2 a 2 )p 2 =c 2 (r 2 a 2 ); prove that the radius of curvature 
 
 isv^ 2 a 2 )(r2 a 2 )- 
 
 (32.) The equation to the involute of the circle is 
 
 ad -\-asec~ 1 l~\=(r 2 a 2 )^ ; prove that its radius of curva- 
 \o// 
 
 ture is p, and that its evolute is a circle whose centre is the 
 origin, and radius a. 
 
 (33.) The equation to the hypocycloid is $t-f yt=a$ ; 
 show that the equation to its evolute is 
 
 (34.) Referring to example 22, and letting 7? and R t re 
 spectively represent the radii of curvature at the extremities* 
 
146 RADIUS OF CURVATURE, 
 
 of the major and minor axes of an ellipse, prove that the 
 length of the evolute is 4 ( --- \ - 
 
 (35.) R being the radius of curvature, and s the length of 
 
 ds 3 
 
 an arc of a plane curve : show that jR= ^- 
 
 dxd 2 y 
 
 (36.) Considering the earth to be an oblate spheroid, or 
 ellipsoid, 2 a its equatorial and 2 1 its polar diameter, m and m t 
 respectively the lengths of an arc of 1 of a meridian in 
 two given latitudes X and X,, and considering these lengths 
 to coincide with the osculating circles through their middle 
 points show, by reference to Ex. 14, that the 
 equatorial diameter : polar diameter 
 
 (37.) Show how the result of the last example would be 
 modified if one of the arcs of the meridian were measured 
 at the equator. 
 
 (38.) Let AP be a parabola, P any point in the curve, 
 draw the tangent PT, and the normal PG ; through T, the 
 point in which the tangent intersects the axis of abscissae, 
 draw TQ at right-angles to that axis, produce PG to meet 
 TQ in Q j prove that the radius of curvature at P is equal 
 to GQ, and show the centre of the osculating circle. 
 
 (39.) The equation to a curve being x sec2y=0 ; show 
 
 that - = 2x(x 2 1)^, and that the radius of curvature 
 
 . 2 
 is 
 
 4:X 
 
 (40.) If, in the common parabola, a point, determined 
 by x=3a, be taken ; show that the part of the radius of 
 curvature below the axis of x is 12 a. 
 
EVOLUTES. 147 
 
 (41.) If ds represent the small arc between two points 
 (x, y\ (x+dx, y + dy\ in a curve, and R the radius of cur 
 vature, investigate a general expression for that radius, 
 whatever be the independent variable ; that is, prove that 
 
 ds 3 
 
 j?= ;: 5 and thence deduce expressions for JK 
 
 d*xdy d^ydx 
 
 when- x, y and s be severally taken as the independent 
 variable. 
 
 (42.) Show that, if an inextensible thread were applied 
 to the evolute of a curve, and were to be gradually unwound, 
 a fixed point in the thread would describe the involute or 
 original curve. 
 
 (43.) Prove that the tangent to the evolute is the normal 
 to the involute. 
 
 (44.) Prove that, when the radius of curvature is either 
 a maximum or a minimum, the contact is of the third order. 
 
 CHAPTER XVI. 
 
 ENVELOPES TO LINES AND SURFACES. 
 
 Considering the evolute to a curve to be generated by the 
 ultimate intersections of consecutive normals, the evolute is 
 their envelope. 
 
 If f(x, y, a)=0 be the equation to a system of known 
 curves, intersecting each other in points determined by x 
 and y remaining constant whilst the variable parameter a 
 undergoes an infinitely small variation so as to become da, 
 the problem of finding the equation to the envelope resolves 
 
148 ENVELOPES TO 
 
 itself into that of finding an equation involving x, y and 
 constant quantities only, a being eliminated between the 
 equations f(x, y, a)=0, and/(#, y, a + da) = Q. 
 
 If there are several equations of condition involving the 
 parameter, it is expedient to have recourse to the method of 
 indeterminate multipliers, as in example 2. 
 
 This method of finding envelopes may be applied to the 
 determining of the equation to the evolute of a curve. 
 
 Ex. (1.) A series of equal ellipses are so placed that their 
 axes are in the same straight lines, the ellipticities alone 
 being variable ; find the equation to the curve which will 
 touch all the ellipses. 
 
 Let the constant rectangle ab=m 2 , 
 
 . - - aj2 
 
 - + -=.]. the eouation to the ellipse. 
 a 2 b 2 
 
 Here, a and b being variable, we must consider x, y and m 
 constant, and differentiate with respect to a and b. 
 
 Hence 
 
 dl_ . db_ b 
 
 da da a 
 
 b 
 
 a? 
 
 :. 2xy=:ab=.m 2 , the equation to a rectangular hyperbola 
 referred to its asymptotes. 
 
 (2.) A straight line, whose length is /, slides down be 
 tween two rectangular axes x and y ; find the equation to 
 
LINES AND SURFACES. 149 
 
 the envelope of the line in all its positions, that is to the 
 curve to which the line is always a tangent. 
 
 Let a and b be the variable intercepts of the line on the 
 axes, then 
 
 [- =1. the equation to the line, 
 a o 
 
 a? + b 2 =l 2 . Euc. b. i. p. 47. 
 
 Now, a and b being variable, we must differentiate consi 
 dering x, y and I constant. 
 
 2a~ + 2b=0, ada + bdb-0. ... (2) 
 
 do 
 
 Multiply (2) by the indeterminate multiplier X . 
 \ada -f \b db = 0. Add equation (1). 
 
 ~+Xa da+ 
 
 Assume ^ + \a=:0, and -|^-+\5=0, then 
 
 or 
 
 Hence j+ytaa^f, the equation to the locus of the ulti 
 mate intersections of the line. 
 
 (3.) To determine the curve whose tangent cuts off from 
 the axes a constant area. 
 
 o 2 
 
150 ENVELOPES TO 
 
 First, if the axes be rectangular, let a and I be the vari 
 able parts cut off, and m 2 =the constant area. 
 
 x y 
 
 -fy=l, the equation to the line, . . (1) 
 
 =m 2 , the area. . . . . (2) 
 
 Now, differentiating with respect to the variables a and I, 
 considering x, y and m constant, we have from (1) 
 
 x y db 
 
 a 2 I 2 da " 
 
 2m 2 
 
 a 
 
 b 2 x 2m 2 
 
 " da a 2 y 
 
 . db 
 : Ta~ 
 
 2m 2 y 
 
 aiiu. iruin ^ 
 
 2m 2 
 
 v 2 m vy 
 
 a z y ^ 
 2m 2 
 
 x 
 
 
 b 
 
 X \ y x ^ V \ 
 
 \/2 m A/^ 
 
 V^C 
 
 Vy 
 
 ^y ! v/ ^ - 
 
 a b vzm vx 
 
 \/2w/y 
 
 m v/2 m v/2 
 
 o / /o / m m 
 
 .. 2i v xy=m v 2, V xy j=.> sc y = > 
 
 v 2 2 
 
 the equation to a rectangular hyperbola, whose asymptotes 
 are the axes of x and y. 
 
 Secondly, if the axes be oblique, let them be inclined at 
 any angle a, a and b being the parts cut off, and m 2 the 
 area j then 
 
 2m 2 1 db 2m 2 1 
 
 =m 2 , b= -> = : 2> 
 
 2 sina a da sina dr 
 
 db b 2 x 
 
 Also - = ^ as in the first case. 
 
 da a 2 y 
 
LIXES AND SURFACES. 
 
 151 
 
 tfx 2m 2 
 Hence -=- 
 
 2m 2 ?/ 
 
 b 2 = : > 
 ic sin a 
 
 7 _ /- 
 v x V si 
 
 sn a 
 
 a=- 
 
 2m 2 
 
 b sin a v 2m vy v sin a vy v sin a 
 
 ic y xvyv sina yvxv sina 2v ic^sina .. 
 
 .-. (__ = _ ____j _ _ = -__ _l, 
 
 x v 2m v y m v 2 
 
 .*. xy=-rr 5 the equation to a hyperbola whose asymp- 
 
 2 sm a 
 
 totes are the oblique axes Ax, Ay. 
 
 (4.) Determine the equation to the curve which touches 
 all the curves included under the equation 
 
 a? 
 
 ?/=#tan0 -- - s > the variable being 6. 
 4 h cos 2 
 
 Differentiating with respect to 0, considering a?, y and 7* 
 constant, 
 
 sinfl 
 
 1 8^# 2 cos0sin0 
 ~^ 
 
 cos0 
 
 X* 
 
 Hence y=2h -7-7 h=k -7^? the equation required. 
 4A 4A 
 
152 ENVELOPES TO 
 
 If, in this problem, we consider h to vary as well as 6, 
 and if some constant area m 2 =A 2 sin 3 cos0 ; then we have 
 
 ! and h= 
 
 4m cos 2 6 4m cos 
 
 4 m 
 
 Differentiating with respect to 0, considering x, y and m 
 constant, 
 
 ""2 8 S P 
 
 0=#sec 2 -tan~*0.sec 2 0, - tan*0 = l, 
 
 4m 2 om 
 
 8m 64m 2 . 512m 3 
 
 f\ o X/^WA^.- ^ r*rT o 
 
 Q^ 2 27^ 
 
 Whence by substitution in (1) we have 
 _64m 2 x 2 512m 3 64m 2 128m 2 
 9# 4m 
 
 192m 2 -128m 2 64m 2 
 
 (5.) Two diameters of a circle intersect at right-angles ; 
 find the locus of the intersections of the chords joining the 
 extremities of the diameters, while the diameters perform a 
 complete revolution. 
 
 Let AB, Ab be two semi-diameters at right- 
 angles, 2 the diameter of the circle, A the 
 origin of co-ordinates, r=AP the line joining 
 the origin and point of intersection of the 
 chords. Then 
 
 AB a B111 ^ lJl 111 ^ ^2 ^2 
 

 I 
 
 LINES AND SURFACES. 153 
 
 Now, this problem is the same as that of determining the 
 curve to which the chord at its middle point shall be con 
 
 stantly a tangent; and y = mx + rV m 2 -f-l is the equation 
 to a straight line, r being the perpendicular upon it from 
 the origin. 
 
 Differentiating this equation with respect to m, consider 
 ing x, y, and r constant, 
 
 in <m*+l r m 2 +l r 2 
 
 0=# + r - ---- > -- = -- > - -=-, 
 
 x 
 
 
 1 r 2 I r 2 r 2 -a? a? 
 
 Hence = 
 
 a?-\-?/ 2 =r 2 => the equation to a circle, whose radius 
 
 is ^j and whoSe centre coincides with that of the original 
 
 circle. 
 
 (6.) If(tf-a) 2 + (y-6) 2 + * 2 =r 2 , and <* 2 + J 2 =c 2 ; deter 
 mine the equation to the envelope of the system of spheres 
 denned by these two equations. 
 
 Differentiating with regard to a and I, considering x, y, # 
 and c constant, we have, 
 
154 ENVELOPES TO 
 
 Multiplying the last equation by the indeterminate multi 
 plier X, and adding, we have 
 
 a=Q, ...(1) \b+y 6 = 0, ...(2) ; whence by 
 
 eliminating X we have -=T 
 
 Again (1) \a 2 + ax-a 2 =0, (2) \b 2 + by-b 2 =0, 
 
 ax+by . 
 .. X=l -- ^- -> ...(3) 
 c 2 
 
 (fix I 2 x , . a: 
 
 But aa?-|-iv= -- -- =(# 2 -f-& 2 ) - 
 
 a a a 
 
 Also ^ 2= 
 
 Hence X=l + ^ +y Substituting in (1), (2), 
 
 
 
 a- (a?+ffi=-(x- a }, b- (x 2 +y 2 )* = -(y - b), 
 
 c c 
 
 C 
 
 or 
 
 =r 2 -c 2 is the equation to the 
 envelope of the system of spheres. 
 
 (7.) Two straight lines /z and v, of variable length, are 
 drawn at right-angles to the axis of x, one of them v passing 
 

 LINES AND SURFACES. 155 
 
 through the origin of co-ordinates : now if they vary in such 
 a manner that the rectangle contained by them is a constant 
 quantity equal to I 2 ; determine the curve to which the 
 straight line passing through their upper extremities is 
 always a tangent. 
 
 Let^Z>=^, C=p, A=2a, AN=x, NP=y. Then 
 
 PN_BC _AD 
 
 ~~~ 
 
 :. y>AT=vAT+vx, and y.AT+2ay=p-AT+ px, 
 (y-v)AT= v x, (y-p)AT=px-2ay, 
 
 y-v y-p 
 
 vx ux%ay 
 Hence - =C - -> 
 
 . J0 
 y-v y-p b-_ yp 
 
 2 
 
 i or 
 
 where p alone is to be considered variable. 
 Differentiating with respect to p, we have 
 
 a?v 2 2a 2 v 2 
 Hence, by substitution, J - 2- + 1 2 (x - 2 a), 
 
 a 2 y 2 = b 2 (2 ax x 2 ), or 
 
 b 2 
 y 2 = (2axx 2 ), the equation to an ellipse, referred to 
 
 the vertex. 
 
156 ENVELOPES TO 
 
 (8.) If a series of parabolas be included under the equa 
 tion y 2 =a(xa), a being the variable parameter; show 
 that they will all be touched by the two straight lines de 
 termined by the equations y =-}--#, i/=-x, and draw 
 
 2i 2i 
 
 these lines. 
 
 (9.) Show how the method of determining envelopes may 
 be applied to finding the evolute of a curve, and apply it to 
 determine the evolute of the ellipse, whose equation, re- 
 
 a? y 2 
 
 ferred to the centre, is +-^-=1. 
 a 2 b 2 
 
 Equation to evolute (aa)*+(fy3)*=(tf 2 2 ). 
 (10.) Prove that the curve which touches all the straight 
 
 lines determined by the equation y=ax-\ > where a is 
 
 variable, is the common parabola. 
 
 (11.) A system of ellipses, with coincident but variable 
 axes, is subject to the condition that a 2 4-& 2 =ra 2 , a and I 
 being the major and minor axes; determine the curve which 
 shall be the envelope of the system. 
 
 (12.) If shot be discharged from a cannon with a con 
 stant velocity, but at various angles of elevation, they 
 will describe the parabolas included under the equation 
 
 x 2 
 y = ax (\-\-a 2 )) a being the variable parameter. Show 
 
 T C 
 
 that the curve which will touch all these parabolas is itself 
 
 x 2 
 a parabola whose equation is y=c 
 
 TtC 
 
 (13.) Considering the envelope to be formed by the inter 
 sections of straight lines ; show that the problem " to deter 
 mine the equation to the envelope " is the inverse of the 
 
LINES AND SURFACES. 157 
 
 problem " to determine the equation to a tangent to a 
 curve." 
 
 (14.) If /?, be a perpendicular of constant length from the 
 origin upon the straight lines defined by y=:ax-\-p, ( 2 + 1) 2 ; 
 show that the envelope of all these lines is a circle whose 
 radius is/?,. 
 
 (15.) If a surface be produced by the continued intersec 
 tion of planes represented by the equation \- -f-=l, 
 where abc = m 3 j a, b, c being variable, and m 3 constant ; 
 
 / Tft \ 
 
 prove that the equation to the surface is xyz\-^-\ 
 
 \ o / 
 
 (16.) A straight line, cutting from two straight lines 
 which meet in any angle, two segments whose sum is a, is a 
 tangent to a curve ; prove that that curve is a parabola, and 
 trace it. 
 
 (17.) If on one side of a horizontal straight line AR an in 
 definite number of parabolas of equal area be described from 
 a common point A, with their axes perpendicular to AM, 
 the equation to this system of parabolas is ai/ = 2oL5a%x tf 2 , 
 where a is variable ; prove that the curve which will touch 
 them all is an equilateral hyperbola whose equation is 
 
 2 5 
 xy=. a 2 , AR and a perpendicular to it from A being its 
 
 asymptotic axes. 
 
158 
 
 4 
 
 CHAPTER XVII. 
 
 MISCELLANEOUS EXERCISES. 
 
 (1.) Prove the ordinary rules for differentiation. 
 
 (2.) Explain the difference between explicit and implicit 
 functions. 
 
 (3.) Define and illustrate the terms " limit," " differen 
 tial," " differential coefficient." 
 
 (4.) Explain the difference between algebraic and trans 
 cendental functions. 
 
 (5.) Investigate the differentials of u=sinx, ?j = 
 
 (6.) Prove Taylor s Theorem, and from it deduce Stirling s 
 or Maclaurin s Theorem, and the Binomial Theorem of 
 Newton. 
 
 (7.) If y = e*smx , show, by means of the theorem of 
 
 Leibnitz, that -=2^ sin (x + n~}- 
 ax \ 4/ 
 
 (8.) In what manner may the value of a fraction be 
 determined when its numerator and denominator vanish 
 simultaneously ? 
 
 (9.) If u=.f(x] ; show that u is a maximum or minimum 
 when an odd number of differential coefficients becoming =0, 
 the differential coefficient of the next succeeding order is 
 negative or positive. 
 
 (10.) Deduce the equation to a straight line, ymx-\-l>, 
 and show that the equation to a perpendicular to it is 
 
MISCELLANEOUS EXERCISES. 159 
 
 (11.) Show that the equation to a straight line, which 
 intersects the axis of x at a distance a from the origin of 
 co-ordinates, and the axis of y at a distance b from that 
 
 . y x 
 
 origin. iSyH si. 
 o a 
 
 (12.) Show that the equation to a tangent to a curve, re- 
 
 ^/?/ 
 
 ferred to rectangular co-ordinates, is (y ) y)= (x,x). 
 
 ClX 
 
 (13.) If AT and AD be the intercepts of the tangent on 
 
 dx 
 the axes of x and y respectively ; prove that AT=y x, 
 
 and AD = y x--i and determine the equation to the 
 dx 
 
 normal. 
 
 (14.) Determine the differential expression for the sub- 
 tangent, subnormal, tangent, normal, perpendicular on tan 
 gent, and the tangent of the angle which the tangent makes 
 with a line from the origin. 
 
 (15.) If =/(#, y) ; prove that du=( j dx+(- \ dy, 
 
 \CLdO/ \(jL f U t 
 
 . , d*u d 2 u 
 and that - = -- 
 dydx dxdy 
 
 (16.) Ifu=f(y, z), where y, z, and consequently u, are 
 
 functions of x : show that du=l J dy 4- ( ) dz. 
 
 \dyt \dzl 
 
 (17.) Determine the conditions upon which a function of 
 two independent variables is a maximum or minimum. 
 
 (18.) Determine the differential expression for the area of 
 a plane curve, and if s be the length, and -j-=p j prove 
 
160 MISCELLANEOUS EXERCFSES. 
 
 (19.) If S be the surface and V the volume of a solid 
 generated by the revolution of a curve round its axis ; show 
 
 dV d/S 
 
 that = Try 2 , and = 2 7r$/ 2 ( 1 + p 2 )* 
 dx dx 
 
 (20.) If r and r, be the radii of the greater and smaller 
 ends of the frustrum of a right cone, and a the slant height \ 
 prove that the area of the frustrum is TTO, (r + r t ). 
 
 (21.) If r be the radius vector, p the perpendicular on 
 the tangent, and the angle swept out by the revolution of r 
 
 1 ^7 2 1 
 
 round the pole Si show that =u 2 + ( ) , where z<=-: 
 
 p 2 \a&l r 
 
 and that = 
 
 dr 
 
 (22.) If in polar curves p be the length of the perpendi 
 cular upon the tangent ; find the value of p in the circle, 
 parabola, ellipse, and hyperbola. 
 
 (23.) Define the rectilinear asymptote and the asymptotic 
 circle. 
 
 (24.) Define conjugate points, double points, cusps, and 
 points of contrary flexure, and show that a curve is concave 
 
 or convex to the axis according as y and -=-^ have the same 
 
 or different signs. 
 
 (25.) Prove that, in spirals, the curve is concave or convex 
 
 towards the pole, according as is positive or negative. 
 
 (26.) If A be the area, and 5 the length of a plane curve ; 
 
 dA dA 1 . ds 
 
 prove that y and =. T 
 
 dx dt) 2 dti 
 
 . ds 
 
 and = 
 dr /V2 
 
MISCELLANEOUS EXERCISES. 161 
 
 (27.) Prove that, in spirals, the subtangent =r 2 
 = f- -- -> and show how to draw the asymptote to a 
 
 spiral. 
 
 (28.) Explain what is meant by the osculating circle; 
 and show that the evolutes of all algebraic curves are recti- 
 iicable. 
 
 (29.) Explain the theory of the different orders of contact 
 of plane curves ; point out the exceptions to the rule that 
 every curve is cut by its circle of curvature, and show how 
 these exceptions apply to the ellipse. 
 
 (30.) Explain the difference between Taylor s and Mac- 
 laurin s Theorems, and point out the circumstances under 
 which the former sometimes fails. 
 
 (31.) Investigate Lagrange s* Theorem, and apply it to 
 determine a general law for the inversion of series by means 
 of the equation x = ay -f by 2 -}- c?/ 3 -f- dy^ + &c. 
 
 (32.) Apply Lagrange s Theorem to the determination 
 of the four first terms of the devolopment of y m , when 
 y/ a-\- xy n \ and find the general term in the expansion of 
 
 A >m in a series of powers of cos ft, when x-\- =2 cosO. 
 
 tJC 
 
 (33.) If u=^J2 - (*gr -y) Y~^ x bein S the inde " 
 
 * Ify=z + x$(y), and if u=f(y), /and ^ being any functions what 
 ever, then 
 
 d? r X* 
 
 p 2 
 
162 MISCELLANEOUS EXERCISES. 
 
 pendent variable ; show that, when x becomes cos 9, and 6 is 
 
 79 
 
 made the independent variable, u=l-^+y\ cosec 2 (9. 
 
 (34.) Explain exactly the mode in which the following 
 curves are generated, construct them, and thence derive 
 their equations : namely, the circle, parabola, ellipse, hyper 
 bola, cissoid of Diocles, conchoid of Nicomedes (superior and 
 inferior), cycloid, epicycloid, lemniscata of Bernouilli, quadra - 
 trix of Dinostratus, involute of the circle, catenary, tractory, 
 elastic curve, witch of Agnesi, curve of sines, cardioid, tri- 
 sectrix, logarithmic or equiangular spiral, spiral of Archi 
 medes, hyperbolic or reciprocal spiral, lituus, parabolic 
 spiral. 
 
 (35.) Show what kind of curves are included under the 
 equations ?/ 2 = mx + nx*, r = a sin n 6, r=a cos 6 + 1, r =a6 r> , 
 r=asmnd + b sin nfi + c sin n ,,0 -f &c. respectively. 
 
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