Production Note Cornell University Library pro- duced this volume to replace the irreparably deteriorated original. It was scanned using Xerox soft- ware and equipment at 600 dots per inch resolution and com- pressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell's replacement volume on paper that meets the ANSI Stand- ard Z39.48-1984. The production of this volume was supported in part by the Commission on Pres- ervation and Access and the Xerox Corporation. 1991.AN ELEMENTARY TREATISE CURVES, FUNCTIONS, AND FORCES. VOLUME FIRST; CONTAINING ANALYTIC GEOMETRY AND THE DIFFERENTIAL CALCULUS. By BENJAMIN PEIRCE, A. M., University Professor of Mathematics and Natural Philosophy in Harvard University. NEW EDITION. BOSTON AND CAMBRIDGE : JAMES MUNROE AND COMPANY. M DCCC LII./ Entered according to Act of Congress, in the year 1841, by James Munroe & Company, In the Clerk’s Office of the District Court of the District of Massachusetts BOSTON: THURSTON, TORRY, AND EVERSON, Printers, 31 Devonshire Street.CONTENTS. BOOK I. APPLICATION OP ALGEBRA TO GEOMETRY. CHAPTER I. GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTI- TIES ...................................1 CHAPTER II. ANALYSIS OF DETERMINATE PROBLEMS . . . 18 CHAPTER III. position............................28 CHAPTER IV. equations of loci ...... 52 CHAPTER V. classification and construction of loci . . 87 CHAPTER VI. equation of the first degree .... 90 CHAPTER VII. equation of the second degree . . . . 108XV CONTENTS CHAPTER VIII. SIMILAR CURVES.......................140 CHAPTER IX. PLANE SECTIONS OF SURFACES .... 143 BOOK II. DIFFERENTIAL CALCULUS. CHAPTER I. FUNCTIONS................................163 CHAPTER II. INFINITESIMALS...........................172 CHAPTER III. DIFFERENTIALS............................179 CHAPTER IV. COMPOUND AND ALGEBRAIC FUNCTIONS . . . 190 CHAPTER V. LOGARITHMIC FUNCTIONS....................196 CHAPTER VI. CIRCULAR FUNCTIONS ......................200 CHAPTER VII. INDETERMINATE FORMS 207CONTENTS V CHAPTER VIII. MAXIMA AND MINIMA .....................213 CHAPTER IX. CONTACT................................220 CHAPTER X. CURVATURE..............................240 CHAPTER XI. SINGULAR POINTS........................248 CHAPTER XII. APPROXIMATION .........................295BOOK I. APPLICATION OF ALGEBRA TO GEOMETRY.CUEYES AND FUNCTIONS. BOOK I. APPLICATION OF ALGEBRA TO GEOMETRY. CHAPTER I. GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUAN- TITIES. 1. In the application of Algebra to Geometry, usually called Analytic Geometry, the magnitudes of lines, angles, surfaces and solids are expressed by means of letters of the alphabet; and each problem, being put into equations by the exercise of ingenuity, is solved by the ordinary processes of Algebra. The algebraical result is finally to be interpreted geo- metrically ; and this geometrical interpretation of an algebraical expression is called the geometrical con- struction of that expression. The geometrical con- struction of the results is, then, the last operation in the solution of problems; but it is convenient, on account of its simplicity, to begin with the consid- eration of it. We begin with the easiest cases and proceed to the more difficult ones, and we regard each l2 ANALYTICAL GEOMETRY. [β. I. CH. I. Sum and difference. Negative sign. letter as representing a line, so that every algebraical expression of the first degree will denote a line; whence it is called linear. 2. Problem. To construct a -f- b. Solution. Take (fig. 1) AB a, BC=zb; and we have AC = AB + BC = a + b; so that A C is the required value of a -f- &· 3. Problem. To construct a — b. Solution. Take (fig. 2) AB = a, and from B, in the opposite direction, BC=b; we have then AC = AB—BC=a — b, so that AC is the required result. 4. Corollary. If a were zero, the preceding solution would become the same as to take from A (fig. 3) in the direction AC, opposite to AB, AC=b; so that the negative sign would only be indicated by the direction of AC. In order to generalize the preceding con- struction we must, then, adopt the rule that The geometrical interpretation of the negative sign is} that it indicates an opposite direction. 5. Problem. To construct an algebraic expression consisting of a series of letters connected together by the signs -f· and —.$9.] geometrical construction. 3 Product aud Quotient. Surface and Solid. Solution. Collect into one sum, by art. 2, all the letters preceded by which sum may be denoted by a; and col- lect into another sum all the letters preceded by —, which sum may be denoted by δ; and the value of a — b may then be constructed by art. 3. 6. Corollary. If a letter is preceded by an integral numerical coefficient, it may be regarded as a letter repeated a number of times equal to this integer. 7. Problem. To construct a b. Solution. The parallelogram of which the base is a, and the altitude is £, is equal to the product a which accord- ingly represents a surface ; and this conclusion is a general one, that is, A homogeneous algebraical expression of the second degree represents a surface. 8. Problem. To construct a be. Solution. The parallelopiped of which the base is the parallelogram a by and the altitude is c, is equal to the pro- duct aic, which accordingly represents a solid; and, in general, A homogeneous algebraical expression of the third degree represents a solid. 9. Problem. To construct %. o Solution. Make (fig. 4) the right angle ABC, take AB = a BC = b, and join AC. The angle ACB is, by trigonometry, that angle whose tangent is4 ANALYTICAL GEOMETRY. [β. I. OH. I. Angle. To render homogeneous. 10. Corollary. If we had taken AC = l·, CL the angle ACB would have been the angle, whose sine is -, and, in general, A homogeneous algebraical expression, whose de- gree is zero, represents the sine, tangent, 6pc. of an angle. 11. Scholium. Since no other magnitudes occur in Geometry but angles, lines, surfaces and solids, all al- gebraical quantities which represent geometrical mag- nitudes must be either of the 1st, the 2d, the 3d, or the zero degree ; and since dissimilar geometrical magni- tudes can neither be added together, nor subtracted from each other, these algebraical expressions must dlso be homogeneous. If, therefore, an algebraical result is obtained, which is not homogeneous, or is of a different degree, from those just enumerated; it can only arise from the circumstance, that the geometrical unit of length, being represented algebraically by 1, disappears from all algebraical expressions in which it is either a fac- tor or a divisor. To render these results homogene- ous, then, and of any required degree, it is only necessary to restore this divisor or factor which rep- resents unity. 12. Problem. To render a given algebraical expres- sion homogeneous and of any required degree. Solution. Introduce 1, as a factor or divisor, repeat- ed as many times as may be necessary, into every term where it is required.*U] GEOMETRICAL CONSTRUCTION. 5 To render homogeneous of any degree. 13. Examples. a2J _l c _1_ ^2 1. Render----- —r--homogeneous of the 1st degree. e "j— η- -4ns.---\ -75-------- 1. e + Λ2 2. Render —^ homogeneous of the 2d degree. Z2 -\- m3 Ans. (υ4^ + (υ2^ + (1)3^· 1 .12 -j- m3 3. Render —^ ----homogeneous of the 3d degree. (l)6q + (l)^g + ^A« K\fde— (1)3 a λ2 1 b 4. Render —homogeneous of the zero degree. Ans, *2+l. δ l.c— 5. Render a b homogeneous of the 1st degree. a b ΊΓ' 6. Render ab c-\-d — e2homogeneous of the 1st degree. ab c Ans. (I)2 • d——· 1 14. Scholium. By the preceding process, every fraction, which does not involve radicals, may be reduced to a homogeneous form, in which each term is of the first degree; and, although this form is not always that which leads to the most simple form of construction, its generality gives it a peculiar fitness l*6 ANALYTICAL· GEOMETRY. [β. I. CH. I. To render homogeneous. Fraction. for the general purposes of instruction, where the artifices of ingenuity are rather to be avoided than displayed. 15. Examples. 1. Reduce the fraction of example 1, art. 13, to a homo- geneous form, in which each term is of the first degree. /a?b Ans. \(1)2 , , ( , Λ*2 \ + c + T)^(e+T) 2. Reduce the fraction of example 2, art. 13, to a homo- geneous form, in which each term is of the first degree. / d3 . he\ β /Z*2 m3 \ Ans. 3. Reduce the fraction of example 3, art. 13, to a homo- geneous form, in which each term is of the first degree. / be b4 5h2 \ /de \ ** {a+T+w^(T + ay 4. Reduce the fraction of example 4, art. 13, to a homo- geneous form, in which each term is of the first degree. \ / d2 ' M?+*)+(-t)· 16. Problem. Solution. We have cl b To construct — · c c : a = b : a b that is, the given fraction is a fourth proportional to the three lines, c, a, and b. Find, then: by Geometry, a fourth proportional to$ 19] GEOMETRICAL CONSTRUCTION. 7 Monomial. the lines c, a, and b; and this fourth proportional is the required residt. Qyl 17. Corollary. The value of is a third propor- tional to c and a. 18. Problem. To construct any monomial which denotes a line. Solution. If the monomial is not of the first degree, reduce it to the first degree by art. 12. It is then of the form. abed... a! b* d .. . ab c d — X t^x-· λ' h· /*' Construct first —, and let m be the line which it a' represents. The given quantity becomes m c d c‘ Let, again, ml = m c ΊΓ’ a/so 771" = m' d ~~d~' &c. awe? the last line thus obtained is plainly the required result. 19. Examples. 1. Construct the line a b. Ans. m 2. Construct the line a δ c. ulns. *»= a b z Γ a δ X’ = ab. m'= me —-=abc.8 ANALYTICAL GEOMETRY. [β. I. CH. I. Any expression not involving radicals. 3. Construct the line a5. m! a m" == ■ 1 ’ a5. a , Ans. m — —, m! m a ~ΊΓ5 m" a m!" — n5 i - <&b a &2 m b Ans. - S3 d ’ d ’ m'. 1. a2 b ~df~' „ ~ A . .. 2ab . 2a. b . m. 1 5. Construct the line —-—. Ans. m~---,m — efg e f g 6. Construct the line —. a 7. Construct the line Λ _2ab 'efg' Ans. m = ill=zi a a λ 1-4 Λ Ans. m = — = ~. 20. Corollary. By this process each term of an algebraic expression, which does not involve radicals, is reduced to a line; and if the expression does not involve fractions, it may then be reduced to a single line by art. 5; if it does involve fractions, the nume- rator and denominator of each fraction is, by art. 5, reduced to a single line, and each fraction, being then of the form is constructed like example 7 of the preceding article, and the aggregate of the fractions is then reduced to a single line, by art. 5. Any algebraic expression, which represents a line, and does not involve radicals, may therefore be con- structed by this process.$22.] GEOMETRICAL· CONSTRUCTION. 9 Any expression free from radicals. 21. Examples. , ^ t . (Pb + c + d* 1. Construct the tine-----—-----. e 2 Solution. Let m = — ψ = d*b m' — — = a1 m"=-j = h* and the fraction becomes m -(- c -f- w! ^ e -f- wi" ’ A — m c m' B = € -f- Ml", 1.4 let now and the line represented by is the required line. B 2. Construct the line represented by the fraction of exam- ple 2, art. 13. Ans. m* — d3y m11 — h e, m,u — Z2, mxv — m2, A = a -(- m! -f- m!\ B — m,u + mXY , and the required line is the fourth proportional to J5, 1, and A. 3. Construct the line represented by the fraction of exam- ple 3, art. 13. Ans. Let m ~ h c, m1 = cZ5 Λ2, m" — d e, A — a m m', B = m'' — a, and the required line is the fourth proportional to B, 1, and A.10 ANALYTICAL GEOMETRY. [B. I. CH. I. Radicals of the second degree. 4. Construct the line represented by the fraction of exam- ple 4, art. 13. Ans. m — a2, ml = d2, B — c — m', and the required line is the fourth proportional to i?, 1, and A. 5. Construct the line represented by the polynomial of ex- ample 6, art. 13. · Ans. m = ah c, m! = e2, and A = m -}- h — m! is the required line. 22. Problem. To construct the line \/{a b). Solution. Since */{ab) is a mean proportional be- tween a and b, the required result is obtained by con- structing, geometrically, this mean proportional between a and b. 23. Corollary. The expression a/A = ^(I.A) may be constructed by finding a mean proportional between 1 and A. 24. Corollary. The square root of any algebraical expression, which does not involve radicals, may be constructed by finding, as in art. 20, the line A, which this algebraic expression represents, and then construct- ing s/A as in the preceding article. By the repeated application of this process, *any algebraic expression may be constructed which repre- sents a line, and which does not involve any other radi- cals than those of the second degree.GEOMETRICAL CONSTRUCTION. 11 § 27. Radicals of the second degree. 25. Examples. 1. Construct the line \/(α + b — c — e.) Ans. A — a b — c — e, and \fA is the requited line. 2. Construct the line \Z(«2 + « »*)· -4ns. i = a2 -f « and \Λ1 is the required line. 3. Construct the line r ^ ·. e2 —V(c2 — fn) Ans. m= /^/a ft, —A), m"=e2, ml"—*/{&—/n), .4 z= m', B mm" — m,,\ A and the line — is the required line. B 4. Construct the line s/a. .Ans. 7» = \/a, and is the required line. 26. Scholium. When the expression whose square root is required is easily decomposed into two factors, it is immediately reduced to the form */{ab) and con- structed, as in art. 22. 27. Examples. 1. Construct example 2, art. 25, by decomposing the quantity under the radical sign into two factors* Solution. cfiam-==l a(am). Let b = a -|- m, and the line */(a b) is the required line.12 ANALYTICAL GEOMETRY. [β. I. CH. I. Square root of sum and difference of squares. 2. Construct ^/(a2 ae -1- am — an) by decomposing the quantity under the radical sign into two factors. Ans. h — a e m — n, and \/(a required line. 3. Construct \/(a -[-a2— a3) by decomposing the quan- tity under the radical sign into two factors. Ans. b — 1 -}-a — a2, and \/(a l·) is the required line. 4. Construct \/(a2 — £2) by decomposing the quantity under the radical sign into factors. Ans. c= a-\-b,e = a — and ^/(c e) is the required line. 28. Scholium. Example 4 of the preceding article may also be solved by constructing a right triangle, of which a is the hypothenuse and h a leg, and \Ζ(«2 — 62) will be the other leg. 29. Corollary. In the same way \/(a2 + &9) ls hypothenuse of a right triangle, of το hieh a and b are the legs. 30. Corollary. By combining the processes of the two preceding articles, any such expression as V(«2 ψ _ C2 _ e2 &c. may be constructed. For if we take m = \/(a2-(-J2), m'nn \/(m2 — c2), m" = \/(m'2 — e2), m'" = -f- Λ2), &c. # we have m2 = a2 -|- wi' = \/(m2 — c2) = \/ (a2 —|— Z>2 — c2), m'2 = a2-f-£2 —c2; or§33.] GEOMETRICAL CONSTRUCTION. 13 Construction of radicals. m" = — e2) — V^(a2 -|- δ2 — c2 — e2), or m//2 =zz a2 -j- J2 — c2 — e2; m'" — VK'2 + Λ2) = V(«2 + δ2 — c2 — e2 + A2), &,c. 31. Corollary. The square root of the sum or dif- ference of any expressions, which involve no other radicals than those of the second degree, may a/so he constructed by the preceding process. For if either of these expressions is constructed by the processes before given, it may be represented by A ; and, if we denote rfA by m, we have mQ= A, so that each expression is reduced to the form of a square, and the whole radical is reduced to the form of the preceding article. 32. Examples. 1. Construct example 1, art. 25, by the process of art. 31. Ans. m — \/a, m! — m" — s/c, m!n = \/e, and the line -f- m'2 — m"2— m'"2) is the required line. 2. Construct example 2, art. 25, by the process of art. 31. Ans. m' — \/(a m), and \/(a2 + m'2) is the required line. 3. Construct the line s/(a2-|-be — e3-f-h) by process of art. 31. Ans. m — s/(b c), m! ~ \/e3, m" = ^/Λ, and the line V(«2+ ^2 — m'2-}-»»"2) is the required line. 33. Proble?n. To construct an algebraical expression which represents a surface. 214 ANALYTICAL GEOMETRY. [b. I. CH. I. Surface. Solid. Angle. Solution. Let A be the line which is represented by this algebraical expression, and since we have A — l.A, the required surface is represented in magnitude by the parallelogram, whose base is 1, and altitude A, or by the equivalent square, triangle, &c. 34. Problem. To construct an algebraical expression which represents a solid. Solution. Let A be the line which is represented by this expression, and since we have A = {\)*. A, the required solid is represented in magnitude by the parallelopiped, whose base is the square (1 )2, and whose altitude is A. 35. Problem. To construct an algebraical expres- sion, which represents the sine, tangent, fyc. of an angle. Solution. Let A be the line which is represented by this expression, and since we have the required angle is found by art. 9 or 10. 36. Scholium. The construction of all geometrical magnitudes being, by the three preceding articles, reduced to that of the line, we shall limit our con- structions hereafter to that of the line.§38.] GEOMETRICAL· CONSTRUCTION. 15 Equations of first and second degree. 37. Problem. To construct the root of an equation of the first degree with one unknown quantity. Solution. Every equation of the first degree may, as is proved in Algebra, όβ reduced to the form Ax + M=0; whence and this value of x may be constructed by art. 18. 38. Problem. To construct the roots of an equation of the second degree with one unknown quantity. Solution. The equation of the second degree may, as is shown in Algebra, be reduced to the form Ax* + Bz + M=0. If we divide this equation by A, and put B M a~2A’ m~ A' it becomes χ^-\-2αχ~\~ιη — 0. The roots of this last equation are x — — fli\/(a2 — m). Case 1. When m is positive and greater than a2, the 7'oots are both imaginary, and cannot be con- structed. Case 2. When m is positive and equal to a2, each root is equal — a, which needs no farther construc- tion.16 ANALYTICAL GEOMETRY. [β. I. CH. I. Quadratic equation. Case 3. When m is positive and less than a2. Let, in this case, b = s/m or i2 = m. The roots become x — — a db s/ (a2 — δ2), which are thus constructed. Draw (fig. 5) £Ae two indefinite lines DAD' and AB perpendicular to each other. Take AB — b; /rom- B as a centre, a radius BC=a, describe an arc cutting DAD' in C. Take CD=CD'=:BC=a, awe? /Ae required roots, independently of their signs, are and JZX Demonstration. For 4 C = \/(£C2 — ^^2)= V(«2 —*2) and — = — CD + AC = — a +V(«S— — AD' = — CD' —AC =—a — yV — δ2). Case 4. TFAew m is zero, ?Ae roo/s are .τ = 0 and # = —2a, which require no f urther construction. Case 5. TFAew m is negative, so ?Aa? — m is positive. Let h — */— m, or Z>2 = — m. The roots become x — — a± \/(a2 + ), which are thus constructed.§39.] GEOMETRICAL· CONSTRUCTION. 17 Quadratic equation. Draw (fig. 6) the two lines AB and AC 'perpendicu- lar to each other. Take AB = a, and AC~b; through BC draw the indefinite line BCD'. Take BD = BD' = AB=a, and the required roots. independently of their signs, are CD and CD'. Demonstration. For BC- xf(AB* + A&) = V(a2 + 3») and CD = — BD -t-BC=z — α + Λ/(«2 + ^2) — CDf=—BD—BC= — a — V(2 + &2)· 39. Scholium. Radicals of a higher than the second degree, and roots of equations of a higher than the second degree, do not usually admit of geometrical construction. 2*18 ANALYTICAL GEOMETRY. [b. I. CH. II. Solution of determinate problems. CHAPTER II. ANALYSIS OF DETERMINATE PROBLEMS. 40. Geometrical problems are of two classes, deter- minate and indeterminate. Determinate problems are those, which lead to as many algebraical equations as unknown quantities; and indeterminate problems are those, in which the number of equations is less than that of the unknown quantities. 41. The solution of a geometrical problem consists of these three parts ; First, the putting of the question into equations; Secondly) the solution of these equations; Thirdly, the geometrical construction of the alge- braical results. The last of these processes has been discussed in the preceding chapter, but it must be observed that much skill is often shown in arranging the construction in such a form, that it may be readily drawn and be neat in its appearance. The second process is exclusively algebraical, and the first process, the putting into equations, is a task which, necessarily, requires ingenuity, and can only be taught by examples. One great object is to obtain the simplest pos- sible equations, and such as do not surpass the second de- gree. It is not unfrequently the case, that, when a question admits of several solutions, two or more of these solutions are connected together in such a way, that the same quan-§42.] DETERMINATE PROBLEMS. 19 Division oi line. tity, being obviously common to them, should, on this ac- count, be selected as the unknown quantity. 42. Examples. 1. To divide a line AB (fig. 7) into two such parts, that the difference of the squares described upon the two parts may be equal to a given surface. Solution. Let the magnitude of the given surface be equal to that of the square whose side is AC, and let Ώ be the point of division·, AD being the greater part. Let a = AB, b — AC, and x = AD, we have then BD = a — x ; and the equation for solution is or Hence a;2 — (a — a?)2 = δ2, 2 ax — α2 := δ2, δ2 + a2 Ψ , , *= 2ΪΓ = ^ + 5“· Construction. Let E be the middle of AB, Draw the indefinite line EB' in any direction whatever. Take EB' = EB = J a, EC' = EC"=iAC=:}>b. Join B'Cand through C" draw C"D parallel to BO, D is the point of division required. Demonstration. We have EB*: EC'= EC" . ED, or = ED;20 ANALYTICAL GEOMETRY. [b. I. CH. II, Rectangle inscribed in triangle. whence ED = iia-f-Aa= —, * 2 2 a and AD = ED 4- AE = ~ A-ha. 2a 1 2. To inscribe in a triangle ABC (fig. 8), a rectangle DEFG whose base and altitude are in the given ratio m :n. Solution. Let fall the perpendicular AIH. Let BC = b, AH=k, DE = HI=x, Al = AH—HI=h—x; and, since n: m = DE : EF, we have EF — —. n But the triangles AEF and ABC are similar, and their bases are, therefore, proportional to their altitudes, that is, BC:EF=AH:AI, or b: m x = h : h — z. Hence we find, by algebraical solution, nbh 7 7 / mh x= ——------ =bh~(-------- mA -f- nb \ n Construction. Find a fourth proportional to w, m, and A, and denote it by h\ and then x is obviously a fourth pro- portional to h( A, δ and λ. The following simple form has been obtained by geome- ters. Draw AK parallel to BC, and take AK=h‘ Join KCy and ED is the required altitude.§42.] DETERMINATE PROBLEMS· 21 Line of given length intercepted between parallels. Demonstration. For since h\ or its equal AK, is a fourth proportional to », m, and h, we have n : m — h : AK, or AK : AH = m : n. If we let fall the perpendicular KL upon BC, we have the quadrilaterals CFED, CAKL, which are formed of similar triangles; they are therefore similar, and their homologous sides give the proportion JF.E : ΖλΕ = AST : KL (or .4H) = m : n ; so that FE and DE are in the required ratio. Corollary. If the ratio m : n were that of equality, the rectangle would be a square. By making, then, AK equal to AH, and completing the construction as before, a square is inscribed in the given triangle. 3. To draw through a given point A (fig. 9) situated between two given parallels BC and DE a line HI, which may be of a given length a. Solution. Since the point is given, its distances from the parallels must be given, which are AF— b, AG=c; let AH=zx·, we shall leave it as an exercise for the learner to find the value of x, which is __ a b X“b+c* Construction. The value of x is a fourth proportional to h -j- c, a, and £, and may be easily constructed. The following form is quite simple. From any point G in the line DE as a centre, with a radius equal to a\ de- scribe an arc cutting BC in K. Join GK, and the line22 ANALYTICAL GEOMETRY. [B. I. CH. II. Circle tangent to given line. drawn through A parallel to GK is obviously of the same length with GK, and is, therefore, the required line. Corollary. The problem is impossible when the length a is less than GF or its equal b c. 4. To draw a circle through two given points A and B (fig 10), and tangent to a given line DC, Solution, Join AB, and produce AB to meet DC at D, Let C be the point of contact, and let DA~a, DB = b, DC = x; we have, by geometry, DA : DC = DC: DB, or a : x — x : b ; whence x = ± \f{a b). Construction, Find a mean proportional between a and b, and take DC or DC* equal to it, and C or C' is the point of contact, these two values corresponding to the two different circles BCA and BC'A. Instead of finding the mean proportional by the ordinary process, we may find it by drawing any arc AEB through A and B, and the tangent DE to this arc is, by geometry, the mean proportional between DA and DB. Corollary. The problem is impossible if a and b are of opposite signs, that is, if A and B are in opposite directions from D, one being above the line and the other below it. Corollary. If either a or b is zero, as in fig. 11, where DA — a — 0, the problem is reduced to that of finding a circle which passes through the given point B, and is tangent to a given line CA at a given point A. Construction of this case. Erect OA perpendicular to AC. Join AB, and at the middle E of AB erect the per-§42.] DETERMINATE PROBLEMS. 23 Division of a line. pendicular £0; 0 is the centre. The demonstration of this construction is left as an exercise for the learner. Corollary. If a and b are equal, as in fig. 12, the problem is reduced to finding a circle which touches a given line DA at a point A, and also touches another given line DC. Construction. Take DC = DO = a — DA, and the point O or O', the intersection of the perpendicular 0^40', with the perpendicular CO or C'O', is the centre of the required circle. 5. To divide a given line AB (fig. 13) into two such parts, that the sum of the squares described upon the two parts may be equal to a given surface. Solution. Let the given surface be twice the square whose side is AC, and let D be the point of division. Let also E be the middle of the line, and let BE zz: AE — a, AC = b, DE == x. The value of x will be found to be x = dc \/(£2 — a2), so that £ is a leg of a right triangle whose hypothenuse is b and other leg a. Corollary. The problem is impossible when b is less than a, and also when *> a, or b* — cP>a*, or J2 > 2 a2, or 2 J2 > 4 a2, 2 i2> (2 a)2;24 ANALYTICAL GEOMETRY. [b. I. CH. II. Line divided in extreme and mean ratio. that is, when the given surface is greater than the square of the given line. 6. To divide a given line AB (fig. 14) at the point C in extreme and mean ratio. Solution. Let AC be the greater part, and let AB — a, AC ~ x, CB — a — x, we are to have a : x = x: a — x, whence we find a?2 -|-ax — a2 = 0, and x = J a (— 1 ± \/5). Construction. The roots of the equation x*-\- ax — a2 = 0, being constructed by case 5, art. 38, give the usual con- struction of this problem. 7. Through a given point C (fig. 15) to draw a line BCD, so that the surface of the triangle ABD intercepted between the lines AB and AD may be of a given magnitude. Solution. Let the given surface be double that of the given rhombus AEFG. Draw CH parallel to AD, and Cl parallel to AB. Let AI — CH=a, AHz=z CI=l·, AE = c, AD ~ x, AB = y. We have surface of triangle = J x y sin. A = 2 c2 sin. A, whence x y = 4 c2. The similar triangles BHC, BAD, give BH: HC=BA :AD, or y — h:a=y:x- whence x y = a y b χ.§42.] DETERMINATE PROBLEMS. 25 Given length intercepted. The solution of these equations gives X = y — «&)) y=^C±V(ca — a 5)) which are easily constructed. Corollary. The problem is impossible when c2 is less than α δ ; that is, when - c2 sin. A <^ab sin. ^4, or when the rhombus AEFG is less than the parallelogram AHCI. 8. Through a given point C (fig. 16) to draw a line BCD, so that the part BCD intercepted between two given lines AB and AD may be of a given length, the point C being at equal distances from the two given lines. Solution. Draw CH and Cl parallel respectively to AB and AD^ and they are obviously equal to each other. Let. then AH—A1— CH— Cl—a, BD = hy AD = AB = y. From triangle ABD, we have a?2y2 — 2x y cos. i = and from similar triangles BIC and BAD, a; y = a (z + y). / As these equations are symmetrical with regard to z and y they are simplified by putting xy = t·, 326 ANALYTICAL GEOMETRY. [β. I. CH. II. Given length intercepted. and become t — as s2— 2 a (1 -{- cos. .4) s = Ψ; whence s and t, x and y are found. The following solution is, however, much neater. Join A C, and let the angle ACD be the unknown quantity, and put c = AC, CAD = \A — A\ ACD = $>, ADC == 180° _ (φ + A'), ABC = φ — A', sin. ADC = sin. (φ 4~ A1) ; and, by trigonometry, sin. ADC:sin. DAC = AC: DC, sin. (φ -f- -4') : sin. A/ = c: DC; Hence DC = C^1P- A‘ sin. (φ + A') Also sin. ABC : sin. BAC = AC : jBC, sin. (φ — A'): sin. A! =. c: BC; c sin. A' jBC = b=BD=BC+DC = sin. (t. The coordinates of the new origin referred to the first origin and axis must be known ; let them be AAX = a, and AXAC = β ; the inclination of the two axes must also be known, and let it be *. Produce CLAt to Awe have A1A/C = and A A1 A x = 180° — a; AAXA' = ΑλΑΤ— Α,ΑΑ' = * — β. also§50.] POSITION. 31 Distance of two points. AAxB=z 180 ° — (BA1C1+AAlA) = 180° — (φχ + a —μ) cos. A A XB = — cos. (φ t -j- a, — p). The triangle AAXB gives, then, AB2=zAA2 + A1B2—2.AA1 .A,B.cos. AA,By r2 = a2 + r\ -f- 2 ary cos. (φ1 -f- » — β) ; r = V [a2 -f-r\ 2 a rx cos. (φ1 -j- «■— /3)] (1) and AB : AXB = sin. (AAXB) : sin. BAAX, or r : rt = sin. {φχ -|- * —β) : sin. {φ — β) ; v whence sin. (φ — β) = -A . sin. (Pj + * —/s), (2) and equations (1) and (2) are the required formulse, 49. Corollary. If the new origin is in the former axis, and if the axes coincide, we have <* = 0, β — 0, and equations (1) and (2) become r = V(a*+r? + 2a r/cos. φ1 ) (3) r sin. φ = y-. sin. φι. (4) 50. Problem. To find tne distance of two given points from each other. Solution. Let B and Bf (fig. 20) be the two points whose coordinates are respectively r and r' φ·. The triangle BAB' gives BB'2 = r2 -\-ri2 —2 r r* cos. ( — *. The right triangles ΑχΑΑ' and BAP' give AXA' = PP' = AAX . sin. AXAA‘ = a sin. (/3—a), A4/ = -4.4 j . cos. AXAA* = a cos. (β — a!); PP' = PP + PP' = V + λ sin. (/3 — a), -4P' = P'-4' A1 A = x -f- a cos. (/3 — *); -4P2 = (AP*)2 -f- (PP')2, r2 = x2 -f- 2 a x cos. (β — a) + a2 cos.2 (β — *), + 3/2 + 2ay sin. (β — ») + a2 sin.2 (/* — *), x2 4-y2 4-2a[icos. (/b—a.) 4-y·sin. (/3—*)]4“a2i r= \/(^24“^2Η“α2Η“2«[*· cos.(/3—a)4-y sin. (/3—*]); (9) BP· tang. JL1P = —, tang. (

/^=^+^ sin. (=1; sin. γ ΛΡ = 44' -f- 4P = 44' 4- 4,P' = 44' + 4, R + RP1, sin. (γ— u) +y, sin. (γ — ft) or a; =z a -f- sin. γ ; (25) BP = PP' + P B = Λ ,4' -f- P'B' + P'P, , . sin. α + y. sin. ft or « = 54--ί--------(26) ^ sin. γ v ' and (25) and (26) are the required values of x and y. 71. Corollary. have If the original axes are rectangular, we -γ = 90°, sin. γ = 1, and formulas (25) and (26) become X=z «4-®! cos. «4-y, COS. /3, (27) y — 6 4-*! sin. sin. ft. (28) 72. Corollary. If the new axes are rectangular, we have ft = 90° 4- a, sin. ft — cos. a, sin. (y — ft)= sin. (γ — a — 90°) = — cos. (γ — «); and formulas (25} and (26) become , z, sin. (y —*) —y, cos. (y —«) y = a+----------5577.-------’ (29) x% sin. tt 4- y cos. * *=*+-——· (3°) 4*42 ANALYTICAL GEOMETRY. [β. I. CH. III. Point in space. Rectangular axes. 73. Problem. To determine the posit ion of points in space. Solution. The most natural method of determining the position of a point in space is to determine the position of some plane passing through the point, and then to determine the position of the point in this plane. For this purpose some fixed axis AC (fig. 28) is assumed, and some fixed plane CAD passing through this axis. The plane CAE, which passes through the point B and the axis AC, is deter- mined by the angle EAD, which it makes with the fixed plane CAD. The position of the point B in the plane CAB is determined by the radius vector AC and the angle BAC, which this radius makes with the axis. The same method may be adopted for.any other points B', B", &,c., which are not given on the figure, as they would only render it con- fused. We may, then, denote these radii vectores of the points B, B', &c. by r, r', &,c. ; the angles which these radii make with the axis AC by φ, φ' &c. ; and the angles which the planes in which they are contained make with the fixed plane by 0, 0', &c. 74. A system of rectangidar coordinates in space has been adopted similar to those in a plane, and possessing the same practical advantage of simplicity. For this purpose three planes XA Y, YAZ, andX4Z (fig. 29) are drawn perpendicular to each other, and the rectan- gular coordinates of a point are its distances from these planes. Thus, if the point B is taken, and the perpendicu- lars BP, BQ, and BR are drawn perpendicular to the given planes, these distances are the rectangular coordinates of B. If these coordinates are given, the point B is determined, by taking AL = BR AM=zBQ,nndAN=:BP,§76.] POSITION. 43 Projection of a point. and drawing planes through the points L, M and N parallel respectively to the given planes, and their intersection B is the required point. The intersections AX, iY3 and AZ of the three given planes are the axes. If the coordinates of the point B are x — AL = BR z=z NQ — MP, y — AM =z BQ = LP = NR, z — AN — BP — MR — LQ, the axis AX is called the axis of x, AY is called the axis of y, and AZ the axis of z ; the plane XA Yis called the plane of x y, the plane XAZ is called the plane of x z, and the plane YAZ is called the plane of yz. The coordinates of B, B", &-c. are, in the same way, xl — AL', y' = B Q', z' = L Q' and x" = AL", y" = B"Q", z"= L Q!', &c. 75. The foot of the perpendicular let fall from a point upon a plane is called the projection of the point vpon the plane. Thus the projection of B upon the plane of x y is P, the coordinates of which are x and y ; the projection upon the plane of x z is Q, the coordinates of which are x and %; the projection upon the plane y z is R, the coordinates of which are y and z. 76. Problem. To transform from rectangular coor- dinates to the polar coordinates of art 73, the origins being the same, the polar axis being the axis of x, and the fixed plane the plane of x y.44 ANALYTICAL GEOMETRY. [b. I. CH. III. Polar coordinates transformed to rectangular. Solution. Let AX, AY, AZ (fig. 30) be the rectangular axes. Let XAD be a plane passing through the point B. The values of AL = x, BQ = y, and LQ = z, are to be found in terms of AB = r, BAX = φ, and BAY— 0. Join BL. In the right triangles ABL and BLQ, we have the angle LBQ z=z 0, because the sides LB and BQ are parallel to AD and AY; we also have a; = AL =. AB cos. BAL = r cos. φ, (31) BL = AB sin. BAL = r sin. φ; y = BQ = BL cos. LBQ = r sin. φ cos. 0, (32) z — LQ — BL sin. LBQ z=z r sin. φ sin. 0. (33) 77. Problem. To transform from the polar coordi- nates of art. 73, to rectangular coordinates, the origins being the same, the polar axis being the axis of x, and the fixed plane the plane of x y. Using the figure and notation of the preceding article, the values of r, φ, and 0, are to be found in terms of x, y, and z. They may be immediately found from equations (31), (32), and (33). The sum of the squares of these equations is -J_ ^2 — r2 (cos.2 φ -f- sin.2 φ cos.2 0 -f- sin2 φ sin.2 0) = r2 [cos 2 φ -j- sin.2 φ (cos.2 0 -(- sin.2 0)] = r2 (cos.2 φ -f- sin.2 φ) = r2; because 1 = cos.2 0 sin.2 0 = cos.2 φ -(- sin.2 φ.POSITION. 45 §79.J Distance of two points in space. Hence r — s/(oft -1- z-). and (31) gives x x cos. Φ — — — ——------------------- ψ r \Z(x2 + y*+z*) The quotient of (33) divided by (32) is sin. & cos. Θ — tang, 6 = (34) (35) (36) 78. Problem. To find the distance apart of two points. Solution. Let jE?, B (fig. 31) be the points, and P, P' their projections upon the plane of x y. Join PP', and draw BE parallel to PP1. Since P, P' are two points in the plane YAX, we have, by equation (23), pp/2 = (xf—a?)9 + (y'—y)2; and in the right triangle BEB\ BE = BP'—BP = z' — z, PP'2 = PE2 + BE* = PP'2 + PE*, = (*'—*)3 + (y'—y)*+(*'—z)2; BB' = */[(*'— — y)2 + (z'— *)2]. (37) 79. Corollary. If one of the points, as P', is the origin, we have x! — 0, y’ — 0, z' ~ 0, and formula (37) becomes AB = which agrees with equation (34). (39)46 ANALYTICAL GEOMETRY. [b. I. CH. ΙΠ. Projection of a line. 80. The line PPwhich joins the projections of the two extremities jB, B' of the line BB\ is called the projection of the line BB' upon the plane of xy. 81. Corollary. If the angle B'BE, which is the inclina* tion of the line BB‘ to its projection or to the plane of y x, is denoted by the right triangle B'BE gives BE ~ BB' cos. B'BE, or PPf =z BB' cos. λ ; that is, the projection of a straight line upon a plane is equal to the product of the line multiplied by the cosine of its inclination to the plane. 82. If planes BPL, B'P'Lare drawn, through the extremities jB, B' of a line BB', and perpendicular to an axis AX’ the part LU of the axis intercepted be- tween these planes is called the projection of the line upon this axis. ξβ. Corollary. Since LL! = AJJ — AL z=z x' — a?, the projection of a line upon an axis is equal to the difference of the corresponding ordinates of its extremi- ties. The projection of the radius vector AB is AL, or the corresponding ordinate of its extremity. 84. Corollary. It follows from equation (37), that the square of a line is equal to the sum of the squares of its projections upon the three rectangular axes.POSITION. 47 *87.] Sum of the squares of (he angles made by a line with the axes. 85. Corollary. If the inclination of the line BB' to the axis AX is denoted by-’γ, we have, by drawing LS parallel to BB’ to meet the plane B'P'L1 in S, LS = BB, SLLf = LLf= LS1 cos. SLL\ LB— BB1 cos. φ; that is, the projection of a straight line upon an axis is equal to the product of the line multiplied by the cosine of its inclination to the axis. 86. Corollary. If ψ is the inclination of the line to the axis of y, and oj its inclination to the axis of z, its projections upon these axes are, respectively, BB1 cos. ψ, and BBf cos. ω; so that, by art. 84 ? BB 2 = BB/2 cos.2 φ -f- BB/3 cos.s ψ BB'2 cos.2 w, or, dividing by BB'2, 1 = cos.2 Φ -{- cos.2 ψ cos.2 o>; that is, the sum of the squares of the cosines of the angles which a line makes with three rectangular axes is equal to unity. 87. A different system of polar coordinates from that of art. 73, is often used upon account of its symmetri- cal character. It consists in determining the direction of the radius vector by the angles which it makes with three rectangular axes.48 ANALYTICAL· GEOMETRY. [b. I. CH. III. Rectangular transformed to oblique coordinates. 88. Corollary. If the angles φ, ψ, and ω denote the angles which the radius vector makes with the axes of x, y, and z, we have, by arts. 83 and 86, x — r cos. φ, y — r cos. ψ, z = r cos. w; (39) cos.2 φ -j— cos.2 ψ cos.2 ω — 1, which will serve to transform- from rectangular to polar coordinates in the system of the preceding article. 89. Oblique coordinates are sometimes used similar to oblique coordinates in a plane; thus, if the axes AX, AY, and AZ (fig. 29) had been oblique to each other, and the other lines drawn parallel to the axes, the point B would be determined by the oblique coordinates AL — x, LQ — y, QB=zz: and, in the same way, for other points B', Bn, &c. 90. Problem. To transform from rectangular to oblique coordinates. Let AX, AY, AZ, (fig. 32) be the rectangular axes, and A1Xl, A1Y1, AlZl, the oblique axes. Let the coordi- nates of the new origin be AA‘ = a, ΑλΑ" =: b, A'A" = c ; and let the inclination to the axes AX, AY, and AZ, of those axes A1X1 be, respectively, a, a/, a"; let those of the axis Ai Yj be β, β', β" ; and those of the axis AfZ x be γ, γ', γ" ; these angles must be subject to the condition of art. 86; that is, COS 2 a —|— cos.2 a' COS.2 a!( z=l 1, pos.2 β cos.2 β1 cos.2 β,( — 1, cos.2 y -|- cos.2 yf -j- cos.2 yn = 1.§91.] POSITION. 49 Oblique transformed to rectangular coordinates. The values of the rectangular coordinates AL — x, BQzzzy, LQ — z, are to be found in terms of AXLX == ? BQX — yi? Lx Q, :=: Let _L' and Q' be the projections of the points L± and Qt upon the axis of x. Since A'L', L'Q', and Q L are the projections, respectively, of ALLn L1Q1, and Q1B, upon the axis of x; and A1Ll, QjH, and L1Q1, are respectively parallel to the axes of a?x, yx, and zx, and, therefore, inclined to the axis of x by the angles a, /3, and γ, we have A'L1 z— A1L1 COS. U=1 x1 cos. a, Q'Z. = Q^B cos. /3 = y x cos. /3, L'Q' = L1Q1 cos. y z=z zx cos. γ; so that AL = ^4.4' + 4'Z/ + Q'L + LQ' gives *=« + xt cos. u -}- yx cos. /* + zx cos. y. (40) In the same way we might find y — h -\-x i cos. a! 4~ y 1 cos. & -\-zx cos. Y7» (41) z = c 4- cos. u" 4- y! cos. /3" 4- zx cos. γ" ; (42) so that equations (40), (41), and (42) are the required equa- tions. 91. Corollary. If the new axes are also rectangular, equations (40), (41), and (42) may still be used, but the angles α, /3, γ, a/, /3', γ', af\ β", and γ", will be subject to cer- tain conditions, which are thus obtained. Let r be the radius vector drawn from A x to B^and let the angles which r makes 550 ANALYTICAL GEOMETRY. [β. I. CH. III. Angle of two lines. with the axes AX, A Y, AZ, A1X1y A1 Y,, A1Zl, be respec- tively !-{- cos. β cos. ψχ + cos. ycos. ωχ Ϊ (43) which expresses the angle φ made by two lines, one of which is inclined to the three axes of x1} ylf z1} by the angles β, y ; and the other by the angles (53) (54) (55)52 ANALYTICAL GEOMETRY. [β. I. CH. IV. Loci. Angles. CHAPTER IV. EQUATIONS OF LOCI. 94. When a geometrical question regarding position leads to a mi tuber of equations less than that of the unknown quantities, it is indeterminate, and usually admits of many solutions; that is, there are usually a series of points which solve it, and this series of points is called the locus of the question, or of the equations to which it leads. 95. The equation of the locus of a geometrical ques- tion is found by referring the positions of its points to coordinates, as in the preceding chapter, and express- ing algebraically the conditions of the question. 96. Scholium. Instead of denoting angles by de- grees, minutes, &c., we shall hereafter denote them by the lengths of the arcs which measure them upon the circumference of a circle whose radius is unity, and shall denote by tt the semicircumference of this circle, which is nearly 3.14159*26. The angle of 90°, or the right angle, is thus denoted by ^ 7Γ, the angle of 180°, or two right angles, by tt, and the angle of 360°, or four right angles, by 2 97. Corollary. The arc which measures an angle 0 in the§98.] EQUATIONS OF LQCI. 53 Circle. circle whose radius is jR is R 0, because similar arcs are proportional to their radii, and 0 is the length if the radius is unity. 98. Examples. 1. Find the equation of the locus of all the points in a plane, which are at a given distance from a given point in that plane. This locus is the circumference of the circle. · Solution. Let the given point A (fig. 33) be assumed as the origin of coordinates, and let R = the given distance. If the polar coordinates of art. 44 are used, we have for each of the required points, as M, r = R; (56) so that this equation is that of the required locus. Corollary 1. Equation (56) is the polar equation of the circle whose radius is i?, and centre at the origin. Corollary 2. Equation (56) may be referred to other polar axes by arts. 48 and 49. Thus for the point Ax, for instance, for which AAX = a = — R equation (3) becomes r = s/ (R2 + r\ — 2 R rx cos. φχ) which substituted in (56) gives, by squaring and reducing, r\ — 2 Rrx cos. ^ = 0; or we may divide by r1? since rx is not generally equal to zero, and the equation is rx = 2 R cos. (57)54 ANAHTTICAL GEOMETRY. [b. 1. CH. IT. Circle. which is the polar equation of a circle whose radius is jR, the origin being upon the circumference, and the line drawn to the centre being the axis. Corollary 3. Equation (56) may, by art. 60, be referred to rectangular coordinates; and equations (11) being sub- stituted in (56), and the result being squared, we have *2 + y% — £2 (58) which is the equation of a circle whose radius is R, referred to rectangidar coordinates, the origin of which is the centre. Corollary 4. Equation (58) may, by art. 66, be referred to any rectangular coordinates. Thus the axes A2 X2, A2 Y2, for which the coordinates of A are A2A' and AA\ so that . a — — A2A,=z — a\ l·-— A A' — — V give x — x2 — a'*y — y2 — K which, substituted in (55), give (x2-a')* + (ya-h')* = R*, (59) which is the equation of a circle, referred to rectangular coordinates, the radius of the circle being R, and the coordinates of the centre a! and b'. Corollary 5. For the point A x we have a’ = R,b'=z 0, so that for this point (59) becomes (*,-*)» +*?=», x\ — 2 R x1-\- R? y'f z=[R* yi = % R Xii or (60)§98.] EQUATIONS OF LOCI. 55 Sphere. which is the equation of a circle, whose radius is R, referred to rectangular coordinates, the origin of v)hich is upon the circumference, and the axis of x is the diameter, 2. Find the equation of the locus of all the points in space, which are at a given distance from a given point. This locus is the surface of the sphere. Solution, Let the given point be assumed as the origin of coordinates, and let R == the given distance. If polar coordinates are used, we have for each of the re- quired points r=zR; (61) which is, therefore, the polar equation of a sphere, whose radius is lt: and centre at the origin. Corollary 1. Equation (61) may be referred to rectan- gular coordinates, by art. 77 ; and if equation (34) is sub- stituted in (61), and the result squared, we have x* + y*+z2=R*; (62) which is the equation of the sphere whose radius is R, referred to rectangular coordinates, the origin of which is the centre. Corollary 2. Equation (62) may, by art. 93, be referred to any rectangular coordinates, and the substitution of equa- tions (53, 54, 55) in (62) gives (A + af + (Vl + ψ + (zl + cf = IP, (63) which is the equation of a sphere referred to rectangu-56 ANALYTICAL GEOMETRY. [B. I. CH. IV. Ellipse. lar coordinates, the radius of the sphere being R) and the coordinates of the centre being —a, —b, and —c. 3. Find the equation of the locus of all the points in a plane, of which the sum of the distances of each point from two given points in that plane is equal to a given line. This locus is called the ellipse, and the given points are called its foci. Solution. Let F and Ff (fig. 34) be the foci, let F be the polar origin, let the line FF' joining the foci be the polar axis, and let 2 c = FF' = distance between the foci, 2 A = the given length ; where the length A is not to be confounded with the point A of the figure. If, then, we put in equation (6) r' = FF' = 2c, we have for the distance MF‘ of each point M from F\ MF' \/(r2 -|- 4 c2 — 4 cr cos. φ); so that FM-J- MF' =■ 2 A = r -f-\/(r2 -(-4 c2 — 4 c r cos. φ) λ/(^2 + 4 c2 — 4 c r cos. φ) = 2 A — r, and squaring and reducing 4 c2 — 4 cr cos.

A, ' A or cos. ψ">—. c A If then we take cos. φ0 = —, φ must be confined to the limits φ0 and — = A + c — c = A = \ CC‘ = AC. A is called the centre of the hyperbola. Corollary 7. If we put J 2 fc=' = -7=t=' + a· — — c — A. Hence CO =FC—FC'=2A; 6*66 ANALYTICAL GEOMETRY. [β. I. CH. IV. Conjugate axis of hyperbola. Eccentricity. (76) and (77) become B2 (78) c cos. φ — A B3 (79) r = —--------------. A c cos. φ If this value of B2 is compared with that of the ellipse of corollary 4, we see that it is, in form, the negative of it. 2 B is called the conjugate axis of the hyperbola, and is laid off upon \he liue BAB1 drawn through the centre perpendicular to the transverse axis, taking and this, substituted in equations (76) and (77), gives AB = AB' = B. Corollary 8. If we put e is called the eccentricity of the hyperbola. We have c = e A, A( 1—e2)__ A(e*— 1) 1 — ecos. φ ecos. φ—1 (80) A (ea— 1) (81) e cos.

/(AE* — ^C2) = V(c2 — 42) = B. Corollary 10. The equation of the hyperbola may be referred to rectangular coordinates by arts. 59 and 60. But since equation (76) differs from the equation (64) of the ellipse only in regard to the value of c, this equation may be referred to the rectangular axes CACf and BAB\ by the very same formulas as in corollary 6 upon the ellipse, and we shall have {A* — cs) «2 + A* y* = A* (A% — c% or, substituting J92, — Β2χ* + Α2 y2= — A*B2, which, divided by —A2 B2, is x2 y3 (84)68 ANALYTICAL GEOMETRY. [β. I. CH. IV, Hyperbola referred to oblique axes. With regard to equation (77), since it may be deduced from equation (76), by changing c into —c, or —c into c, it may be referred to rectangular coordinates by the same process, and the corresponding result may be deduced by changing in that for (76) c into —c. Since, however, = (—c)2, the result is the same in both cases. Equations (84) and (S5) are, then, the equations of both branches of an hyperbola referred to rectangular coordinates, the centre of the hyperbola being the or igin, the transverse axis being the axis of x, and the conju- gate axis being the axis of y. Corollary 11. If we wished to find the point where the curve meets the axis of y, we should have for these points £ — 0, so that the corresponding value of y would be y = v(-^-) = V- B* = ± Β V-h which is imaginary, and there are no such points. Corollary 12. The equation of the hyperbola may, by art. 7i, be referred to oblique axes. If the origin remains at A, the result is the same as that of corollary 9 for the ellipse, by changing B2 into —B2. By this change (70) becomes (il2sin.9«—J52cos 2α)α?2 _|_2(^l2sin. asin .β—l^cos. acos./s)^,^ + (A2 sin.2 fi — B2 cos 2β)y2 = — A2 E*. (86)§98.] EQUATIONS OF LOCI. 69 Hyperbola referred lo conjugate diameters. Corollary 13. If and β are so taken, that A2 sin. cc sin. β — B2 cos. * cos. β = 0, (87) B2 or tang, a tang, β —(88) the axes are said to be conjugate to each other; and equa- tion (86) becomes (.A2 sin.2»—B9 cos.2»)xf -f (J.2sin.2/3—i^cos .^)y f =—A2B2. (89) Corollary 14. It may be proved precisely as in corollary 11 for the ellipse, that a line drawn through the centre to meet the curve at both extremities is bisected at the centre, whence it is called a diameter. If such a diameter is as- sumed for the axis of and if we denote it by A\ we have A'=__________________________· \^(B2 cos.2 »—A2 sin.2 »)5 and if we take B‘ = we have __________AB____________ \/ (A2 sin.2 β — B2 cos.2 β)’ A2 sin.2 a, — B2 cos.2 » — — A2 B2 ~~ A'2 ’ A2 sin.2 β — B2 cos.2 β A2 B2 ~B2 ’ which substituted in (89) give, by dividing by y]___ A'2 B 2 ~ ’ —^i2 B2, (90) or — B'2 x\ + A'2 y\ — — A 2 B 2, (91) which are the equations of the hyperbola referred to conjugate diameters.70 ANALYTICAL· GEOMETRY. [B. I. CH. IV. Parabola. Polar equation. 5. Find the equation of the locus of all the points in a plane so situated, that the distance of each of them from a given point is equal to its distance from a given line. This locus is called the parabola, the given point its focus, and the given line its directrix. Solution. Let the given point F (fig. 37) be assumed as the origin of polar coordinates, and let the perpendicular BF to the given line BQ be produced to X, and let FX be the polar axis. Let which is the polar equation of the parabola, the origin being the focus, and the axis the perpendicular from the directrix. Corollary 1. If equations (67) and (92) are compared together, it is evident that (92) is what (67) becomes, when Corollary 2. For the point A where the curve meets the axis we have BF = 2P. 1 — cos. φ 9 (92) Φ = *·, cos. φ= — 1 r=FA = l2P=zP.§98.] EQUATIONS OP LOCI. 71 Parabola referred to rectangular axes. The point A is called the vertex of the parabola, and is just as far from the focus as from the directrix. Corollary 3. The equation of the parabola may be re- ferred to rectangular coordinates by arts. 59 and 60. If we take the vertex A for the origin, we have a —FA —P α = 0, β = 7Γ, sin. (Q — a) — 0, cos. (β — a) r= — 1 ; whence (9) becomes r — s/{x^-\-y2 —2Pa?-f- P2), and the projection of r upon the axis is r cos. φ — χ — P. Now equation (92), freed from fractions, is r — r cos. φ —2 P, in which, if we substitute the preeeding values, we have V(y2 + x2 — 2 P x + P2) — +P = 2P */(b2-\-x2—2Px-\-P2) = P-\-x·, which squared and reduced gives y* = 4Px; (93) which is the equation of the parabola referred to rec- tangular coordinates, the origin being the vertex, and P its distance from the focus. Corollary 4. The equation of the parabola may be re- ferred to oblique axes, by art. 71. If the axis of xt is taken parallel to x, we have72 ANALYTICAL GEOMETRY. [β. I. CH. IV. Parabola referred to oblique axes. a = 0, sin. = 0, cos. « = 1, and (27) and (28) become x = a -(- λ x -|- y 1 cos. /3 y = * + yis'n·/3; which, substituted in (93), give t/2 sin.2 P+(2h sin. £ — 4 P cos. β) y x + h<* — 4Pa = 4Px1. (94) Corollary 5. If the new origin is taken at a point Ax upon the curve, we have, by equation (93), & = 4 Pa, which reduces (94) to ^ y \ sin.2 β -f- (2 l· sin. β —4P cos. β)y1 = 4P x^ (95) Corollary 6. If the inclination β is taken so that 2 l· sin. β — 4 P cos. β = 0, or 2 P tan. p = -y- , (96) (95) becomes y\ sin.2 p — i PXj (97) 2 4P ^sin.2^ (98) And if we put p- p 1— sin.2 p ’ (99) (98) becomes y\ — 4 Pj a: j. (100)§98.] - EQUATIONS OF LOCI. 73 Prolate ellipsoid of revolution. The axes determined by the equation (96), are said to be conjugate to each other, and (100) is the equation of the parabola referred to conjugate axes. 6. To find the equation of the surface described by the revolution of the ellipse about its transverse axis. This surface is called that of the prolate ellipsoid of revolution, which is the included solid. Solution. Let CMC' (fig. 34) be the revolving ellipse. If we use the notation of the 3d problem and solution, and let F be the origin of the polar coordinates in the system of art. 73, and the axis of revolution the polar axis, it is evident that the distance FM of each point from the origin, or any other point of the axis, remains unchanged during the revo- lution of the ellipse. The value of r is then independent of 0, and depends only upon the angle ]= V(y\ + *? )· But MR and AR are the same with coordinates AP and M.P, or x and y of the point M in the plane of the ellipse ; so that, for this point, which, substituted in the equation of the ellipse, give γϋ ,|2 «r2 iHs+i·-1* <104> which is the equation of the oblate ellipsoid of revo- lution referred to its centre as the origin, the axis 2 B of revolution being the axis of xx. 8. To find the equation of the surface formed by the revolution of the hyperbola about either its transverse or its conjugate axis. This surface is that of the hyperboloid of revolution.76 ANALYTICAL GEOMETRY. [b. I. CH. IT. Hyperboloid and paraboloid of revolution. Solution. By reasoning exactly as in the preceding solu- tion, we find #2 t/2 %2 A*~ (105) for the equation of the hyperboloid of revolution re- ferred to its centre as the origin, the transverse axis 2 A being the axis of x and also the axis of revolution, and we find «2 /g2 -^ + 2-2+ϋ = 1 « for the equation of the hyperboloid of revolution re- ferred to its centre as origin, the conjugate axis 2 B being the axis of x, and also the axis of revolution. 9. To find the equation of the surface generated by the revolution of the parabola about its axis. This surface is that of the paraboloid of revolution. Solution. By reasoning exactly as in the preceding solu- tions, we find y2 +22 ==4px (107) for the equation of the paraboloid of revolution re- ferred to its vertex as origin, the axis of revolution being the axis of x. 10. To find the equation of the straight line in a plane. Solution. Let AB (fig. 39) be the line, let any point A§98.] EQUATIONS OP LOCI. 77 Straight line. in it be assumed as the origin of polar coordinates, and let the polar axis be AX, which is inclined to BA by the angle BAX = x. For every point M of this line we have, then, φ = MAX = λ ; so that Φ — λ (108) is the polar equation of a straight line, which passes through the origin, and is inclined to the axis by the angle x. Corollary 1. The equation of the axis is Φ= 0. Corollary 2. The straight line may be referred to rec- tangular axes by art. 60, and if the axis of x is that of AX, (11) gives tang, (109) or y — z tang, χ; (110) which is the equation of the straight line, which passes through the origin, and is inclined to the axis of x by the angle λ. Corollary 3. For the axis of x λ= 0; so that y = 0 is the equation of the axis of x. (Ill)78 ANALYTICAL GEOMETRY. [β. I. CH. IV. Straight line. In like manner x = 0 (112) is the equation of the axis of y. Corollary 4. The straight line may be referred to any oblique coordinates by art. 71. But since the new axes may be situated in any way whatever with regard to the former ones, the generality of the result is not diminished by limit- ing the original position of the line to that of the axis of y, corresponding to equation (112). Thus, if the new origin is at the point A19 we have a = A,A = A,lA^ and (27) becomes cos. cos. β = —a. Now —a is the value of A"Al counted from At, or it is the perpendicular let fall upon the line from the new origin, and if we put p = —a, we have cos. a, + y. cos. β = p ; * (113) which is the equation of a straight line passing at the distance p from the origin, and β being the angles which the perpendicular to the line makes with the axes of Xj and yv Corollary 5. Equation (113) may be applied to the case in which the new axes are rectangular, when “ — β = £*- μ — α — cos. β = sin. » and cos. » = — sin. £,§98.] EQUATIONS OF LOCI. 79 , Straight line. and (113) becomes cos. a,. xx -(- sin. a . yx = p, (114) or — sin. β . xx + cos. β . y x = P (115) cos. β ,y1 — x± sin. β -|- p yx = tan. β -(- p . sec. /3, in which β is the angle made by the line itself with the axis of xx. In (fig. 40) let AB be the line, we have in the right trian- gle AXPB, formed by letting fall the perpendicular ΑτΡ, A1P = p, PAtB = β AXB = AXP sec. PA1B = p sec. /3, and if h ~ AXB — p sec. β y1=z xx tang, β -J- A, (116) which is the equation of a straight line inclined to the axis of x j by the angle β, and cutting the axis of yx at a height h above the origin. Corollary 6. The equation of the straight line may be obtained for any polar coordinates by art. 47, or more simply by art. 61, applied to the axis of y; in this case we have, as in corollary 4, a = —p, and (12) substituted in (112), gives r cos. (φ u) = jt?, (117) which is the polar equation of a straight line passing at the distance p from the origin, the perpendicular upon it being inclined to the axis by the angle —80 ANALYTICAL GEOMETRY. [β. I. CH. IV. Straight line in space. 11. To find the equation of a straight line in space. Solution. If a point in the line is assumed as the origin, and such rectangular axes of z, z, that the straight line makes with them the angles the polar equations of the line in the system of art. 87, are -0 l[ <- II ε II (118) Corollary 1. It must not be forgotten that entirely independent of each other, but are restriction of art. 86, 2, μ, v are not subject to the cos.2 λ -|- cos.2 μ -j- cos.2 v = 1. (119) Corollary 2. The equations of the axis of x are P = 0, Ψ = ^5τ, « = (120) those of the axis of y are Φ = % π, Ψ = 0, <*> =z i π ; (121) those of the axis of z are Ψ = £ ω = ο. (122) Corollary 3. Equations (39) become, by substituting in them the values of — JR2 cos.2 0) = */[R?— (R— y)2] = */(2Ry — y2); which, substituted in (130), gives x=R«—V(2Ry — ^) Λ _ * + V(2 Ry — y9) ------β-------* which, substituted in (131), gives ,= K-R«.(i±^a|»=j3.), (132) which is the equation of the cycloid, fo// is wo/ 50 cow- venient for use as the combination of the two equations (130) and (131). 14. A line revolves in a plane about a fixed point of that line, to find the equation of the curve described by a moving point in that line, which proceeds from the fixed point at such a rate, that its distance from the fixed point is proportionate to the wth power of the angle made by the revolving line with the fixed line from which it starts. This curve is called a spiral. Solution. Let the fixed point A (fig. 42) be the origin, and the fixed line AB be the polar axis. Let M be the moving84 ANALYTICAL GEOMETRY. [b. I. CH. IV. Spirals. point, which, after the line has revolved completely round once, has arrived at M'. Let R = AM, we have, by condition, r : R = φη : (2 *■)% or γ(2*·)" = Rq>n, (133) for the equation of the spiral\ Corollary 1. If η = 1 and R = 1, equation (133) becomes 2 7r r = φ, (134) which is the equation of the spiral of Conon or of Archimedes. Corollary 2. If n = — 1, (133) becomes (2 tt)-1 r = R φ~*, or v

and lhis equation expresses that the two lines are parallel. 112. Corollary. If the two lines are perpendicular, we have ϊ— tang. I = oo = i; or the denominator of (149)«*nust be zero ; that is, AA1 + BB1==0; (152) and this equation expresses that the two lines are per- pendicular. 113. Corollary. In case the two lines are parallel, their distance apart must, by art. 109, be M ν(Λ2 + £3Ί+ν(Λ? + ζ??)· 114. Problem. To find the coordinates of the point of the intersection of two straight lines in a plane. Solution. Let the coordinates of the point of intersection be x0 and y0, and let the equations of the line be the same as in the. preceding article. Since the point ot intersection94 ANALYTICAL GEOMETRY. [b. 1. CH. VI. Intersection of two lines. is upon each line, its coordinates must satisfy each of their equations, or we must have A Xq + B yQ -f- M = 0, ■^ixo + ·®ι3^ο+ ^1 = 0 ; from which the values of x() and y0 are found to be _BMl — BXM Xq AB1 —A^B' A,M— AM, Va~ ABl — A1B ’ (153) (154) 115. Corollary. If the equations of the line had been given in the form corresponding to (115) — sin. a . x -|- cos. ou y = p — sin. αχ . x -f- cos. »1 y =2 pt we should have found «* p COS. αχ—px COS. u P COS. u1---Pj COS. a sin. a 1 cos. a—cos. & χ sin. a, sin. —a) » sin. tt —ρΛ sin. a· y»= Bin. (.;-») · (155) (156) 11β. Corollary. The values of x0 and y0 (153) and (154) would be infinite, if their denominators were zero, that is, if we had ABX — Aj B = 0, or by (150) if they were parallel, in which case they would not meet, and there would be no point of intersection. 117. Problem. To find the equation of a straight line, which makes a given angle with a given straight line.§ 120.] LINEAR LOCUS. 95 Line inclined to given line. Solution. Let the given angle be J, and the equation of the given straight line — sin. a . x -f- cos. a, ,y and let that of the required straight line be — sin. cos. u1. y — p1, in which ) + (y — y') = 0, (162) which is the required equation, a being indeterminate, be- cause an infinite number of straight lines can be drawn through the same point in different directions. 121. Corollary. If this straight line is also to pass through another point, the coordinates of which are a?" and y'', we also have this condition corresponding to (160) — sin. . x" -J- cos. a, .yu — p, from which and (160) the values of p and a are to be found. The difference between the last equation and (160), divided by cos. u and by xtl — xis tans-*=!fe!’ which substituted in (162) gives, by transposition, νμ—ν', y-y'=^rz^(x~T') (164)§ 124.] LINEAR LOCUS. 97 Parallel and perpendicular to given line. for the equation of a straight line, which passes through the two points whose coordinates are x,' yf and xn, y 122. Corollary. If the straight line of art. 120 has also to make a given angle with the straight line whose equation is — sin. x -}-eos. u1y — pi, we have, by art. 117, a =!+«!, which substituted in (162) gives, by transposition, y — y' = tang. (1+ (x — x') (165) for the required equation. 123. Corollary. If the two lines of the preceding article are to be parallel, we have 1= o, and (165) becomes y — y'~ tang. *1 (x — a?'). (166) 124. Corollary. If they are to be perpendicular, we have I-i”, tang.) »±) = — cotan. and (165) becomes y — y' ——cotan. (g— a;'), (167) which is, therefore, the equation of the perpendicular let fall from the point, whose coordinates are x’ y‘ upon the straight line, whose equation is — sin. tt1 a?-|- cos. u1 y = ρΛ. 125. Corollary. The perpendicular, let fall from the 998 ANALYTICAL GEOMETRY. [β. I. CH. VI. Length of perpendicular to line. origin upon the straight line, which is drawn through the point y* parallel to the line — sin. u . x -J- cos. & . y —p is p' = — sin. a xf -f- cos. a . y'. But p is the perpendicular let fall from the origin upon the given line ; and therefore the perpendicular let fall from the point a?7, ?/', upon the given line, is the difference of these two perpendiculars, or it is pQ — p — p1 = p -|- sin. ec . xf — cos. u. y7. (168) 1*26. Corollary. The perpendicular upon the line, which is inclined to the axis of x by the angle *, is itself inclined by the same angle to the axis of y, and by the angle ^ ^ -f- a to the axis of x. The projections of the perpendicular let fall from the point yl to this line upon the axes of x and y are then p0 cos. (J 7Γ -|- a) — — p0 sin. * and p0 cos. a, and the coordinates of the foot of the perpendicular are, by (168), x0 — %* — ρ^ sin. u z= xf — p sin. u — xf sin.2 a -(- y1 sin. a cos. a = x7 cos.2 a ~|- y* sin. u . cos. a — p sin. a, (169) and y0 =!/' + Pocos· * nn y* -|~ p COS. a -(- x' sin. a . cos. a — y' COS.2 u, —z a?7 sin. u. cos. a -j- y* sin.2 » -}- p cos. a. (170) 127. Problem. To reduce the general equation of the first degree in space to its simplest form. Solution. Let the general formulas (40, 41, 42) for transformation from one equation of rectanglar coordinatesLINEAR LOCUS. 99 § 128.] Linear locus in space. to another in space be substituted in the general equation (143) Ax + By + Cz+M= 0, the result is (A COS. co + B COS. CO1 + c COS. co") Xx + (A COS. β + B COS. βι + C cos. β") y x + (A cos. γ + B cos. y' + c cos. γ") %x + Aa + Bh+ Cc + M— 0, (171) in which α, /3, γ, /3', y\ β", y" are subject to the six conditions (44-49). Let now the position of the new origin be assumed at such a point, that its coordinates a, ft, c satisfy the equation Aa + Bb+Cc + M—0, (172) and let the angles β and γ be subject to the two conditions A cos. β + B COS. β1 + C. cos.. r = 0 (173) A cos. γ + B cos. yf + c . cos. yN = 0. (174) By this means equation (171), divided by A cos. co + B COS. CO1 + C COS. co'\ is reduced to = 0. (175) 128. Corollary. Let A cos. co -|- B cos. co1 -j- C cos co'f ·=. L, 0?6) and if (176) is multiplied by cos. (173) by cos. /3, (174) by cos. γ, the sum of the products, reduced by means of equations (47, 50, 51), is A ~ L cos. co. (177)100 ANALYTICAL· GEOMETRY. [β. I. CH. VI. Equation of plane. In the same way we find B~L cos.*', (178) C — L cos. (179) The sum of the squares of (177, 178, 179) is, by art. 86, A* + j5-2+ C2 = L* (180) L = s/(A* + &+ C2), (181) whence cos. a z= — L· A L B COS. W I= „ c COS. cc" — -Τ- Α V(^2 + £2+C2) B V(^a +£* + c2) c V(^2+-B2+ C2)* (182) (183) (184) 129. Corollary. Since is iAe equation of the plane yxz,, ^Ae focws of the general equation (143) ο/’ the first degree in space is a plane, the perpendicular to which is inclined to the axes by angles, which are determined by equations (182-184). 130. Corollary. Since the intersection of two planes is a straight line, the locus of two equations of the first degree is a straight line. 131. Corollary. If the equation (143) is divided by L, and the values (182-184) substituted in the result, it becomes COS. * X -f- COS. a' y -(- cos. dJ' % M L’§ 133.] LINEAR LOCUS. 101 Angle of planes. which, compared with (128), leads to the conclusion that is the length of the perpendicular let fall upon the plane from the origin. 132. Problem. To find the angle of two planes. Solution. Let their equations be Ax + By+Cz + M = 0 Ax% -f- Bxy -f- Cxz -j- Mt = 0, and let u p γ, a1 βχ γ1 be the angles which the perpendic- ulars to them make with the axes of #, y, z ; and let I be the angle of the planes. The angle I is also that made by the perpendiculars to the planes, so that, by (43), COS. I — COS. a COS. « χ + COS. β COS. β j + cos. γ cos. γ15 (185) and by equations (181-184) 133. Corollary. If the planes are parallel, their per- pendiculars are parallel, and make equal angles with the axes, so that M (186) * = β = βί, r = rn or A_A± Β__Βλ C _ C_x L~ Lt' L-'Lf L~LX (187)102 ANALYTICAL GEOMETRY. [β. I. CH. VI. Perpendicular planes. A _ £ _C__ L 01 A,~ and their distance apart in this case is Mj________________________ Af Τ' (188) (189) 134. Corollary. If the planes are perpendicular, we have I = 90°, cos. 7=0, and (185) and (186) give COS. O'cos. O-J + cos. β cos. β t + cos. y cos. y x — 0 (190) AAt+BB^CC^ 0. (191) 135. Corollary. Since sin. i=V( i — cos.2 7), we have, by (186), sin.2 — I — ____(ΑΑ,+ΒΒ, + CC.f (A* + W-\-C2) {A* + Bl + C\) (Az+m+VKAt+Bt+Cy—iAA^+BB^CCLY (A*+Bs+C>){Ai+B*+C2) ( A*B\—2AA1 BB^AzBt+AtCl—ZAA,^ CC, \ I -\-A* (P+mez—2BB1CC1+B\ C2 > {Α*+Β*-\-02) pf-fT^a-jlcf)$ 138.] LINEAR LOCI. 103 Angle of line and plane. We also have, by (181-184), sin2 r (ABl-AiB)*+(AC1-AlC)*+lBC1-BlCF ■ — : L^L\ " = (eos. u cos. /3j — cos. β cos. »1)2 -(- (cos. a cos. y x — cos. γ cos. u χ )2 + (cos. β cos. yx — cos. y cos. βχ )*· (1»3) 136. Problem. To firld the angle which a line makes with a plane. Solution. If *, /3, γ are the angles which the line makes with the axes, and those which the perpendicular to the plane makes, and I the angle made by the given line with the plane, the angle which the line makes with the per- pendicular to the plane will be the complement of I, and we shall have sin. I = cos. u cos. a t -|- cos. β cos. β 1 + cos. γ cos. yt (194) cos,21 = (cos. a cos. β J -cos. β COS. cc x )2 + (cos. a cos. y x — cos. γ cos. a. χ )2 + (cos. β cos. y χ — cos. y cos. β 3 )2. (195) 137. Corollary. If the line is parallel to the plane, we have sin. 1=0 = cos. * cos. * x -f- cos./3 cos./3 χ -f- cos. γ cos. yl. (196) 138. Corollary. If the line is perpendicular to the plane, we have * = *!» /3 = /3n y = Vi.104 ANALYTICAL GEOMETRY. [B. I. CH. VI. Perpendicular lo plane. 139. Problejn. To find the equation of a plane, which passes through a given point. Solution. Let /3, y be the angles which the perpen- dicular to the plane makes with the axes, and let x', y\ z‘ be the coordinates of the point, and p the perpendicular let fall upon the plane from the origin. The equation of the plane is cos. ou . x + cos. β : y -f- cos. y . z = p, and since the point is in this plane, its coordinates satisfy the equations of the plane, and we have cos. a . x1 cos. β . y' -|- cos. γ . z' = p ; and if the value of p thus obtained is substituted in the equation of the plane, it gives cos. » (# — x') -}-COS. β (y — y-|- cos. y {z—%’) = 0, (197) in which <*, /3, y are arbitrary. 140. Corollary. The distance of this plane from another plane parallel to it, and which passes at the distance px from the origin, is p —p j = cos. a,. xf + cos. β . y1 -j- cos. y. z*—pXJ (198) which is therefore the length of the perpendicular let fall from the point x\ yl, z', upon the plane, whose equation is cos. & . x -|- cos. β . y -|- cos. y . z =px. 141. Examples involving Linear Loci. 1. To find the locus of all the points so situated in a plane, that m times the distance of either of them from a given line,k HI.] LINEAR LOCUS. 105 Examples of linear loci. added to n times its distance from another given line, is equal to a given length. Solution. Let the first given line be the axis of #, and let the intersection of the two lines be the origin, and cc the angle which these lines make with each other. Then, if a?, y are the coordinates of one of the points of the locus, we have y = the distance from the first line, and if pQ is the distance from the second line, we have, by (168), p0 = x sin. cc — y cos. cc. If, then, l is the given length, l = m y + n p0 l =z m y n x sin. » — n y cos. cc — n . sin. u . x (n cos. cc — m) y = — Z; so that the required locus is, by art. 108, a straight line, inclined to the first given line by the angle /3, such that tang, β = n sin. cc n cos. cc — ίϊι and which, by art. 109, passes at a distance from the inter- section of the two lines equal to _______________l____________________________l____________ ^/[w2sin.2a-J-(wcos. cc—m)2] \/(n2-[”^2—2 mw cos.«)' Scholium. The line thus obtained satisfies, throughout its whole length, the algebraical conditions of the problem, but not the intended conditions. For at those points, where the value of y or that of p0 is negative, Z is no longer the absolute sum of m y, and n p0 but their difference.106 ANALYTICAL· GEOMETRY. [B. I. CH. VI. Examples of linear loci. Corollary. When m = n we have cos. a—1 —2 sin.2 A a cotan. β = —:-------= —r----------:— sin. a, 2 sin. £ * cos. £ * ,s = 90° + J *, and the distance from the point of intersection becomes _________l____________ l n */(2 — 2 cos. u) 2 n sin. £ a,' 2. To find the locus of all the points so situated in a plane, that the difference of the squares of the distances of either of them from two given points in that plane is equal to a given surface. Ans. If 2 λ nr the distance of the two given points apart, and if the given surface is a parallelogram, whose base is a and altitude δ, the required locus is a straight line, drawn perpendicular to the line joining the given points, and at a distance equal to £ b from the middle of this line. 3. To find the locus of all the points, from either of which if perpendiculars are let fall upon given planes, and if the first of these perpendiculars is multiplied by m, the second by Wj, the third by w2, &c., the sum of the products is a given length Z, Ans. If cos. & x + cos. β y -f- cos. γ z = p cos. x + cos. β 1 y + cos. yx % — &c., are the given planes, the required locus is a plane, whose equation is (m cos. α-fwij cos. cos. /3-f-mA cos. βj+&c.)y -|-(m cos. cos. yL -rf-&c.) % = -\-ηι1ρί -)-&c. tang. A *§ Ul.] LINEAR LOCTJS. 107 Examples of linear loci. or if the letter S. is used to denote the sum of all quantities of the same kind, so that S. m = m -f- m1 + &c. the equation of this plane may be written S. m cos. &. x -(- S.m cos. β. y+s . m cos. y — S.mp> Scholium. This result is subject to limitations, precisely similar to those of example 1. 4. To find the locus of all the points, whose distances from several given points is such that if the square of the distance of either of them from the first given point is mul- tiplied by «ip that of its distance from the second given point by wi2, &e., the sum of the products is a given surface V. The quantities »ιχ, m2, &,c. are some of them to be negative, and subject to the limitation that their sum is zero. Ans. If Zj, yx, %1 is the first given point, a?2, y2, z2 the second one, &c., and if S is used as in the preceding exam- ple, we have 5 . mx = 0, and the required locus is the plane whose equation is 2 x S . m v xx ,-\-2y S.m iyl.-\-2zS.mlzl = S. »,(*; + *?)·—V.108 ANALYTICAL GEOMETRY. [β. I. CH. VII. Reduction of the equation of the second degree in a plane. CHAPTER VII. EQUATION OF THE SECOND DEGREE. 142. The general form of the equation of the second degree in a plane is A x2-\-B xy -j- Cy2D xEy M =0, (199) and that of the equation in space is A x2 -j- B x y + Cy2-f- Dxz-\-Eyz-{-Fz% + Hx + Iy K z M — 0. (200) 143. Problem. To reduce the general equation of the second degree in a plane to its simplest form. Solution. I. By substituting in (199) equations (18) and (19) for transformation from one system of rectangular co- ordinates to another, the origin being the same ; representing the coefficients of x\, y\9 zt, and y1 by A19 B19 Dl9 and E1; and taking u of such a value that the coefficient of xxyx may be zero; (199) becomes Αχ x\ + + Εχ Vi + M = 0. in which we have Ax — A cos.2 «< -f- B sin. u cos. a, -|- C sin.2 Bx = A sin.2 a — B sin. a, cos. a C cos.2 a D1 = Ό cos. a -J- E sin. u Ex — — D sin. a, E cos. (201) (202) (203) (204) (205)§ 144] QUADRATIC LOCUS. 109 Reduction of quadratic equation, and a satisfies the equation 2 (C — A) sin. a cos. a -}- B (cos.2 u — sin 2 a) = 0. (206) II. If, now, we substitute the formulas (20) for transposing the origin in (201); using x2 and y2 for the new coor- dinates ; take the coordinates a and b of the new origin of such values* that the coefficients of x2 and y2 may be zero; and denote the sum of the terms which do not contain %2 or y2 by Mt ; (201) becomes A1x% + B1yl + M1=Q, (207) in which Mx =A1a* + B1 b* +Dla + El b -f M, (208) and a and b satisfy the equations 2 A i cl D j zz 0 (209) 2BX b + Ex =0. (210) The form (207), to which the given equation duced, is its simplest form. is thus re- 144. Corollary. If we take L, L' such that L = 2 A cos. sin, a 0211) L' = 2 C sin. a-\- B cos. a, (212) these values substituted in (206), and the (202) give double of 2 A j zz L cos. cc -j- Lf sin. u (213) 0 = V cos. a — L sin. «. (214) The product of (213) by cos. a, diminished by that of (214) by sin. <*, reduced by means of the equation sin.2 a, —|— cos 2 u = 1 10 (215)110 ANALYTICAL GEOMETRY. [b. I. CH. VII. Reduction of quadratic equation. is, by (211), 2 At cos.» = L = 2 A cos. -B sin. a, (216) or 2(^!—.4) cos. u — 2? sin. a = 0. (217) The product of (213) by sin. u added to that of (214) by cos. u is, by (215) and (212), 2 At sin. u — Ώ = 2 C sin. » + B COS. Ur (218) 2 (Ax — C) sin. a — jBcos. a = 0. (219) The product of (217) by 2 {Ax — C) added to that of (219) by B is, when divided by cos. *, 4:(At — A) (Ax — C) — B2 = 0, (220) from which equation the value οΐ Ax may be deter- mined, that is, if we put X instead of A1, At is a root of the quadratic equation 4 (X — A) (X— C) — B2 = 0; (221) the roots of which are X=i(A+C)Ai i\/(B2 + A2—2AC+C2) = i(A+C)±:iA/[B* + (A — C)*]. (222) 145. Corollary. If we take Lx and L\ such that L1 = 2 A sin. u — B cos. u (223) L\=2Ccos. a — B sin. a, (224) we find that by changing A to C, C to A and B to —B, At (202) becomes Bx (203), 206 remains unchanged except in the reversal of its sign, L (211) becomes L\ (224) and L*§ 146.] QUADRATIC LOCUS. Ill Inclination of axis. (212) becomes Lx (223). But, by the same changes, (220) becomes 4(Bx — C) (B, — A) — B2=0, (225) so that Βλ is determined by precisely the same equation with which Ax is determined in (220), and is a root of the equation (221). The sum of the roots of the equation (221) is by (222) = (A+C)9 (226) and the sum of (202) and (203) is reduced by means of the equation sin.2 « -f- cos.2 «= 1, (227) to Al + B1=A + C-9 (228) and, therefore, Ax and are the two roots (222) of the equation (221). 146. Corollary. The value of & may be obtained from the equation (217), which gives . Sin. a tang, a —---------= COS. a 2 (Ax — A) B (232) or it may be obtained directly from (206). If we substitute in (206) sin. (2«) = 2 sin. a cos. «,cos.2 a = cos.2a —sin.2a, (233) it becomes (C — A) sin. 2 a -f- B cos. 2 « = 0 ; (234)112 ANALYTICAL GEOMETRY. [b. I. CH. VII. Case of two lines. whence (235) 147. Scholium. The values of Ax and Bx (222) are always real as well as that of « (235), and those of Dj and Ex (204) and (205), and therefore the transformation from (199) to (201) is always possible. But the equations (209) (210) are impossible if Ax and Bx are both zero, while Dx and Ex are not zero, or if either Ax or Bx is zero, while the corftsponding value Dx or Ex is not zero; so that in these cases the transformation from (201) to (207) is impos- sible. 148. Scholium. The values of Ax and Bx cannot both be zero, for, in this case, the quadratic terms would disap- pear from (201), and (201) could not, then, by art. 100, be a reduced form of a quadratic equation. 149. Scholium. If either Ax or Bx were zero, the cor- responding root of (221) would be zero; that is, this equation would be satisfied by the value and if we take A x for the root which vanishes, we have, 1=0, which reduces it to 4 AC — JB2 0 ; (236) by (232) tang, a, — — 2 A (237) But 1 1 sec. α> λ/( 1 -f- tan·2 a) cos. « = (238)§ 150.] QUADRATIC LOCUS. 113 Case of two lines. whence cos*α = a/(B2 + 4 Αη 2 A sin. u — cos. cc tang, a ~-77-——— V(B^ + < _ DB—2AE 1 ~ V(^ + 4 A*) 5 so that Dl will also vanish, only when DB=2AE; and in this case (201) becomes Bi y\ + Et Vi + M — 0; which gives Vl ~ 2i, so that the required locus is the combination of two lines drawn parallel to the axis of x1 at the distances from it equal to these two values of yunless these values are imaginary or equal, in the former of which cases there is no locus, and in the latter the given equation is the square of the equation of the line. 150. Scholium. If the values of A, B, C satisfy (236), so that one of the roots of (221) is zero, and if this one is taken for Αλ, we have for the other root, by (222), B1 = i(A+ C) + £ when§ 154. J QUADRATIC LOCUS. 115 Ellipse. Point. 4 AC is greater than B2; and they are of opposite signs if 4 AC is less than B2. 152. Corollary. When B2 is less than 4 ACy and, con- sequently, Ax and B1 are of the same sign, we will put that sign being prefixed to Mx, which renders the first mem- bers of these equations positive. If then (207) is divided by ± the quotient is 153. Scholium. If Mx were zero, the equations (253) would be absurd, but in this case equation (207) would be in which both the terms* of the first member have the sign, so that the equation can only be satisfied by the conditions which represents the origin of the axes of x2 and y2. Hence, and by art. 143, the locus of the given equa- tion is, in this case, the point whose coordinates are the values of a and b (209) and (210). 154. Scholium. If Mx were of the same sign with Ax and J5j, the upper sign would be used in equations (253) and (254), the first member of (254) would then be the sum of three positive quantities, and could not be equal to zero. The given equation has) then, no locus, in this case. Az Mx A\' * ~~ B\' A, __ 1 Bt __ 1 (253) (254) Ai xl + Bi yl — 0, (255) x2 — 5, y2 — 0 (256)116 ANALYTICAL GEOMETRY. [b. I. CH. VII. Hyperbola. Two lines. 155. Scholium. When Ml is of the sign opposite to that of Ax and Br, the lower sign must be used in equations (253) and (254), and (254) becomes, by transposition and omitting the numbers below the letters, which are no longer necessary, which is of the same form with the equation (69) of the ellipse. 156. Corollary. When J52 is greater than 4 AC^ and, consequently, Ax and Βλ are of opposite signs, we will put those signs being prefixed to M1, which render the first members of these equations positive. If, then, (207) is divided by d= Mx, the quotient is 157. Scholium. If Mx were zero, the equations (258) could not be used, but in this case equation (207) would be the second member of which is real, because Ax and Bk are of opposite signs. (257) ± Mt ~ A\ ’ ± Μ, B% ’ (258) (259) Ai xl + Bi V% = 0, which, multiplied by A^ gives — — Ai B, y\; or, extracting the root, A1 x2 —■ i \^( 2*i) V2> (260)$ 159.] QUADRATIC LOCUS. 117 Hyperbola. Ellipse. The locus of the given equation is then the combina- tion of the two straight lines represented by the two equations included in (260), each of which passes through the origin of x2 and y2. 158. Scholium. If M1 is not zero, equation (259) may, by omitting the numbers below the letters and transposing the terms, be written in one of the forms x* y B* (261) or il (262) and the second of these equations becomes the same as the first by changing x, y, A, B into y, x, B, A respectively. Equation (261) is of the same form with equation (85) of the hyperbola. 159. Theorem. The equation (257) is necessarily that of an ellipse. Proof. To prove this, it is only necessary to show that each point of its locus, is so situated, that the sum of its distances from two fixed points is always of the same length. By comparing the equation (257) with the solution of exam- ple 2, art. 98, it is apparent that, since all the points of the ellipse satisfy the equation (96,) they are in the required locus ; so that if, conversely, all the points of the required locus are in the ellipse, the two fixed points must be in the axis of a? at a distance c from the origin such that c = ± */(A* - B\ and that the given length must be 2 A. (263)118 ANALYTICAL GEOMETRY. [b. I. CH. VII. Ellipse. Now the distance r of the point a?, y from one of these fixed points is, by (23), but since y2 = B2 we have r = V[(* —c)2 + y2]; J32®2 (264) A2 and c2 = A2 — B2, r = \/(a?2 — 2 c x -J- c2 4“ y2) B2x2 — s/i^x1 — 2 cx — A2 — —J~ ) = ^(5*-®ββ + ^9) / c a? . \ , c a? — ^42 = ±(t-j)=±— (365) Now of the two signs + and —? that must be used which gives the distance r positive. But we have for and Hence c A and x A c — λ/(A2 — B2) ,=v(a>-^). : x A2 or c x — A2 0; so that the lower sign must be used in (265), which gives cx r — A — A ’ (266) ves (267)§ 160.] QUADRATIC LOCUS. 119 Hyperbola. so that for the distance from one of the fixed points we have B2) rt = A- and for the distance from the other (268) (269) whence rx -{- r2 = 2 A; (270) that is, all the points of the required locus belong to the ellipse. 160. Theorem. The equation (261) is necessarily that of an hyperbola. Proof. The proof is the same as in the preceding theo- rem, except that the word difference is to be used for swm, the sign of B2 Is to be changed, and in the value of r (265) the upper sign is to be used, where c and x are both positive or both negative. For, since the values of c and x are C = ± V (42 + B2) andi = ± V ( A* +^r) we have, when c and x are of the same sign, C*z=V(^2 + B2).V(^2+^) c x A2 or c x — A2 0; whence ri=~7 — A. (271) A But if c and x have opposite signs the product c x is nega- tive, so that120 ANALYTICAL· GEOMETRY. [b. I. CH. VII. Parabola. whence r2 — rx = 2 A; (273) that is, all the points of the required locus belong to the hyperbola. 161. Theorem. The equation (251) is necessarily that of a parabola. Proof. Omitting the numbers written below the letters, we have only to show that the distance of each point of the locus from that point of the axis of a?, whose distance from the origin is p, is equal to its distance from that line which is drawn parallel to the axis of y, and at the distance — p from it. Now since the distance of the point a?, y from the axis of y is a?, its distance from the line parallel to it must be and its distance r from the fixed point must be r=z*/[(x— p)2 + jr2] = m/ (a?2 —2px -|-p2 +4px) = s/ (a?2 -f- 2p x = *+*, (274) which is the same as the distance from the line; alt the points of the locus of equation (251) are then upon the same parabola. 162. Theorem. In different ellipses which have the same transverse axis, the ordinates which correspond to the same abscissa are proportional to the conjugate axes. Proof. Let the common transverse axis be 2 A, the dif- ferent conjugate axes 2 B. 2 B&c., and let the ordinates,$ 166.] QUADRATIC LOCUS. 121 Ratio of ordinates in ellipses and circles. which correspond to the same abscissa a?, be y, yx &c. we have A2 y2 = B2 {A2 — x2) A2y2 = B2 (A2 — x2), whence, by division, A2 y2 : A2 y2 = B2 (A2 — x2) : B\ (A2 — x2) or y2 :y\ = B2 : B\, or extracting the square root y : yt = B : Bx = 2 B: 2 Bt. 163. Corollary. Since the ellipse, whose conjugate axis is equal to its transverse axis, is a circle, the ordinate of an ellipse is to the corresponding ordinate of the circle, described upon the transverse axis as a diameter, as the conjugate axis is to the transverse axis. 164. Corollary. In different ellipses which have the same conjugate axis, the abscissas which correspond to the same ordinate are proportional to the transverse axes. 165. Corollary. The abscissa of an ellipse is to the corresponding abscissa of the circle, described upon the conjugate axis as a diameter, as the transverse axis is to the conjugate axis. 166. Corollary. It may be proved in the same way that in different hyperbolas, which have the same transverse axis, the ordinates which correspond to the same abscissa are proportional to the conjugate axes; 11122 ANALYTICAL GEOMETRY. [β. I. CH. VII. Ratio of ordinates in hyperbolas. and that in different hyperbolas, which have the same conjugate axis, the abscissas, which correspond to the same ordinate, are proportional to the transverse axes. 167. Corollary. Understanding, by an equilateral hyperbola, one in which the axes are equal, the ordi- nate of any hyperbola is to the corresponding ordinate of the equilateral hyperbola, described upon its trans- verse axis, as the conjugate axis is to the transverse axis, and the abscissa of the hyperbola is to the corres- ponding abscissa of the equilateral hyperbola, described upon its conjugate axis, as the transverse axis is to the conjugate axis. 168. The term abscissa is often applied, in regard to the ellipse and hyperbola, to denote the distance of the foot of the ordinate from either of the extremities of the transverse axis. Thus the abscissas of the point M (fig. 38) of the ellipse are CPz=zAC — AP—A — x and CP =zAC' + AP=A + x. The abscissas of the point M (fig. 36) of the hyperbola are CP = AP — AC=x — A and C P = AP-\- AC' = x + A. 169. Theorem. The squares of the ordinates in an ellipse or hyperbola are proportional to the products o f the corresponding abscissas, the term abscissa being used in the sense of the preceding article.§ 170.] QUADRATIC LOCUS. 123 Ratio of ordinates in ellipse and hyperbola. Proof. I. The product of the abscissas for the point a?, y of the ellipse is, by the preceding article, (A -f- %) (A — x) =: A2 — x2; and this product for the point x', y' is A2 — x'2. But, by equation (68), we have A2 y2 = A2 B2 — B2 x2 A2 y'2 — A2 B2 — B2 xf2 ; whence A2 y2 : A* ya = A2 B2 — B2x2 : A2 B2 — B2^2, or, reducing to lower terms, y2 : y'2 = A2 — x2 : A2 — x which is the proposition to be proved. II. In the same way, for the hyperbola, the products of the abscissas for the points a;, y, and a?;, yl are x2 — A2 and x'2 — A2. But, by equation (84), A2 y2 = B2 x2 — A2 B2 A2yi2 = B*x,2 — A2B*, whence y2 : y‘2 = x2 — A2 : x'2 — A2. 170. Theorem. The squares of the ordinates in a parabola are proportional to the corresponding abscis- sas.124 ANALYTICAL· GEOMETRY. [b. I. CH. VII. Angle inscribed in a semiellipse. Proof. For the point x, y we have by (93) ^=z4Pr, and for xf, y‘ y1- — 4 P x\ whence j/2:y/2z=4Pi:4Pi/ziai:z/, which is the proposition to be proved. 171. Problem. To find the magnitude of an angle, which is inscribed in a semiellipse. Solution. Let CMC1 (fig. 45) be the semiellipse, whose semiaxes are A and B, let I be the required angle CMC1, λ the angle MCX, β the angle MC'X, and x\ y' the coordi- nates of the point M. Because the line MC passes through the point x\ yf and the point C, whose coordinates are and, because the line MC' passes through the point xl, yf and the point C', whose coordinates are y = 0, x = AC = A we have, by art. 121, (275) y — 0, x = — A we have (276) hence * tang. I = tang (» — fi) = γ tang a — tang, β 1 -j- tang- » tang, β 2 Ay' — —Az + y®'§ 174] QUADRATIC LOCUS. 125 Supplementary chords. But, by (68), and, therefore, tang. I = · 2 AB* y> 2ABa \A* — B*) y* ~ (A* — £2) y‘' ^2?7^ 172. Corollary. The product of (275) and (276) gives by the substitution of **=%(**-**) IP tang, a . tang, β = — (278) which is the condition that must be satisfied by the two angles * and /3, in order that two lines CM and CM, drawn from the two points C and C, may meet upon the curve. Two such lines are called supplementary chords; so that (*278) is the condition which expresses that two chords are supplementary. 173. Corollary. If equation (278) is compared with (72), it is found to be identical with it; so that the condition that two chords are supplementary is identical with the condition that two diameters are conjugate. If then a given chords as CM, is parallel to a given diameter BXAB\, the chord C'M, supplementary to CM, is parallel to the diameter CXACx, conjugate to BxABlv 174. Problem. To draw a diameter, which is conju- gate to a given diameter. M*t26 ANALYTICAL GEOMETRY. [b. I. CH. VII. Chords bisected by diameter. Solution. Let BXAB'^ (fig. 45) be the given diameter. Through C draw the chord CM parallel to B1AB> λ ; join C'M, and the diameter C1AC\, which is drawn parallel to CM, is, by the preceding article, the required diameter. * 175. Problem. To find the magnitude of the angle formed by two chords drawn from a point of the hyper- bola to the extremities of its transverse axis, which are called supplementary chords. Solution. The solution is the same as that of art. 177, except in regard to the sign of B2, which being changed gives for the required angle I 2 AB2 taDg'1 ~ (4* -f- W) y' ‘ (279) 176. Corollary. The corollaries of arts. 172, 173, and the construction of art. 174, may then be applied to the hyperbola, and equation (88) is the condition that two chords are supplementary. 177. Theorem. The chords which are drawn parallel to the conjugate of any diameter of an ellipse or hyper- bola are bisected by it. Proof. For each value of x there are two equal values of y, one positive the other negative, which are, in the ellipse, y = ± — »a), and, in the hyperbola, y — άι^ */(x~ — A*) ;§ 180.] QUADRATIC LOCUS. 127 Parameter. so that if for the value of x equal to AP (fig. 46), the line MPMΊ is drawn parallel to the conjugate diameter, and if PM, PM' are taken each equal to the absolute value of y, the points Μ, M! are upon the curve, and the chord MM', which joins these points, is bisected at P. 178. Corollary. The same proposition and proof may be applied to the parabola, using the word axis instead of diameter. 179. Corollary. The chords drawn perpendicular to either axis of an ellipse or hyperbola, or to the trans- verse axis of the parabola, are bisected by this axis. 180. Problem. To find the length of the chord drawn through the focus of the ellipse, the hyperbola or the parabola, perpendicular to the transverse axis; this chord is called the parameter of the curve. Solution. I. Represent the parameter of the ellipse by 4 p\ and its half or the ordinate is 2 j?, the corresponding abscissa being, by example 3, art. 98, c = aS(A2—B2) or c2 = A2 — B2. Hence the equation of the ellipse gives 2Ap = B*/{A2 — c2) = B2 4 p — 2B* ~T' II. In the same way in the hyperbola we should find the same values of 2 p and 4jv.128 1 ANALYTICAL GEOMETRY. [b. I. CH. VII. Tangent. III. In the parabola whose equation is y2 = 4 p x the abscissa for the parameter is p; at which point y2 ~ 4 J52, y ~2p parameter = 4p. 181. Corollary. In the ellipse or hyperbola, we have A: B= B:2p or 2 A :2 B— 2 B :4p·, so that the parameter is a third proportional to the transverse and conjugate axes. 182. Theorem. The line drawn through either ex- tremity of a diameter of the ellipse or hyperbola, parallel to the conjugate diameter, is a tangent to the curve. Proof For the two values of y are equal to zero at the point, so that neither of these lines has only one point in common with the curve. 183. Problem. To draw a tangent to the ellipse or hyperbola at a give?i point of the curve. Solution. Join the given point C1 to the centre A. Through the extremity C of the transverse axis draw the chord CM'parallel to AC1. Join C'M', and the line drawn through Cx parallel to C'M' is, by arts. 173 and 182, the required tangent. 184. Scholium. The drawing of tangents to these curves will be more fully discussed in a subsequent chapter.§ 185.] QUADRATIC LOCUS. 129 Reduction of quadratic equation in space. 185. Problem. To reduce the general equation of the second degree in space to its simplest form. Solution. I. Substitute the equations (40, 41, 42) in (200), making a = 0, bz=z 0, c = 0 ; so that the direction of the axes may be changed without changing the origin. If we represent the coefficients of y?, zj, y1? zl by Ax — A cos.2 ct + B COS. CL cos. CL1 + C COS.2 cl' + B COS. CL COS. cl" + E COS. CL7 COS. CLU -|- F COS.2 cl" Bi = A cos.2 β-\-Β cos. β cos. β' -j- C cos.2 β7 + -D cos. β cos. β77 + E cos. β1 cos. β" -f- jPcos.2 β" Ci = A cos.2 y -|- B cos. y cos. y' -f- C cos.2 y' + D cos. y cos. y" + E cos. y' cos. r + F cos.2 y" Ηγίζζ H cos. a -j- I COS. a! -[- K COS. cl" Ii = H cos. /3 -I cos. /37 - K cos. β" Κλ = Η cos. y -}-1 cos. y' -|- K cos. γ77, and take *, /3, γ, βγ', α", /37', γ> t0 reduce the coeffi- cients of χχ yu XX yx Zx to zero ; that is, to satisfy the equations 0 = 2 A cos. a cos. /3-f-2 C COS. a7 cos. /37-}-2 F cos. cl" cos. β" +#(COS. a COS. /B'-f-COS. cl1 COS. /3)-j_2)(COS. cl COS. β" -j-C0S.a//C0S./3)-|-JE(c0S. cl* COS. /3/7-|“COS. cl" COS./37) (280) o = 2 A cos. a COS. y-j-2 C cos. cl> cos. y7-f-2 Fcos. a77 cos. y" +JB(cos. a cos. y7-|-cos. a7 cos. y)+P(cos. a cos. y" -f-cos.a/7cos.y)-j-£(cos. a7cos. y/7-J-cos. «/7 cos. γ7) (281)130 ANALYTICAL GEOMETRY. [b. I. CH. VII. Quadratic equation in space. 0 = 2A cos. β cos. y-j-2C cos. β' cos. y‘-\-2F cos. βη cos. γ" -j- B (cos. β cos. γ cos· P'cos· 7)4” & (cos* P cos· y" 4-cos. ^"cos.y^.Eicos. β' cos. y" 4~cos. 0"cos. γ'), (282) which, combined with the six equations (44-49), complete- ly determine the values of these quantities, equation (200) becomes Ai%i-\-B(283) II. Substitute the equations (53-55) for changing the origin to the axes a?2, y2, z2, and (283) becomes Ai x\ Bl y\ + G\ z% + (2 A1 a+ Hj x2 + (2 Bl b + I,) y2 + (2 Cj c + K,) z2+M1 = 0, (284) M^A^+BJz+C^+H^+IJ+KiC+M, and in which if a, δ, c are taken to satisfy the equations 2A1a+H1 =0 (285) 2 Bx b + I1 = 0 (286) 2C1c+i1z:0 (287) (284) becomes Ax x% + Bx y\ -j- Cj %\ = Mt 0. (288) 186. Corollary. Tf we take L z=z 2 A cos. a, + B cos. 4« D cos. a11 (289) JJ = 2 C cos. a1 + B COS. a + E COS. «" (290) Ln—- 2 F cos. a" 4- D cos. a E cos. (291) These values may be substituted nn (280), (281), and the double of the value of A x, and they give$ 186.] QUADRATIC LOCUS. 131 Quadratic equation in space. 2 A ± = L COS. a -(- JJ cos. -|- L" COS. of* (292) 0 — L cos. β -j- It cos. βι -j- L1' cos. β" (293) 0 — L cos. γ + Lf cos. y* +. L" cos. yt!, (294) If (292) is multiplied by cos. a, (293) by cos. β, and (294) by cos. γ, the coefficient of L in the sum of the products is by (47) unity, while those of U and LH are by (50) and (51) zeroj so that this sum is by (289) 2 A x cos. a = Lz=2i cos. a -f- B cos. -f-D cos. <*", (295) or 2 {A j — A) cos. u — B cos’. — B cos. a·" = 0. (296) If (292) is multiplied by cos. (293) by cos. β\ and (294) by cos. γ', the sum of the products is, by (48,50, 52, 290) , 2 A J COS. aJ zzr L1 nz 2 C cos. cos.« -(- 22 cos. a/;, \ 297) or 2 (A j — C) cos. — 2? cos. * — 22 cos. a," — 0. (298) 4f (292) is multiplied by cos. a", (293) by cos. /3·", and (294) by cos. γ", the sum of the products is, by (49, 51, 52, 291) , 2 A jCos. L"z=l 2 Fcos.a "+D cos. a- 22 cos.a', (299) or 2(^—2?) COS. a"----D COS. C*-E COS. C(j — 0. (300) If (296) is multiplied by 4 (ij — C) (J^ —F) — 222, (298) by 2 B (Ax — F) + D £, (300) by 2 D (A1 — C) -}- BE, the sum of the products divided by 2 cos. * is 4(Λ,-Λ) (A,-C) (A1-F)-F?(A1-A) — B3(At — F)—D2(Al — C)—BDE = 0, (301) from which the value of Ax may be found.132 ANALYTICAL GEOMETRY. [b. I. CH. VII. Quadratic equation in space. 187. Corollary. Since the value of Bx is obtained from that of A x by changing into /3, p\ /3'', and since by this same change and that of p, p\ p" into γ, γ', γ", and also by that of γ, γ', γ" into a, «·', a", (280) is changed into (282), and (281) into (280) ; it follows that these same changes may be made in the equations from (295) to (301), and (301) will become 4 (Bt-A) (.Bx - C) {B, - F) - & (B, - A) — IP (Bj — F) —Ifi (jBj — C) — BDE = 0, (302) from which Bx may be found. 188. Corollary. Since the value of Cx is obtained from that of HA, by making the same changes as in the preceding article, and since, by these changes (282) is changed into (281), and (280) into (282); it follows that these changes may also be made in the equations obtained by the preced- ing article, (302) will thus become 4 (C1 —A) ω, — C) (C1 — F) — E^(C1—A) — BHC1—F) — iyHC1 — C) — BDE = 0, (303) from which Cx may be found. 189. Corollary. Since the equations for determining A^ jBjjCj differ only in the letters which denote the unknown quantities, and since these equations are of the third degree, it is evident that A^ B1% Gx are the three roots of the equation of the third degree, 4(X— A) (X—C) (X—F) — E* (X—A) -BHX — F)-IP(X—C) — BDE = 0. (304)§192.] QUADRATIC LOCUS. 133 Reduction of quadratic equation. 190. Scholium. Every equation of the third degree has at least one real root, so that one at least of the three quantities Bx, C1 must be real. If we assume this one to be ΑχΛ the corresponding values of cos. cos. a', cos. a", determined by equations (296,298, 300), and the 1st of art. 90, are also real; so that equations (280) and (281) are satisfied without assigning any values to /3, μγ, γ', γ". If (282) is not also satisfied, let its second member be represented by 11,, and equation (2(f0), instead of being reduced to the form (283), will become x\ -{- B, y2j -J- Cx 2f -f- Dj j/j 2, + *! + Ι\ Vx + 2j -j- M— 0. If now the same transformation is effected upon this equa- tion, so as to transform it to the axes of a?2, y2, z2, the equa- tion for determining A2, i?2, C2 would be obtained from (304), by changing A, B, C, D, £, F into At, 0, 0, H,, Cj, (304) thus becomes 4(J[-^1)(Z-51)(X-C1)-D2(^-^i) = 0, (305) the roots of which are and J=J(B1-+C1)±4V[D? + (B1 —Cx)»] (306) which are all real, so that the given equation can always be transformed to the form (283), and all the roots of (3Θ4) will be real. 191. Scholium. If either Ax, 1?,, or Cx is zero, one of the equations (285-287) is impossible, unless the corres- ponding value of JET,, 71? or Kx is zero, and in this case the second transformation of § 185 is impossible. 192. Scholium. The three roots Al^B1^C1 cannot all be 12134 ANALYTICAL GEOMETRY. [b. 1. CH. VII. Cases of quadratic locus in space. zero at the same time; for in this case (283) would be linear, and would not be a reduced form of a quadratic equation. 193. Scholium. If Ax and H1 are both zero, the values of b and c can be taken to satisfy equations (286) and (287), and (283) is then reduced to Bt Vl + Cxz* + Mx = 0. (307) • 194. Scholium. If A1 is zero and Hh is not so, b and c can satisfy equations (286) and (287), and a can be taken to satisfy the equation ^ = 0, so that (283) is then reduced to Bx Vl + Ct + Hx x2 = 0. (308) ♦ 195. Scholium. If Ax and Bx are zero, c can be taken to satisfy equation (287), and if either Hx or Ix is not zero, a or b can be taken to satisfy the equation Mt = 0, so that (283) is then reduced to c. *1 + H1x2 + 11 y2 = 0. (309) But if both Hx and Ix are also zero, (283) becomes Cx z* + M1= 0. (310) 196. Scholium. If the values of Ax, C1? and Mx have all the same sign, (288) is impossible, and there is no locus.§200.] QUADRATIC LOCUS. 135 Cases of quadratic locus iu space. 197. Corollary. If Ax^ BC1 have all the same sign, which is the reverse of M^let J.2, U2, C2 be so taken, that i_—_4i JL A»- Mx'B* L· JL /οι ιί Mt'C%~ MS K ’ and the quotient of (288), divided by —- M.x, is -1 = 0. A% 2 I ft I *2 "r Bl^ 3 Cl (312) 198. Corollary. If two of the quantities A,x B1, C, have the same sign with M\, while the other one, which we will assume to be Ax, has the reverse sign, we will take 1 Ax 1 _BX 1 _ Cx A\~ Mx'B%~ MS C%~ MS (313) and the quotient of (288), divided by — Mt, is *2 y% zl A\ eg (314) 199. Corollary. If of the quantities Ax, B,, C,, one, namely Cx, has the same sign with Mx, while the other two have the reverse sign, we will take _1_ ___ Aj^ 1 __J?L 1 _ Ct Aj~ ~MS B'i~ Mx ’ C\ Mx ’ and the quotient of (288), divided by — M l, is 3 >yl A%^B% ~Ji — l eg, = 0. (315) (316) 200. Corollary. The values of 2 A,,, 2 15 2, 2 C2 are136 ANALYTICAL GEOMETRY. [B. I. CH. VII. Cases of quadratic locus in space. called the axes of the surface in either of the three last articles, so that the three different values of which are found from equation (304), are the semi- axes. 201. Scholium. If Mx is zero, the equations (311), (313), and (315) are impossible, but in this case (288) becomes Α,χΙ+Β, £+0^1 = 0. (317) 202. Scholium. If Ax, Βλ, and Ct have all the same sign, (317) is only satisfied by the values x2 ~ 0, y2 =: 0, %2 — 0, (318) so that the origin of x2) y2, z2 7S 171 this case the re- quired locus. 203. Corollary. If of the three quantities,^, Bn Cn one, as C1? is negative, while the other two are positive, we will take ^and (317) becomes AV Bi Ci~ # (319) (320) 204. The form of a surface is best investigated by examining the character of its curved sections, which§ 205.] QUADRATIC LOCUS. 137 Examples of quadratic loci. are made by different planes. The farther investiga- tion of the surfaces, represented by quadratic equations, will, therefore, be reserved for Chapter IX. 205. Examples involving plane quadratic Loci. I. To find the iociis of all the points in a plane, which are so situated with regard to given points in that plane, that if the square of the distance of each point from the first given point is multiplied by m7, the square of its dis- tance from the second given point by m7/, &c., the sum of the products is equal to a given surface V. Solution. Let the given points be, respectively, a;7, y7; a?'7, y", &c. The distances of the point x, y from these points is given by equation (23), and we have, by the cftiditions of the problem and using S, as in art. 141, . m7 (a: — a:7)2 S. m‘ (y — y7)2 T, or S. m7. a:2 -j- S . m‘. y2 — 2 S+m' xl. x — 2 S .mJ y*. y + S.mf (a;72 + y'2) — V— 0. This equation is already of the form (201), and may be reduced to the form (207) by making __ S. m' xf S ,m* y* a S. m* ’ S ,m' „ _-[(S.m7z7)2+(Sm7.y')2] 1 ~ ■ S. mJ + S.m'(x'2 + y72) — V. 12*138 ANALYTICAL· GEOMETRY. [β. I. CH. VII. Examples of quadratic loci. We have then for the axes, by (253), -^2 — λ/ S. m1 B0 so that the locus is a circle, the coordinates of whose centre are — a and — Z>, and whose radius is A2. Corollary. — M1 and S . mf must be both positive or both negative. 2. To find the locus of all the points in a plane, which are so situated with regard to given lines in the plane, that if the square of the distance of each point from the first given line is multiplied by mi, the square of its distance from the second line by m2, &>c., the sum of the products is equal to a given surface V. Solution. L^t the given lines be respectively sin. α,χ x — cos. cc1 y — —p1 sin. u2 x — cos. <*2 y — —p2, &c. The distances of the point a;, y of the locus from these lines is given by equation (170), and give, by the conditions of the problem, and using S as before, S .m1 (sin. »x . x — cos. ^ which, developed and compared with equations (199-256), give A1 and Bx as the roots of the equation 4 X?—4 S.mx X-{-4 S.mx sin.2 ccx Smxcos.2 ax—(Smx sin.2 ux)%— 0, and to find tan. 2 cc S. m1 sin. 2 α,χ S .m1 cos. 2»/ and the values of a, 5, Mx may be found by equations (208-210).§ 205.] QUADRATIC LOCUS. 139 Examples of quadratic loci. 3. To find the locus of the centres of all the circles which pass through a given point, and are tangent to a given line. Ans. A parabola of which the given point is the focus, and the given line the directrix. 4. To find the locus of the centres of all the circles, which are tangent to two given circles. Ans. When the locus is entirely contained within the given circles, it is an ellipse of which the foci are the two given centres, and the transverse axis is the difference of the two given radii, if both the contacts are internal; but the transverse axis is the sum of the radii if one of the contacts is internal while the other is external. Otherwise, it is an hyperbola, of which the foci are the two given centres, and the transverse axis the difference of the two given radii, if the contacts are both external or both internal, and their sum, if one of the contacts is external and the other internal; and it may be remarked, that the contact with either of the given circles is external upon one branch of the hyperbola, and internal upon the other.140 ANALYTICAL GEOMETRY. [b. I. CH. VIII· Similar ellipses. CHAPTER ΥΙΠ. SIMILAR CURVES. 206. Definition. Two curves are said to be similar when they can be referred to two such systems of rectangular coordinates, that if the abscissas are taken in a given ratio, the ordinates are in the same ratio. 207. Corollary. If the given ratio is m : w, and if the co- ordinates of the first curve are z, y, the corresponding ones of the second curve must be n x ny m 5 m ’ so that if these values are substituted for the coordinates in the equation of the second curve, the equation obtained must be that of the first curve. 20S. Theorem. Two ellipses or two hyperbolas are similar, if the ratios of their axes are equal. Proof. I. Let the semiaxes of the two ellipses be A, B and A\ B', we have, by hypothesis, A : A' = B : B\ and the equations of these ellipses are (68), z2 A» ~ B* 4-1-1 • — JL — i · R/2 — 1 5 A*~ B‘2§ 209.] SIMILAR CURVES. 141 Radii vectores of similar curves. and if, in the second equation, we *ake the coordinates in the ratio equal to that of the axes, that is, substitute for x, — Λ and for y, equation. B'y B ' A'y A ; it becomes identical with the first II. The same reasoning may be applied to the hyperbola ; but it must be observed, that the ratios of the transverse axes must be equal to that of the conjugate axes in the two hyperbolas; and the theorem must not be applied to the case in which the ratio of the conjugate axis of the first curve to the transverse axis of the second is equal to that of the transverse axis of the first curve to the conjugate axis of the second curve. 209. Theorem. The radii vectores, which are drawn in the same direction to two similar curves, are in the same ratio with the corresponding coordinates. Proof. If x and y are the coordinates for the first curve, and x‘ and y' the coordinates for the second curve, taken as in art. 207, we have x __ x1 y ~ y’ so that, by (11), the angle φ — a, which determines the di- rection of the radius vector drawn to the point a?, y, is equal to the angle which determines the direction of the radius vector drawn to the point x\ y\ These radii vectores must, therefore, coincide in direction, and we have for their values r — x sec. (φ — »), r! — xl sec. (φ — *) r x y r' x1 y‘ *142 ANALYTICAL GEOMETRY. [b. I. CH. VIII. Similar surfaces and solids. 210. Similar surfaces may be defined in the same way as similar curves, and are subject to propositions precisely like those of arts. 207 and 209. Similar solids are solids bounded by similar sur- faces.§ an-] PLANE SECTION OF SURFACES. 143 Section of surface by a plane. CHAPTER IX. PLANE SECTIONS OF SURFACES. 211. Problem. To find the section of a surface made by a plane. Solution. I. If the cutting plane is one of the coordinate planes, that of x y, for instance, the points of the section are all of them in this plane, and we have, therefore, for all these points, * = o, so that we have only to substitute zero for z in the equation of the surface to find the equation of the inter- section with the plane of x y. In the same way by putting x = 0 the intersection with the plane of y z is found, and the intersection with the plane of x z is found by putting y = 0. II. For any other plane the intersection is found by transforming the coordinates of the surface, to a system of which the cutting plane is one of the coordinate planes. If the cutting plane is supposed to be the plane of xx y we shall be oliged to put = 0,144 ANALYTICAL· GEOMETRY. [β. I. CH. IX. Section of a surface by a plane. after substituting the equations (40-42) for the transforma- tion of coordinates. But a useless operation is avoided by putting, at once, z1=0 in the equations for transformation. The required equation is, then, obtained by substitut- ing in the given equation of the surface the equations * = «+ xx COS. a, + Vi cos. β (^21) y — b xx cos. a' -|- y1 cos. β' (322) % — c-\-x1 cos. u,n -|- y j cos. β". (323) Iti which a, b, c are the coordinates of a point of the cutting plane which is the origin of xx and a', and /3, /37j βη are the angles which the two axes of x1 y x make with the given axes. 212. Corollary. If the cutting plane is parallel to to the plane of x y, the axes of x1 and yx may be taken parallel to the axes of x and y, and the origin may be taken in the axis of z, so that the equations (321-323) become x — x^y — y±,z — c. (324) If the cutting plane is parallel to the plane of x z, we have in the same way x — x^y = b,z = zx; (325) and if it is parallel to the plane of y z, we have x — a, y — y^z — zx. (326) 213. Corollary. If the cutting plane passes through the axis of x, the axis of x may be taken for that of .r1, and the§ 214] PLANE SECTION OF SURFACES. 145 Section of quadratic surface. origin may remain as it was. In this case equations (321-323) become χ = χχ, y = y1 cos. p'9 % — y1 sin. p·. (327) If the cutting plane passes through the axis of y, and if the axis of y is taken for that of xl9 (321-323) become x = yx, cos. p9 y — x^ z — yx sin. p. (328) If the cutting plane passes through the axis of 2, and if the axis of z taken for that of x19 (321-323) become x = y1coa. p9 y = y1 sin. p9 z = xx. (329) 214. Problem. To find the section of a surface of the second degree made by a plane. Solution. The equation (200) is the most general equation of the surface of a second degree. It may then be regarded as the equation of the surface referred to coordinate planes, of which the plane x y is the cutting plane. By putting 2 = 0, we have then for the required section Ax* + Bxy+Cy2 + Hx + Iy + M=:0. (330) From the discussion (201-262), it follows that if H2 — 4 -4 C < 0 the section, if there is one, is a point or an ellipse. But if IP — 4 AC— 0 it is a parabola, a straight line, or ά combination of two parallel straight lines. But if B2 — 4 J.C>0 it is an hyperbola, or a combination of two straight lines. 13146 ANALYTICAL GEOMETRY. [β. I. CH. IX. Sections of quadratic surface. 215. Corollary. For the section which is parallel to the plane of x y at the distance c, we have by (324) putting H1 = Dc-\-H (331) I1=:Ec +I (332) M1 = Fc*-\-Kc + M (333) Az2 + Bxiyl + Cyl+H1z1 + I1y1+M1=0;(2M) so that this section is in the same class with that made by the parallel plane of x y. so far as it depends upon the value of B* — 4 AC. 216. The values of A1 and Bx depend by (220,231) only upon those of A, B, C, so that the ratios of the semi- axes A2 and B2 must also depend only upon A, U, C, and be the same for all the parallel sections of the quadratic locus. Hence, if one of the sections of a quadratic locus is an ellipse, all the parallel sections must be similar ellipses, except those which are points. If one of the sections is an hyperbola, all the curved parallel sections are hyperbolas; and all those sections are similar whose greater axes are transverse; and also those are similar whose greater axes are conjugate. If one of the sections is a parabola, all the curved sections which are parallel to it are parabolas. In all the parallel sections the axes are parallel. 217. Problem. To investigate the form of the sur- face of equation (312). Solution, The numbers below the letters were only used§ 217.] PLANE SECTION OF SURFACES. 147 Ellipsoid. to distinguish the different axes of coordinates; they may* then, be omitted, and (312) may be written I. The equation of the section parallel to the plane of x y at the distance c from the origin is II. The sections parallel to the planes of a? 2 and y % are easily found in the same way, and it is evident that the sur- face is included by six planes, of which two are drawn parallel to the plane of x y at the distances -|- C and — C* two parallel to the plane of x % at the distances -{- B and — B, and two parallel to the plane of y % at the distances -(- A and III. The section made by any other plane must then be limited, and must therefore be an ellipse or a point, so that this surface is called that of an ellipsoid, whose semiaxes are J., B, C. IV. The section made by a plane passing through the axis (335) (336) which is impossible when c2> C2; it is the point when *1 = itl = 0 c zz C, and it is the ellipse whose semiaxes are £ V(C3-c2),^V(C2-c2) (337) when c2 < C2. — A.148 ANALYTICAL GEOMETRY. [β. I. CH. IX. Ellipsoid. of a;, and inclined by an angle ./3' to the axis of y is, by (327,) x\ . /cos.2/*' sin.2/3' \ ^ + (338) It is, therefore, an ellipse, whose semiaxes are A and 'cos-2β' ■ sin.2p/\_________________BC_____________ m ^ C* ) s/ (C2cos.2/3/-^-52sin.2/3/) ’ ^ } ItV or if we substitute for cos.2/3/ its value, the second semiaxis is BC VT c2 + {W — C2) si^T] (340) The ellipsoid may, then, be considered as generated by the revolution of an ellipse about the axis of x, the semiaxis of the ellipse, which corresponds to the axis of a:, remaining constantly A, and the other axis changing from B to the value (340). The sections made by planes passing through the axes of y and % may be found in the same way. 218. Corollary. If we have B=C the semiaxis (339) becomes equal to B, so that the ellipse retains the same value of its second axis as well as of its first, during its revolution; and the ellipsoid is one of revolution. The sections made by planes parallel to the plane of y z are, in this case, circles. 219. Corollary. If we have A = B= C the revolving ellipse is a circle, and the surface is that of a sphere. 220. Problem. To investigate the form of the sur- face of equation (314).§ 220.] PLANE SECTION OF SURFACES. 149 Hyperboloid. Solution. By omitting the numbers below the letters, (314) may be written x2 y2 z2 (341) I. The section, parallel to the plane of x y9 at the dis- tance c from the origin is, by (253, 261), an hyperbola, of which the semitransverse axis, which is parallel to the axis of z, is A B — >v/(c2+C2)>and the semiconjugate is — ^/(cs+C2).(342) C C II. The section, parallel to the plane of x z, at the dis- tance b from the origin, is an hyperbola, of which the semitransverse axis is parallel to the axis of x, and is ~ \/{b2-{-B2)) and the semiconjugate is ^ s/iW-^B2). (343) III. The equation of the section, parallel to the plane of y z, at the distance a from the origin, is, by reversing the signs, &+&+1 - i = °; (®“> so that when a2 A2 the section is imaginary, that is, none of the surface is con- tained between the two planes drawn parallel to the plane of y z, at the distances -j- A and — A ; so that the surface con- sists of two entirely distinct branches, similar to the two branches of an hyperbola. When a9 — A2 the section is reduced to the point y = 0, z = 0 ; when a2 A29 13*150 ANALYTICAL GEOMETRY. [li. I. CH. IX. Hyperboloid of two branches. the section is an ellipse, of which the two semiaxes are IV. The section made by any plane, which cuts both branches, is evidently an hyperbola, for no other curve of the second degree is composed of two branches. The sec- tion made by a plane, which cuts entirely across either branch without cutting the other, is an ellipse ; for this is the only curve of the second degree, which returns into itself, so as to enclose a space. The section made by a plane, which cuts one branch without entirely cutting across it, and without cutting the other branch, is a parabola ; for this is the only endless curve of the second degree, which consists of only a single branch. This surface is called that of an hyperboloid of two branches. V. The equation of the section, made by a plane passing through the axis of x and inclined, by an angle /3', to the axis of y is, by (327), It is, therefere, an hyperbola, whose semitransverse axis, directed along the axis of xx is A, and whose semiconjugate axis is precisely that of (340). This hyperboloid may, then, be regarded as generated by the revolution of an hyperbola about the axis of xy the semitransverse axis remaining con- stant, and the semiconjugate axis changing in such a way, that its extremity describes an ellipse, whose semiaxes are B and C. VI. The section, made by a plane passing through tbe axis of y, and inclined by an angle β to the axis of x, is, by (328), ^\/ (a2— A%) and ^ \/(a2— -42). (345) (346) (347)§ 222.] PLANE SECTION OF SURFACES. 151 Hyperboloid of two branches. When, therefore, oos.2/3 sin.2/3^A ‘ C2 or tang.2/3<^— (348) JJl the section is an hyperbola, of which the semitransverse axis is AC —777^-------TZ-.—^p-r, and the semiconjugate is B. (349) \/(C2cos2/3—AP&m*p)' J 8 v 7 When O C2 tang 2 fi = 2Ϊ the section is impossible, but every parallel section is a parabola. C2 When tang.2/3^> —. the section is impossible, but there are parallel sections which are ellipses. In the same way and with like results, the sections made by planes passing through the axis of z may be found. 221. Corollary. If we have B — C the semiaxis (340) becomes equal to B, so that the re- volving hyperbola retains the same value of its second axis as at first, and the hyperboloid is one of revolution about the transverse, axis. The sections made by planes parallel to the plane of y z are, in this case, circles. 222. Problem. To investigate the form of the sur- face of equation (316).152 ANALYTICAL GEOMETRY. [β. I. CH. IX. Hyperboloid. Solution. By omitting the numbers below the letters, (316) may be written x2 ■ y2 A2 B2 %*_ C2 -1 = 0. (350) I. The section, made by a plane parallel to the plane of x y, at the distance c, is an ellipse, of which the semiaxes are ^V(c2 + ^)and-^V(c2 + C2). (351) II. The section, made by a plane parallel to the plane of xz, at the*distance Z>, is when b2 < B2 an hyperbola, of which the transverse semiaxis is parallel to the axis of a?, and is A C -g\/(B'2—£2),and the semiconjugate is-\/(B2-b^). (352) When b2 = B2 the section is the combination of the two straight lines When b2 > B2 the section is an hyperbola, whose semitransverse axis id parallel to the axis of z, and is C A —s/{b2 — B2), the semiconjugate is -g>\/(&2— -E2)· (354) In the same way and with like results, the sections made by a plane parallel to that of y z may be found.§222.] PLANE SECTION OF SURFACES. 153 Hyperboloid. III. The curved section made by any other plane is an hyperbola when it consists of two branches, an ellipse when it is limited, and a parabola when it consists of one infinite branch. IV. The equation of the section, made by a plane passing through the axis of x and inclined, by an angle /3', to the axis of y, is x\ A* When or cos.2 /3' sin.2 /3' \ σ~)y cos.2 /3' sin.2 /3' ^ Λ c2 tang.3 μ1 < — 2 1 1 = 0. (355) (356) this section is that of an ellipse, whose setniaxes are A and When _________BC__________ a/(C2 cos.2 fit — jB'2 sin.2 £')* tang.2 /3' 91 (357) (358) the section is reduced to the two parallel straight lines *i = ±A drawn parallel to the axis of y Any parallel section to this one is a parabola. When tang.3(359) XT the section is an hyperbola, whose transverse semiaxis is in the direction of the axis of and is A, the semiconjugate is — -______z____________f360! a/{B^ sin.2 β1 — C2 cos.2 /3') * '154 ANALYTICAL GEOMETRY. [b. I. CH. IX. Hyperboloid of revolution. In the same way and with like results, the sections made by a plane passing through the axis of y may be found. V. The equation of the section, made by a plane passing through the axis of z, and inclined, by an angle β to the axis of a?, is by (329) /COS 2/3 sin.2/3\ X2 + = <361) This section is, therefore, an hyperbola, whose semiconju- gate axis, directed according to the axis of z or x 19 is AB C, while the semitransverse is —77---—— ..—- , (362) \/(B2cos.^+A2sm.^Y v ' AB or the semitransverse axis is ,r-—t aH—· (363) * when the distance c is zero, this ellipse is a point. II. The section made by a plane parallel to the plane of x z, at the distance 3, is an hyperbola, of which the semi- transverse axis, parallel to the axis of 2, is Cb , t . . Ah -g- and the semiconjugate ——. (366) This section becomes the combination of the two straight lines C x = ± A z, (367) when b is zero. III. The section, made by a plane parallel to the plane of y z, at the distance a, is an hyperbola, of which the semi- transverse axis, parallel to the axis of z, is G CL B CL —-T— and the semiconjugate ——. (368) A A , This section becomes the combination of the two straight lines C y = dt B z, (369) when a is zero. IV. The equation of the section, made by a plane passing through the axis of a?, and inclined by an angle β' to the axis of y, is . 2 β1 sin. 2 β1 σ- -) y\ = o. (370)156 ANALYTICAL GEOMETRY. [b. I. CH. IX. Cone. When the condition (356) is fulfilled, this section is re- duced to the point X l — 0, y 1 HZ 0. But every section parallel to this one is an ellipse. When the condition (538) is fulfilled, the section is re- duced to the straight line zz 0; that is, to the axis of yx ; and every section parallel to this is a parabola. When the condition (359) is fulfilled, the section is the combination of the two straight lines fl. A = =t sin.2 β1 C2 cos.2 β \ (371) and every section parallel to this is an hyperbola. In the same way and with like results, the sections made by a plane passing through the axis of y may be found. V. The equation of the section, made by a plane passing through the axis of z, and inclined by an angle β to the axis of #, is by (329) / cos.2/3 \~~A? sin.2 β ) (372) so that this section is the combination of the two straight lines χΛ , .(cos.2 β . sin.2 β \ π = ±Λ^-+-Ί^-)·1'·· i*™) which are inclined at equal angles on opposite sides of the axis of xv§ 227.] PLANE SECTION OP SURFACES. 157 Conic sections. This surface may then be regarded as generated by a straight line which passes through the origin, and revolves about the axis of z, inclined to this axis by a variable angle, whose tangent is -----------_______________. (374) the surface is therefore that of a cone. 225. Corollary. If A and B are equal, the axes (365) are equal, and the section parallel to xy is a circle; and the tangent (374) of the angle which the revolving lines makes with the axis of z% becomes A_ σ so that its value is constant, and the cone is a right cone. 226. All the curves of the second degree may then be obtained by cutting a right cone by different planes; these curves are therefore called cgnic sections. From examining section iv. of art. 224, it appears that the section of a right cone is an ellipse, when the plane cuts completely across the cone, so as not to meet the cone produced above the vertex; it is a para- bola, when the plane is parallel to one of the extreme sides of the cone, so as not to meet it, nor the cone produced above the vertex; it is an hyperbola, when the plane cuts the cone both above and below the vertex. 227. Problem. To investigate the form of the surface of equation (307). 14158 ANALYTICAL GEOMETRY. [β. I. CD. IX. Paraboloid. Solution. By omitting the numbers below the letters, (307) becomes By2 + Cz2 + M=z0. (375) The equation of the section made by a plane parallel to the plane of y % is, then, the same with (375), so that the surface must be a cylinder, of which (375) is the equation of the base. 228. Problem. To investigate the form of the sur- face of equation (308). Solution. By omitting the numbers below the letters, (308) becomes B y2 -f C z2 + H x = 0. (376) I. The section made by a plane parallel to the plane of y % is an ellipse or an hyperbola, and those made by planes parallel to the planes of x y and x z are parabolas. II. The equation of the section, made by a plane passing through the axis of x, and inclined by an angle β' to the axis of y, is (Boos.2 β' Csin.2 fi')y2 -{-Hxt z=0, (377) so that it is a parabola, of which the vertex is the origin, and the parameter <378> The surface may then be considered as generated by the revolution of a parabola, with a variable parameter, about the axis of xx. It is, hence, called a 'paraboloid.§ 230.] PLANK SECTION OP SURFACES. 159 Cylinder. 229. Corollary. If B and C are equal, (378) becomes = (379) so that the parameter is no longer variable, and the paraboloid is a paraboloid of revolution. 230. Problem. To investigate the form of the sur- face of equation (309). Solution. By omitting the numbers below the letters, (309) becomes Cz*-\-Hx + Iy=zQ. (380) Before proceeding to investigate the sections of the sur- face, we may refer it to other axes, which have the same origin, of which the axis of %x is the same with the axis of %, and the plane of a?, y L the same with the plane of xy. In this case, we have a — b — c — 0, a" = p" = y = y1 == 90°, yn = 0 3' z=a, /3 — 90° a, a/ — — 90°, so that (40, 41, 42) become XZZ1 χχ cos. a — yx sin. a, (381) yz=zxx sin. a, +y. COS. a (382) (383) which being substituted give, by taking a to satisfy the con- dition that the coefficient of yl is zero, Cz°+(H COS. a + i sin. ») = 0, (384)160 ANALYTICAL GEOMETRY. [b. I. CH. IX. Two planes. in which « is determined by the equation tang. «=-, (385) The equation of the section, which is made by a plane parallel to the plane of xx 21, is now the same with (384) ; that is, the section is a parabola, and the surface is that of a cylinder, of which the base is a parabola. 231. Problem. To investigate tiie form of the sur- face of equation (310). Solution. When (310) is possible, it is evidently the com- bination of the two equations M *2 = ±V--c-\ (386) each of which represents a plane parallel to the plane of *2 V2·DIFFERENTIAL CALCULUS. BOOK II. 14*BOOK II. DIFFERENTIAL CALCULUS. CHAPTER I. FUNCTIONS. 1. A variable is a quantity, which may continually assume different values. A constant is a quantity, which constantly retains the same value. Thus the axes of an ellipse or hyperbola are constants* while the ordinates and abscissas are variables. Constants are usually denoted by the first letters of the alphabet, and variables by the last letters, but this notation cannot always be retained. 2. When quantities are so connected together that changes in the values of some of them affect the values of the other, they are said to be functions of each other. Any quantity is, then, a function of all the quantities upon which its value depends; but it is usual to name only the variables of which it is a function. Functions are denoted by the letters λ· ., P., /0., &c.; thus f.(x),F.(x), φ.(χ), 7r.(x),&LC.,/.'(*), &C;f0-(x),fAx)xfAx)> &c. are functions of x.164 DIFFERENTIAL CALCULUS. [b. II. CH. I. Independent variable. Construction of function. /. (x, y), F. (x, y), &c. are functions of x and y. 3. When variables are functions of each other, some of them can always be selected, to which, if particular values are given, the corresponding values of all the rest can be determined. The variables, which are thus selected, are called the independent variables. 4. When a function is actually expressed in terms of the quantities, upon which it depends, it is called an explicit f unction. But when the relations only are given, upon which the function depends, the function is called an implicit function. Thus the roots of an equation are, before its solution, implicit functions of its coefficients; but, after its solution, they are explicit functions. 5. A function of a variable may be expressed geo- metrically, by regarding it as the ordinate of a curve, of which the variable is the abscissa. A function of two variables may be expressed geo· metrically, by regarding it as one of the coordinates of a surface, of which the two variables are the other two coordinates. A function is said to be constructed, geometrically, when the curve or surface, which expresses it, is con- structed. The inspection of the curve or surface, which thus repre-§8.] FUNCTIONS. 165 Algebraic, logarithmic, trigonometric functions. sents a function, is often of great assistance in obtaining a clear idea of the function. 6. Algebraic functions are those which are formed by addition, subtraction, multiplication, division, and raising to given powers, whether integral or fractional, positive or negative. An integral function is a polynomial, which contains only positive integral powers of the variable; and a rational fractional function is a fraction, whose nume- rator and denominator are both integral functions. Every other algebraic function is called irrational, Thus a + a?, a — a?, a x -(- b y, a -|- b x -|- c x2 -|- &c. are integral functions; a a + * x + C X2 x'X 5 a' -(- l·' x -(- c' x2 are rational fractional functions ; and \/a;, a:f, &>c. are irrational functions. 7. Exponential or logarithmic functions involve variable exponents or logarithms of variables. Thus, a*, log. a, &c. are logarithmic or exponential func- tions. 8. Trigonometric or circular functions involve trigo- nometric operations.166 DIFFERENTIAL CALCULUS. [b. II. CH. I. Compound, free, fixed, linear functions. Thus sin. x, tan. z, &c. are trigonometric or circular functions. 9. Compound functions result from several succes- sive operations. Thus log. sin. x is the logarithm of the sine of x. 10. When functions are so related, that the com- pound function formed from their combination is inde- pendent of the order in which the functional operations are performed, the functions are said to be relatively free ; otherwise they are fixed. Thus if the two functions φ and f are so related that the compound function φ ,f. x is equal to the compound function f. φ. a?, these two functions are relatively free, and this con- dition is algebraically * (387) or if we omit the variable, which is often done in functional expressions which involve but one variable, (387) becomes φ ./.=/.?. (388) 11. A linear function is one, which leads to the same result, whether the operation indicated by it is performed upon the whole of a polynomial at once, or upon the different terms of it successively. Such a function is indicated by the equation f.(z±y)=f.x±f.y; (389) and the product mx, that is, m times the variable, is a simple example of such a function.§13.] FUNCTIONS. 167 Repeated functions. _______ 12. Theorem. The compounds of linear functions are linear. Proof. Let f and f* be two linear functions, we are to prove that (* ± y) (390) Now we have from definition f.'(x±y)=f-'x±f.'y, (391) and therefore /·/·' (*±y) =/· (/.'*±/·' y) =/./■' (392) as we wished to prove. 13. When the same operation is successively repeat- ed, the result is called the second, thirds fyc. function. Thus the log. log. x is the second logarithm of x. These repeated functions may be expressed by a notation similar to that of powers; thus log.2 x = log. log. x log.3 x = logi log.2 x = log. log. log. x, &c. f?x=f.f.x f -3x=f ? X =/·/·/· *» &c· Care must be taken not to confound /" (x) with [f(x) or with /(x)n, which have widely different significations; thus, [f(x)]n is the nth power of the function of x, f{x)n is the function of the nth power of x, while fn (x) is the nth function of x. · The common use of a different notation in the case of trigo- nometric functions must, however, cause them to be excepted168 DIFFERENTIAL CALCULUS. [β. II. CH. I. Zero function. from these remarks ; thus, sin." x and sin. xn do either of them denote the wth power of sin. x. Whenever we extend this notation to trigonometric functions, we shall indicate it by en- closing the exponent within brackets; thus we shall denote the second, third, &c. sine of x by sin.t2! x\ sin.i3! a?, &c. 14. By a process of reasoning, precisely similar to that used in the case of powers, it may be proved that we must have f»f.*x=f” + *x; (393) or, omitting the variable j(394) This equation may be adopted as applicable to all func- tional exponents, whether positive or negative, entire or fractional; and the signification of the exponent, when not positive and integral, must, in this case, be determined by the aid of the equation. 15. Problem. To determine the signification of a function, of which Ike exponent is zero. Solution. Equation (393) becomes by making m — 0, ffif." x =/.° + - x ==/.» x ; (395) so that if we put /·" x — y we have f.° y — y; (396) that is, the function whose exportent is zero is the vari- able itself and this function may be represented by unity.FUNCTIONS. 169 § 18.] Negative and fractional functions. 16. Problem To determine the signification of a function, of which the exponent is negative. Solution. Equation (393) becomes, by making or if we put we have f.~nf.nx=f.°x=T; f.nx — y, (397) = (398) that is, if two variables, x and y, are functions of each other, whatever function y may be of x, x is the corres- ponding negative function of y, or. as it is usually called, the inverse function of y. 17. Confusion is likely to arise in the use of negative exponents, unless it is carefully observed that many functions have, like roots, several different values corres- ponding to the same value of the variable. 18. Problem. To determine the signification of a function, of which the exponent is fractional. Solution. We have, from (394) /."·/*/*···=(/")"=/ in which n denotes the number of repetitions off.m. If now n' — m n, m — — n mn n* /."=yVr=/."; 15170 DIFFERENTIAL CALCULUS. [β. II. CH. I. Functions calculated like factors. that is, if m is a fraction, of which the denominator is n and the numerator n', the corresponding function is one which, repeated n times, gives ihen'th repetition of the original function /. 19. Theorem. When the different functions of which a compound function i§ composed are linear and relatively free, they may be combined precisely as if the letters which indicate them were factors instead of functional expressions. Proof. For the two equations (388) and (389), which apply to this case, are the same in form as the two funda- mental equations of addition and multiplication, upon which all arithmetical and algebraical processes are founded. 20. Corollary. The repetitions of the compound func- tions in which f /2, &c., are linear and relatively free, may be effected by means of the binomial and polynomial theo- rems. Thus (/· +/ι·+Λ·)"=/·" + »/·"-7ι· + + «fee. (400) 21. The exponents of functions are so similar to the exponents of powers, that they may be used in a simi- lar way, and called functional logarithms; so that if any function, as is assumed as a base, the functional logarithm of any other function indicates the expo- (/•+/i ·)" =f"+nP~1fi -4 η. n— 1 1.2” «fee. (399)§22.] FUNCTIONS. 171 * Functional logarithms. neat which the base must have to be equivalent to this new function. Thus, if we denote functional logarithms by [/. log.], and if f.n=z and 0 ; and their values must differ infinitely little from these values, when i is infinitely small; that is, ά1 is finite, log. i is infinite, and - is infi- log. i nitely small. But each of these infinitesimals has been shown to be of the order zero, so that there are infinitesimals of the order zero, which are finite, infinite, and infinitely small. 42. Problem. To find the nth power of 1 + h when i is infinitely small and n infinitely great, so that ni — a. (405) Solution. The binomial theorem gives , ,xn - . . . n(n—IV n(n—l)(n—2V0 . β (1+*) =1+nH—>l ~^ 1.2.3----------;* +&c- (406) But n is infinite, and, therefore, n — 1 = η, n —2 = n, &c. (407)INFINITESIMALS. 177 §45.] Neperian logarithms. which substituted in (406) give, by (405), (1 i)r‘ — 1 -j- 7i i—J- = ! + « + n2 i2 172 «2 ΪΤ2 w3 i3 1.2.3 a3 1.2.3 f &c. -(- &c. (408) 43. Corollary. When a = 1 (408) becomes (1 + *)‘=1 + 1 + 1 1.2 1 1.2.3 -f- &c. (409) If we denote the value of the second member of (409) by we easily find e = 2,71828 -f (410) and (409) is (l+i)4=e. (411) 44. Corollary. Since a n = T’ we have (1+ ·')*= (oW)W=l + a + ^ + &c. (412) 45. The number e is the base of Neper’s system of logarithms, and the logarithms taken in this system are called the Neperian logarithms. The Neperian logarithms will be generally used in the course of this work, and will be denoted by log. as usual.178 DIFFERENTIAL CALCULUS. [β. II. CH. II. Exponential infinitesimal functions. 46. Corollary. The log. of (411) is -j iog· (1 + i) = log. e — 1 (413) log. (l + i)z=i. (414) 47. Corollary. If in (412) we put for a the value a = m i, (415) and transpose l to the first member, (412) becomes emi — 1 = m i; (416) all the terms of the second member except the first being omitted, because they are infinitely smaller. 48. Corollary. If h is taken so that b — em or m ~ log. Z>, (417) (416) becomes b{— lzzzilog.Zi. (418) 49. Corollary, If in (416) m zz: 1, (419) (416) is (420)*S1·} DIFFERENTIALS. 179 Differences are linear functions. CHAPTER III. DIFFERENTIALS. 50. The difference of a function is the difference be- tween its two values, which correspond to two different values of the variable. When the difference between the two values of the variable is infinitely small, the difference of the func- tion is called its differential. The letters ^/., &e., z/.0, &,c., &c. placed before a function, denote its differences, and corres- ponding differences are denoted by the same letters. Thus J ,f, ζ, j a:, &c. are corresponding differences of f, (#), fj (a?), &c., and these differences correspond to the difference . x of the variable, so that J-f- x —f· /. (421) J ,fj X —fj (x -j- J . x) —fj x. &c. Differentials are denoted by the letters , „ , _M>R _M"R' D-f-x— MR'D'f‘X~ MR>' But since MWM1’ is an infinitely small portion of the curve, it may be regarded as a straight line, and we have M R _ M"R' MR MR' ’ or D .f.x =z D'.fix; that is, the value of the differential coefficient does not change with that of the differential. II. The differential coefficient is, in general, a finite func- tion, for the ratio M'R : MR, which represents this function, is the tangent of the angle MfMR, by which the curve is inclined to the axis AX. 62. Corollary. We have, by (435) and (437), d .f. x = D .f. x . dz, d2 ,f ,xz= D2 .f.x. dx2) (438) so that if d x is an infinitesimal of the first order, df. x is, by art. 30, of the same order, d2f. x is of the second order, and so on. Differentials may then be regarded as infinitesimals of the same order with their exponents. 63. Corollary. If we put n dx-zz.li, so that n = ~, da? (439)184 DIFFERENTIAL CALCULUS. [B. II. CH. III. Taylor’s and MacLaurin’s theorems. and put d. for 4. in (432), we have (1 . + d.yf.z=f.(x + h); or developing, as in art. 42, and putting D. rr —, D*. = ^ dx’ dx* (440) (441) (l. + AI». + ^ + &c.)/.*=/.(* + i). (442) But, by (434), h2 T)2 1. + AD. + + &c. = eA* D; (443) whence ehD.f.x=f.(x-\-h). (444) 64. Corollary. When, in (442) and (444), we put £ = 0 they become (l. + O.+^+&c.)/.0=/.l (445) e*V-0=/.A, (446) and if we now put x for Λ, we have ('+*D + X-^- + *™)fO=f.x (447) /.0 =f.x. (448) Equation (442) is called Taylor's theorem, of whicli (444) is a neat form of writing, and (447) is called MacLaurin’s theorem. The great use of these the- orems will be seen in the sequel.§68.] DIFFERENTIALS. 185 Increasing or decreasing function. ____ 65. Corollary. If we put, in (442) and (444), h = J. a?, they become, by (431), (\+J .xD.+ &c.)/. x=f. (l.+J.)x=(l* (449) e + (450) 66. Theorem. When the differential coefficient of a continuous function is finite and positive, the function increases with the increase of the variable; but if the differential coefficient is negative, the function de- creases with the increase of the variable. Proof. For the differential coefficient is the ratio between the differential of the function and that of the variable, and is therefore positive when both these differentials are posi- tive, and negative when one is positive and the other negative. 67. Corollary. If the variable increases from any of its values, for which the function vanishes, the func- tion must be positive if the differential coefficient is positive, and negative if the differential coefficient is negative; that is, the function has the same sign with the differential coefficient. The reverse is the case if the variable decreases. 68. Theorem. The greatest value of the differential coefficient of a continuous function, which vanishes with the variable, and extends to a given limit, is larger than the quotient of the greatest value of the function, divided by the corresponding value of the variable ; and the 16*186 DIFFERENTIAL· CALCULUS. [B. II. CH. III. Greatest and least values of differential coefficients. smallest value of the differential coefficient is smaller than the smallest value of this quotient. Proof Let f. denote the given function which vanishes with the variable, let x; be the limit of the variable to which the function is extended, and let A and B be respectively the greatest and the least of the values of the differential coefficient, so that A — Df. x and Of. x — B are positive. But these two quantities are the differential coefficients of the functions A x —f. x and f.x — B a?, both of which vanish when x is zero, and are, therefore, by art. 67, of the same sign with their differential coefficients when x is increasing and positive, and of the opposite sign when x is decreasing and negative. Hence these two func- tions are of the same sign with z, and their quotients, divided by z, must be positive, that is, X X are positive. It follows, then, that A is greater than the quotient of f. x divided by x, and that B is less than this quotient. The truth of this proposition may be exhibited geo- metrically. Thus, if A MB (fig. 48) is the curve which represents f. a?, so that, for AP =. xy we have MP — f.x ; if the curve at M is produced in the straight line ΛΓΓ, we have, by art. 61, B.fxziz tang. MTXy*69.] DIFFERENTIALS. 187 Greatest and least values of differential coefficients. and, by joining AM, _ _. __ MP f.x tang. MAX = = J—. (451) Now, M1AX being the greatest value which the angle MAX has in this curve, it is evident that in proceeding from M1 to A the curve must be inclined to the axis AX by an angle greater than MxAX; so that at ill, for instance, the value of tang. MTX, or of Df. x, is greater than tang. Mx A X, /*. x the greatest value of ; and, therefore, A, the greatest * f, x value of Df. x, must be greater than any value of —. Again, M2AX being the least value of MAX, we see that in proceeding from M2 to A the curve must be inclined to the axis AX by an angle less thanM2AX, so that at Mf, for instance, we have tang. Μ' TX tang. M2 AX; whence it follows as above, that B is less than any value of /•^ x 69. Theorem. If a function /. and its differential coefficient are continuous, and if the function vanishes when the variable is zero, there is, for every value of the variable ar, a value of 0 less than unity, which satisfies the equation f.x = x Df. (θχ). (452) Proof If A and B are respectively the greatest and least values of Df.x, contained between the limits x =, 0, and x — x,188 DIFFERENTIAL CALCULUS. [β. II. CH. III. Ratio of functions. it follows from the fact, that Of. x is continuous, that it must assume every possible value between A and J?, while x varies from 0 to x. But the value of f.x is included between A and J5, and, therefore, there is a value of x' less than x such that or f.x = x 1Of. x'. (453) But since x’ is less than a?, if we put it/ Q z=. — or a/ = 6 x, x we have 0 less than unity, and (453) becomes (452). 70. Corollary. If in (452) divided by ar, we suppose x to be such a function^ of a new variable aij as constantly to increase with the increase of xt and to vanish with a?lf so that x=f1.x1, and take & A, so that °X=fl ·(«!*,)» then 0 x is evidently less than unity, and if we suppose P-=Mx- (452) becomes, by dividing by x, f.x _ df.(ex) <*/·/ι·(Μι) x d(t>x) dfl. (0, Xj) or F.x1 ___dF.(d1xl)____ DjF.^a?,) ~dfl.{»1x1) ~ Ofi· (*j *x) ’ Λ·*ι§71.] DIFFERENTIALS. 189 Ratio of functions. or, omitting the numbers below the letters, which are no longer needed, and put x' = 0 x F.x___DF.(6x) DF.x' f.x Df.(9x) Df. x'' (454) 71. Corollary. If the n first successive differential coefficients of the functions F. and f. were continuous, and all but the nth vanished with the variable, (454) would give F. x DF.x' IF F.x" D ‘ F. x„ Df.x1 ~ IF f.x" ~ IF f.x in which Xy X , X &c. are decreasing, so that if we put (455) we have F. x f.x in which 9 is less than unity, IFF. (9„x) IFf-(·.*)' (456)190 DIFFERENTIAL· CALCULUS. [b. II. CH. IV. Complete differential sum of partial ones. CHAPTER IX. COMPOUND AND ALGEBRAIC FUNCTIONS. 72. Theorem. The differential of a compound func- tion of several simple f unctions, is equal to the sum of its partial differentials arising from allowing each simple function to vary by itself independently of the other simple functions. Proof Let fJ be simple functions, of which φ. is the compound function. We will denote by d/m dy # the partial differentials, supposing f., f‘ respectively to vary by them- selves. We are to prove that d φ. (/., /.') =d,.9. (/., /.') + *,.*. (/., /·')· (457) Now we have, by definition, */. * · (/·, /·') = 9.(f. + d /., /.') - φ. (/., /·'), (458) or, by transposition, f · (/·, /·') -*· /·')· (462) But we have d Φ. (/.,//)= Φ. (/·+<*/·,/'+<*/·')- · (/·,/·'), (463) which, substituted in (462), gives (457) by omitting the last term, because it is a second differential, and therefore an infinitesimal of the second order. 73. Corollary. function, Equation (457) is written by omitting the d , = -(- df,. (464) 74. Problem. To differentiate a f. x. Solution. \#e have, by definition, d. (af. x) = af. {x-\-dx) — af. x — dx) —f. x]z=zad .f.x. (465) 75. Corollary. We have then d .(ax) = adx. 76. Corollary. We have also df. (/- x ·/·' x)=f.'x df. % d/,.(f.x.f.'x) = f.xdf.'x (466)192 DIFFERENTIAL CALCULUS. [B. II. CH. IV. Differential of power. and, by (464), d. (f. x.f.'x) —f· x d f.x -\-f. x d fJx. (467) In the same way, if u and v are functions of x, d. (w v) —u d v -J-1> d u. (468) 77. Corollary. Equation (467), divided by f. x .f.' x, is (469) d· (/·«·/·' *). f.x.f.'x d ·/. x , d x f.x fJx 78. Corollary. From (469), it follows that d.(f.x.f.'x.f."x f.x.f.'x.f."x. )__d.f.x f.x d.f.'x /·'* -f&c. (470) 79. Corollary. If in (470) we have the n functions f. x —// x =/." x = &c. (471) (470) becomes d. (f. x)n n df. x which, freed from fractions, is d (/· *)n = n (/· *)”"1 <*/· * . (472) (473) Hence, to differentiate any power of a function, multi- ply by the exponent and by the differential of the func- tion, and diminish the exponent by unity. 80. Scholium. The proof of (472) and (473), which is given in art. 79, is limited to positive integral exponents, but may be easily extended.§ 80.] COMPOUND AND ALGEBRAIC FUNCTIONS. 193 Differentia) of power. I. Proof of (472) and (473) for fractional exponents. Let n be the fraction ^nd let φ.χ — (f. x)n so that (φ. z)m' — (f. x)m. Equation (472) gives m'd (φ.χ)____m d (jf.x) φ.χ f.x or d (φ.χ)___d (/. x)n__md (f.x)____nd (f. x) φ.χ ~~ (f-xY ~~ m'f.x f.x which includes (472), and consequently (473). II. Proof of (472) and (473) for negative exponents. Let n be negative ♦ .x = (/.«)- = c?Lf φ. x (f. x)m = 1. The differential of which gives, by (470) and (472), άφ.χ d (/. x)m __ d φ. x m df. x φ.χ * (f.*)m ~~ φ7χΓ~*~ f.x ~~ ’ whence d(f.xf mdf.x ndf.x (/·*)“ ~ /·* ~ /.* which is the same as (472), and therefore includes (473). 17 and let so that194 DIFFERENTIAL CALCULUS. [B. II. CH. IV. Binomial theorem. 81. Corollary. Equation (473) gives d. xm — m xm ~1 d x (474) and Ώ .xm ~m xm~l (475) D2. xm — m D. z™— m (m — 1) xm -2 (476) D3. xm— m (m— 1) (m — 2) zm~\ &c. ^77) If now we substitute zm for f. z in (442), we have h2 (x + Ji)m = xm-{-hD.xm + —-D2.zm-{-&c. — χ™ -J-ni xm~~lh —J- -A) xm-2h2-|-&;e. (478) which is the binomial theorem. 82. Examples for Differentiation. 1. Differentiate a xm-f- Ans. The differential coefficient is m a xm~l. df-x 2. Differentiate \/(f· #)· Ans. 3. Differentiate 7T-. f.x & 4. Differentiate --p.—r-. (/. x)n Ans. — ‘W(/·*) .fzJL. '. x)2 gdf.x df-x Ans· ~~(f7x)* (/•*r+I 5. Differentiate 6. Differentiate Ans. fjx.df.x-f-x-df-'x- Ans. ----~Jf.’x)2 f-x fJx CΛ*)" C/·'*)“' (/. x)m ~1 (mfJ x. df. x—m'f ■ * df-'x) § 82.] COMPOUND AND ALGEBRAIC FUNCTIONS. 195 Differential of polynomial. 7. Find the successive differential coefficients of a-^hx-^cx^-^-. . . . M Ans. The first is b-{-2cx...~\-mMxm~1; the second is 2c ... — 1) Mar”-2 *, the ?nth is m (m — I) (m — 2) ... 1 . M;196 DIFFERENTIAL CALCULUS. [b. II. CH. V. Differential of exponential. CHAPTER V. LOGARITHMIC FUNCTIONS. 83. ProblemTo differentiate ax. Solution. We have, by definition, d ax — ax+dx — ax == ax adx — ax = ax (adx — 1). (479) But, by (418), adx — 1 — log. a . d x ; (480) whence d ax = log. a. ax d x, (481) 84. Corollary. Hence D . a* — log. a . ax JP.ax — log. a. D ax =z (log. a)2 ax Dn. ax = (log. a)n a\ (482) and by making a: = 0 a0 = 1 D . a0 =z log. a IP. a0 = (log. a)2 Dn. a0 (log. a)n. (483)§86.] LOGARITHMIC FUNCTIONS. 197 Differential of logarithm. If now we put in (447) /. x = a*, we have a* — 1 log· a · x -f- a~—-----\- &c. (484) 85. Corollary. If we have a = e, we have log. e = 1, eT = D . e* = D5. e* = X)’. e* (485) 1 = ΰ. e° = D5. e° = D". e® (486) e* = 1 + * + ^ + &c. (487) and (487) is the same with (412). 86. Problem. To differentiate log. .r. Solution. We have, by definition, d . log. a? = log. (x -f- d x) — log. x = log.^L^ = ,og. (l +^). (488) But, by (414) and, therefore, d . log. x = —. (489) x 17*198 DIFFERENTIAL CALCULUS. [B. II. CH. V. Development of logarithm. 87. Corollary. Hence D. log. x (490) IP. log. X= D. - =---------— S X 3fi log· *= 7.4! 2.3 _D4. log. x —---------- or* D". log. * = ± 1·2··:_ί!ί--Γ2), (491) the upper sign being used when n is odd, and the lower when n is even. If now in (442) we make /· = log. we have, log.(* + *)=log.* + ^ — — &c. (492) 88. Corollary. If in (492) we put X 2=2 1, we have log. (1 + Λ) = h — 4λ* + £Λ3_*Α4 + &(5. (493) 89. Examples. 1. Differentiate log. [a? -f- s/ (1 -}- a£)]. . d x Arts. - -, ,---. V(1 + ^2) Ans. n°0g 2. Differentiate (log. aO". xLOGARITHMIC FUNCTIONS 199 *89.] Differential of 2d and 3d logarithms. 3. Differentiate log.2 x. Ans. —---------. x log. x d x 4. Differentiate log.3 a?. Ans. —--------------- x log. x . log. 2x 5. Differentiate abx. Solution. Let H II 5^ and we have ahx = ay whence d . abx — d . ay ~ log. a. ay.d y But dy — d . bx = log. b .bx. d x so that d . ahx — log. a . log. b . aPx. h* d x. 6. Differentiate xy. Solution. Equation (464) gives d. xy = dx. xy -f- dy. xy. But by (473) and (481) dx. x? z=z y xy~]dx dy. xy — log. x.xyd y so that d . xy zn y xv~l d x -f- log. x.a? dy.200 DIFFERENTIAL CALCULUS. [β. II. CH. TL Differential of sine and cosine. CHAPTER VI. CIRCULAR FUNCTIONS. 90. Problem. To differentiate sin. x. Solution. We have, by definition, d . sin. x == sin. (x -\-dx) — sin. x. But, by trigonometry, sin. (* + d a:) = sin. x cos. dx + cos. x sin. d a?, cos. d x = 1, sin. d x=z dx, so that d . sin. x ~ cos. x. dx. (494) 91. Problem. To differentiate cos. x. Solution. Substitute in (494), £ w — a?, and it becomes d · sin. (£*■ — a?) = cos. (£*- —*). d (£*·_*). (495) But we have sin. (±π — χ)ζ= cos. a:, cos. (£ *· _ *) = sin. a; * — x)=i—dx, which, substituted in (495), give d . cos. x = — sin. x .dx. (496)§93.] CIRCULAR FUNCTIONS. 201 Development of sine and cosine. 92. Corollary. Equations (494) and (496) give Ό. sin. x = cos. x D2. sin. x — D. cos. x — — sin. x D3. sin. z = _D2. cos. x — — D sin. x — — cos. x D4. sin. x z= D3. cos. x — — D cos. x z= sin. a? .D1. sin. x z= Dn~l. cos.x ~ D”“4. sin. a?, (497) so that the four values of all the successive differential co- efficients of sin. x and cos. x are alternately cos. — sin.z, — cos. z, and sin x. Hence, making x = 0, we have, when n is even, * Dn. sin. 0 = cos. 0 — 0, (498) but when n is odd IT. sin. 0 = D"-1 cos. 0n±l; (499) the upper sign being used when n — 1 is divisible by 4, and the lower sign when η -|- 1 is divisible by 4. These values, substituted in (447), give sin. x=zx- x3 x' 1.2.3 1.2.3.4.5 1.2.3.4.5.6.7 l6 cos. a: 1 1.2+].2.3.4 1.2.3.4.5.6 -|-&c. (500) &c. (501) 93. Problem. To differentiate tang. x. Solution. We have, by trigonometry and example 5, of art. 82, D. t^ng. x — D. sin. x cos.x cos. x D sin. x — sin. x. D cos. x cos.- x202 DIFFERENTIAL CALCULUS. [b. II. CH. VI. Development of tangent. whence, by (494), (496), and trigonometry, cos.2 x sin.2 x_______________________ 1 D. tang, x z= COS.*5 X (502) 94. Corollary. Equations (502) and (496) give .D2. tang, x = 2 D cos. x 2 sin. x______2 tang, x cos.3 x cos.3 x cos.2 x ’ or D2. tang, x — 2 tang, x. D tang, x D3 tang. x = 2 tang, x _D2 tang, x 2 (D tang, a?)2 D4 tang, x =z 2 tang, x Ό3 tang, x 6 D tang, x D2 tang, x D5 tang, x — 2 tang, a; Di tang, x -(- 8 D tang, x D3 tang, x -f- 6 (D2tang. #)2 Dn tang. x — (D. tang, x D. tang, a?)"”1, (503) in which the exponents are to be annexed to the D.; when the exponent is zero, as in the first and last terms, the D. is omitted, and tang, x retained; and as the terms equally dis- tant from the two extremities of the developed series are alike, they may ^e added together. By making x — 0, these equations give D tang, x — 1 D2 tang, x = 0 Z>3 tang, x = 2 D4 tang, x = 0 D5 tang, x zzr 16§97.] CIRCULAR FUNCTIONS. 203 Differential of negative sine and tangent. Hence, by (447), teng. * = * + *3 + 1,2:3!4.5 ** + &C~ = X + i λ;3 + τ2τ x5 + &c. (504) 9 95. To differentiate arc sin. x = sinJ— x. Solution. Let y = sin.t—b a?, or sin. y =z x so that by differentiation cos. y . dy = d x—dys/{ 1 — sin.2y) = dy a/{ 1 — a;2) and D. «».!-■!,= D, = = 0—(505) 96. Corollary. By the same process we should find D . cos. [_1] x — Dare cos. x = — (1 — a;2)—h. (506) 97. Problem. To differentiate tang, f— ]l .r = arc tang. x. Solution. Let y — tang.r_,] x or tang, y zzi a?, so that, by differentiation, col 2 y = dxz=idy sec.3 y = d y(1 + tang 3 y) = d y(1 -f- a:3) and • , D tang. a: r= Dy = =r (1 + ®3)~l. (507)204 DIFFERENTIAL· CALCULUS. [β. II. CH. YI. Development of negative sine. 98. Scholmm. Dy applying Taylor’s theorem, the values of sin.t— i] χ? cosJ-^ and tang.!-1! might easily be found expressed in series; but they may more easily be found by the following process. 99. Problem. To develop sin.^x in a series arranged v according to powers of x. Solution. Suppose the series to be sin.t'-1]x= C0-f- C1x-{-C2x<2-{-Cri ,x3-{- C4o;4“(-&c. (508) in which the number below the coefficient denotes the power of a?, which it multiplies. We find, then, by differentiation, (505), and the binomial theorem, (l_z2)_£_ C1+2C2tf + 3C3*2 + 4C4 =1+^1**+&c· whence C1 = 1 C — -i _____*_— - 3 υ5 — 5*2 4 — 40 and, Cn =z 0 which n is even, and if in (508) we put x = 0 it becomes c0 = o. Hence, by substitution, · sin.[_1] x = a? £a;3 -f- jo v* “f" &c. (509)CIRCULAR FUNCTIONS. 205 § 102.] Development of negative cosine and tangent. 100. Corollary. In the same way, if we put cos. !· ^ x — Cq —|— C j x —|— C2 (510) we find all the coefficients except C0 to be the negative of those for sinJ-1! or, and if in (510) we put 07—0 we find C0 = arc. cos. 0 — J tt, whence COS.t-1] 07 = J 7tr-X-- £ 073 -^ 075 --&>C. (511) 101. Corollary. we have If in (511) we put 07—1 cos.M 1 — — 1 — £ — — &c. (512) and if (512) is subtracted from (511), term from term, it gives cosJ“]l o7z=z (1—#) + i(l — o?3) 4^5-(1—ot·^) —|— &c. (513) 102. Problem. To develop tang.[ 11 x in series, ac- cording to powers of x. Solution. Suppose the series to be tang.H] x = C0 -f- Ci 07 -f- C2 x2 -j- &c. (511) we find, by differentiation, (507), and the binomial theorem, (1 + 072)-1 = Cx +2 C2 x + 3 C3 072-f 4 C4 073 -f &c. IZZ 1 --07^ -f- 074 - 0?Q 4“ &C* whence C1 — 1 C3 — — i £5 — τ and when n is odd Cn z= ± —» n 18206 DIFFERENTIAL· CALCULUS. [β. II. CH. VI. Differentials of circular functions. the upper sign being used when n — 1 is divisible by 4, and the lower sign when n + 1 is divisible by 4; but when n is even or zero, c„ = 0. Hence, by substitution, tang.t-1] x — x — £ a:3 -(- l x5 — γ x7 -f- &c. (515) 103. Examples. 1. Differentiate sec. x. Ans. tang. x. sec. x d x. 2. Differentiate sin.n x. Ans. n sin."”1 x cos. x d x. 3. Differentiate log. cos. x. Ans. — tang, x d x. 4. Differentiate log. s/( . Ans. sec. x d x. ° \1 — sin. x) ____ 1 r , / b + a cos. x \ 5. Differentiate —77—5-----7^ cos.C-1]! —r-7----------- )· \/(«2 — δ2) \ a + b cos. x j Ans. dx 6. Differentiate cot. a?. Ans. cos. x dx sin.2 x *§ 104.] INDETERMINATE FORMS. 207 Value of fraction, when both terms vanish. CHAPTER YII. INDETERMINATE FORMS. 104. Problem. To find the value of a fraction when its numerator and denominator both vanish for a given value of the variable, of which they are both continu- ous functions. Solution. Let the fraction be F. x 7^' (516) and let xQ be the value of a?, for which the terms vanish; let h = x — a?0, or x = x0 -[- λ, (5Γ7) the given fraction becomes then F. (j0 + h) /·(*»+*)’ ( 5 which is a fraction, both whose terms vanish for A = 0; (519) so that in (456), h being the variable instead of a?, we have, if all the differential coefficients of the terms vanish, up to the wth, when λ = 0, F.(.r0+A)_D’.f'.(;Co + ^A) /. (*0+ (520)208 DIFFERENTIAL CALCULUS. [b. II. CH. VII. Fraction of which both terms vanish. which when h — 0 becomes F. x0 __ D\ F.xq f. x0 ~~D\ f.xQ ’ (Ml) so that the value of the given fraction is obtained by differentiating its numerator and denominator, until a differential coefficient is obtained, which does not vanish for the given value of the variable. 105. Examples. 6X ——- 6— x 1. Find the value of —:---, when x = 0. sin. x Solution. We have e° _ e-o __ D. e° — D. e~° __e°+e-° _____1 1___0 sin. 0 D. sin. 0 cos. 0 1 sin. (x^\ 2. Find the value of ———when x = 0. Ans. 0. , sin. x 3. Find the value of —, when x 0. Ans. qd. log x 4. Find the value of , when x z= 1. Ans. 1. x— 1 χη__ 5. Find the value of-----, when x z=z a. Ans. n a”-1. 1___cos x 6. Find the value of----, when x 0. Ans. 7. Find the value of x- x— sin. x 7-^3--, when x z=z 0. Ans.INDETERMINATE FORMS. 209 § 106. Fraction of which both terms are infinite. g* __ g I ___ ^ 8. Find the value of -----------:------, when x = 0. . 2. Λ 1 , , - e* — e~x — 2 sin. x , Λ 9. Find the value of----------------------, when x — 0. x3 Ans. §. 10. Find the value of --*)' when ^ = 0. ilns. m. 11. Find the value of sin. φ sin. m φ , when φ = 7r, and m is an sin. φ integer. Ans. — m cos. m so that when m is even, the answer is —m; and when m is odd, the answer is rn. 106. Problem. To find the value of a fraction, when • both its terms become infinite for a given value of the variable. Solution. Let the numerator of the fraction be Y and the denominator y, the fraction is ~ = -£r· (522) But when y and Y are infinite, their reciprocals, y”1 and Y~\ vanish, and we have by the preceding art., Y _ D . y~l ___ — y-2 D y _ Y2 Dy ~y dTy~x — Y-2 D y y* D Y* and, dividing by Y* y γ Dy Y DY’ °Ty : DY Dy' (523) 18*210 DIFFERENTIAL CALCULUS. [b. II. CH. VII. Product of which one factor is infinite, the other zero. so that the value of the fraction may be found by dif- ferentiating both its terms; and if both the terms of the fraction thus obtained are infinite or zero, the differentiation may be continued, until a fraction is obtained, of which both terms are not infinite or zero; and equation (521) applies to the present case as well as to the preceding one. 107. Examples. 1. Find the value of -, when x = 0. cot. x Solution. We have log. 0 D. log. 0 sin.2 0 . _ — = — 2 sin. 0 cos. 0 — 0. * cot. 0 D. cot. 0 0 2. Find the value of cot. τη ψ , when Φ =z 0. Ans. 1 cot. φ ’ m 108. Problem. To find the value of a product of two factors, when one of the factors become infinite and the other factor becomes zero for a given value of the variable. Solution. Let y and Y be the two factors, and we have for the given product (524) so that it is equal to a fraction, of which both the terms§110.] INDETERMINATE FORMS, 211 Infinite or zero powers. t are infinite, or both are zero, and the value of this fraction may be found by the preceding articles. 109. Examples. 1. Find the value of x« e~x when x oo. A ns. 0. 2. Find the value of x (log. x)n, when x — 0. Ans. 0. 3. Find the value of xn log. x, when x ~ 0. Ans. 0. 4. Find the value of x cotan. x, when x — 0. Ans. 1. 110. Problem. To find the value of a power, when the exponent and the root are both such functions of a variable, that they assume, for a given value of the variable, one of the forms 0°, oo0, or l00. Solution. Let the power be z = Yv (525) and we have, by logarithms, log. z = y log. Y, (526) so that in either of the given cases log. is equal to a product, of which one of the factors is zero, while the other is infinite; its value may therefore be found by the preceding articles; and when its value is found, we have or z — e l°s- z Yy — ey 1os Y. (527) (528)212 DIFFERENTIAL CALCULUS. [β. II. CH. VII. Infinite or zero powers. 111. Examples. 1. Find the value of a?37, when x = 0. Solution. Since 0 log. 0=0 we have, here, 0° = e° = 1. 2. Find the value of (ex—l)37, when x = 3. Find the value of cot.(p <£>, when φ = 4. Find the value of cos. ) sin. A. (Dy-f-^ xD2y) sin. A = Ja D2y sin. A. Hence 0 =.bx — ay, y = 2l·,x = 2a,D^y = ^, (V* and the surface is a minimum. 19218 DIFFERENTIAL CALCULUS, [b. II. CH. VIII. Maxima and minima of product. 9. Find on the circle (fig. 33), referred to the centre A, as the origin of rectangular coordinates, the point ΛΖ, for which the product of the coordinates is a maximum or a minimum. Ans. The product is a maximum when y = x. 10. Find on the ellipse (fig. 35), referred to its centre as the origin, its axes as the axes of coordinates, the point ΛΓ, for which the product of the coordinates is a maximum or a minimum. Ans, The product is a maximum, when the coordinates are in the ratio of the axes. 116. When the function, of which the maxima and minima are to be found, is a product or quotient, the solution is often simplified by finding the maxima and minima of its loga- rithm, which evidently correspond to those of the function. Examples. 1. Find the maxima and minima of the function x~a eb*% when a is positive. Solution, We have log. (#—« ehx)=z — a log. x -f“ l x Ό. log. (ar-e ebx) z=z---— h X A D2.log. (x—aebx) ; x X a Ί so that the value§116.] ΜΑΧΙΜΑ AND MINIMA. 219 Maxima and minima of product. corresponds to a minimum. The corresponding value of the given function is (V)· 2. Find the maxima and minima of the function (x — a) (a? — b) x~2. Arts. There is a minimum, when 2 ab XZH----r-7- · cl b220 DIFFERENTIAL· CALCULUS. [b. II. CH. IX. Orders of contact. CHAPTER IX. CONTACT. 117. When two curves meet or cut at a given point, and, at any infinitesimal distance of the first order from this point, are at a distance apart, which is an infinitesimal of the n + 1st order, they are said to be in contact at this point, and their contact is of the wth order. When one of the curves is of the first degree or the straight line, it is the ordinary tangent· When n is zero, there is no contact, but only an intersection. 118. Theorem. When two curves are in contact, the portion which is intersected between them upon a line, drawn at an infinitesimal distance from the point of contact, and inclined by a finite angle to the directions of the curves, is of the same order of infinitesimals with the distances of the curves apart. Proof Let M'M0M and M\M0M1 (fig. 49) be the curves, M0 the point of contact, MN their distance apart, and MM1 the intercepted portion of the line PM. The line MXN may be regarded as a straight line, and the angle MNMX as a right angle ; so that we have§ 119.] CONTACT. 221 Conditions of contact. MMX = MNx sec. N MMX. Bat sec. NM Mx is finite, as long as the angle MMx N, by which NMMX differs from a right angle, is finite ; and, therefore, by art. 31, MMX is generally of the same order with Μ N. 119. Problem. Find the algebraic conditions which denote that two given curves have a contact of the nth order. Solution. Let the equations of the two given curves M'MjM, (fig. 49), referred to rectangular co- ordinates, be, respectively, y=f.x (535) y=/i·*· (536) Then if A is the origin and M0 the point of contact, we have at this point A. Pq — #0, Pq Mq zzz or f. Xq :=zfi · fm *o f \ · == ^ > (537) and if P P0 = &, we have MMx =/. (x0 + h) —fx . (x0 -f h). (538) If, now, /. a?0—fx . Xq is such a function of x0 that all its differential coefficients inferior to the mth are zero, we have, by (533) and (537), substituting f. —f. for f. “i ~ l. 2 3.. .ίπΡ^' X° * (^39) and if we put IP. (f. x0 1.2.3 Xo = —fi ■ ^ο)__ D-fi -xo ---m 1.2.3 . ...m , (540) 19*222 DIFFERENTIAL· CALCULUS. [b. II. CH. IX. Condition of curves crossing at point of contact. we have MMx = X0.hm, (541) so that if h is of the first order, M Mx is of the mth order. But M Mx is to be of the (n+ir order, and therefore we must have or m = n + l, MMxz=X0.hn+\ -Dn+1 /· Xq — Dn+1 · /i · *0 Λ 1.2.3.....(n + 1) (542) (543) and all the differential coefficients of f. x0—ft. z, inferior to the (n-|- l)st, must be zero ; that is, we must have, when m is less than n -j- 1, Bmf. *o — *fi · *o = (544) or D»/.fl5o = D-/1.x0· (545) 120. In the same way it may be proved, that if the equations of the two curves are expressed in the polar coordinates of art. 45; so that they are r = F. )· (548) Hence, by differentiation, D.y = tang. r = D.y0 = D.f. x0 (549) and the equation of the tangent is y — y0 = D.f.x0(z — x0). (550) AT— x‘ the coordinates of T are y = 0 and x — x', so that PT— x0 —x1 =224 DIFFERENTIAL CALCULUS. [β. II. CH. IX. Tangent. 124. Corollary. When the equation, of the curve is expressed in polar coordinates, as in (546), the tangent can be found by means of (117). Thus we have and by logarithms = i* — τ (552) = £*■ — (T — *=V[i + (i>y)*] (571)§ 137.] CONTACT. 233 Element of curve. and if s is the independent variable, using the notation of art. 129, \=s/[{D.axY + {B.syf]. (572) 136. Corollary. In the same way, if the radii vectores AM, AM? are drawn, and the arc MR described with r as a radius, we have MAM — d, (573) and if MRM is regarded as a right triangle d s — */{d r2 + r2 d Φ2), (574) so that if φ is regarded as the independent variable D s = s/[{D r)2 + r2]. (575) 137. Corollary. The triangle MM.N gives, by (560), a? being regarded as the independent variable, when none is expressed in the formula MMN— τ dx 1 cos. τ — = —— — D .sx — sm. v ds Ds (576) dy D y sin. τ — ■ — D.. y = — cos. v ds Ds * (577) and by (564) and (565) y D.s _ '™s- = -Q^r=yD-vs y D..y y sin.T y COS, V (578) normal = y Ds — y __ jy______ D . , X COS. τ y sin. * (579) 20*234 DIFFERENTIAL CALCULUS. [β. II. CH. IX. Contact of surfaces. 138. Corollary. The triangle MRM1 gives MMR — ε (580) COS. B rz dr D.r _ ΖΓ = “ri—= D.,r ds D.s (581) sin. ε — r άΦ r _ (582) ds Ds ·*φ’ φ being regarded as the independent variable, when none is expressed in the formula. 139. Two surfaces which intersect at a given point are said to be in contact, when their sections, found by any plane passing through this point, are in contact. 140. Problem. To find the algebraic conditions that two surfaces are in contact at a given point. Solution. Let the surfaces be referred to rectangular co- ordinates, and let #0, z0 be the coordinates of the point of contact. Then the sections of the two surfaces, made by any plane, are found by equations (321-323), and since these sections are in contact, the values of D y found on the hypothesis that xl is the independent variable, must be equal for the two surfaces at the point of contact. Hence the values of Ox0, Ό y0, D z0, found by differentiating the equations (321-323) must also be equal for the two surfaces, as well as Dyo_±yo__ D D .r0 d x0 Vo Dzn Dxn dz_0 d #o (583) and (584)§ 142.] CONTACT. 235 Tangent plane. 141. Corollary. If one of the surfaces is the tangent plane, its equation, since it passes through the point of con- tact, is by (197) cos. α (X—x0) cos. fi (y—y0) + cos. y (;—z0) — 0, (585) from which we find by differentiation T>xV COS. ci ^ COS. β (586) D. ZZ COS. Cl cos. y = D.xz0 (587) which substituted in (585), divided by cos. give, for the equation of the tangent plane, (x —a?0) y—y0 z — gq (5S8) 142. Corollary. If all the terms of the equation of the given surface are transposed to the first member, and this first member, which is a function of x, y, z, represented by F, and if V becomes V0 at the point of contact, the equa- tion of the surface is # V = 0. . (589) The differential of this equation being taken on the hypo- thesis that 2 is constant, in order to find D. x y0, gives D. xV0.dx^-\-D .yV^.dij^ — 0, (590) whence, by (586), n _ dy0 D · τ 1 o cos, a xVo~ dx0~ D.,V~ COS. /3 1 _cos. β D'.yVo D-tVo ~~ cos· “ _ D.mV0 (591) (592)236 DIFFERENTIAL CALCULUS. [b. II. CH. IX. Tangent plane. In the same way we find 1 ___cos. y Ό. z VQ D. x 20 cos. a D. x F0’ (593) and these values, substituted in (588), give, by freeing from fractions, D.ZV0 (x-x0) + n.yVa (y-y0) + D.ZV« (z—z0) = 0. (594 This equation of the tangent plane compared with the com- plete differential of (589), which is D.xVQdi0 + D.yV0dyQ + D.£V0dz0 = 0, (595) shows that the equation of the tangent plane may be obtained from that of the surface by changing in the complete differential of the equation of the surface referred to the point of contact dx0, dy0, dzo respec- tively into (x—xQ)) (y—yo), (z—z0)· 143. Corollary. The sum of the squares of (592) and (593) increased by unity is cos.2 a4"cos-2/3+cos.2 γ (D-x Fo)9+( D ·α^ο)2+(_Ρ·_ζ Fn)2 ( -----Γ"" (D-zVof ( } or by (47)* and putting L = V [(D .. F0)2 + (D. , F0)2 +(D.Z F0)2] (597) cos. * — B.XV0 o COS. β =-------y—± L0 ’ D.ZV0 I) . z Vc, = L0 cos. a. D. yVo = L0 cos. β , D.zVo = L0cos. y. (598) (599) cos. y = U (600)§ 146.] CONTACT. 237 Normal to surface. 144. The perpendicular to the tangent plane drawn through the point of contact, is called the normal to the surface. 145. Corollary. The angles *, /3, y are, by (128), the angles which the normal makes with the axes; hence, by (124) and (598-600), the equations of the normal are y—yo z — z0 (601) COS. a cos. β cos. y X — Xo 1 § 1 S'. 1 z — so (602) -Β.,ν- to N 146. Examples. 1. Find the equations of the tangent plane and of the normal to the sphere. Solution. If the sphere is that of equation (62), we have V= x* + + z2 — R* =0 D . * V0 =z 2 x0 D . y Fo = 2 yQ D.zV0z=2zq L0 = 2 \/(^o + zo) R cos. » = ~, cos. β = ji, cos. y = The equation of the tangent plane is, by reduction, XQZ-\-y0y-\-ZdZ=: JR2.238 DIFFERENTIAL· CALCULUS. [β. II. CH. IX. Tangent and normal to ellipsoid. The equations of the normal are, by reduction, x y z so that by (123) it passes through the origin and is radius. 2. Find the equations of the tangent plane and of the normal to the ellipsoid of equation (335). Ans. The equation of the tangent plane is, by reduction, , y0y | -02 A2 B2 * C2 0. The equations of the normal are Wjl_i) = Wjl_A=0*(—Λ. \ r0 / \ Vo / \ *0 ) 3. Find the equations of the tangent plane to the cone of equation (364). Ans. The equation of the tangent plane is *o * , yQy gog o A2 B2 C2 “ ’ so that it passes through the origin. 4. Find the equations of the tangent plane and of the normal to the cylinder of equation (375). Ans. The equation of the tangent plane is By0 y + Czoz + 0, so that it is perpendicular to the plane of y z. The equa- tions of the normal are x — x0 Cz0y — By0z=(C—B)y0 z0.§ 146.] CONTACT. 239 Tangent and normal to paraboloid. 5. Find the equations of the tangent plane and of the normal to the paraboloid of equation (376). Ans. The equation of the tangent plane is 2 Byo y+2 c *o + #(* + *„) = o. Those of the normal are 2 By0 (x — ®o) = H (y—^0) Cz0y — By0 z = (C — B)y0z0. 6. Find the equations of the tangent plane and of the nor- mal to the cylinder, of which the base is a parabola, and the equation is (384). Ans. Put H cos. * + 1 sin. a — — 4 p C, and omitting the numbers below the letters in (384); the equation of the tangent plane is z0z = 2p(x + x0), so that it is perpendicular to the plane of xz. The equa- tions of the normal are so(* — (606) and D2y0 = D2^, so that by (603) and (606) - o II o * (607) which substituted in (605) gives ds0 = Qd*09 (608) or omitting the cyphers below the letters, ds Ds _ ? = -*—= tt" = Dvs. dv Dv (609) 149. Corollary. Equations (609), (571), (576), (577), (606), and (565) give . . -P·»*____&-.·)· =[i + (D.zy)*f sin .2vD2.xy D2.ty jD2.sy (610) (sec. τ)3 N3 ds _ D2,xy y3 D2.xy dr (611) 150. Corollary. When the equation of the curve is given in polar coordinates, the radius of curvature may be found by means of equations (557) and (558). For these give d ε — sin.2 e d .( - \r d φ 21 (612)242 DIFFERENTIAL CALCULUS. [B. II. CH. X. Radius of curvature. ά,τ 1 e __ d ψ —|— d ε — d$ d

/ (613) (614) 151. Examples. 1. Find the radius of curvature of the ellipse. Solution. Equation (69) of the ellipse gives 9 (Α4ν2+Β4χψ * A4 B4 m.xy — B2x A?y A2 ϊ/3 A2 N3 _ A2 B2____________ B4 (A2 sin.2 τ - B2 cos.2 This value of the radius of curvature is the same also for the hyperbola. 2. Find the radius of curvature of the parabola of equa- tiomof B. 1. § 180. Ans (ya~Mff2)g _ _ 2p ' 4 p2 4j?2 cos.3t* 3. Find the radius of curvature of the cycloid. Ans. 4 R sin. J d = 4 R cos. r = 2 N. 4. Find the radius of curvature of the spiral of Archi- medes. Ans. 2 (2 + .(628) 159. Corollary. The evolute of the evolute is called the second evolute·, and the evolute of the second evo- lute the third evolute, and so on. If, then, Qn is the radius of curvature of the nth evolute, and τη the angle, which the tangent to this evolute makes with the axis of a, (625) and (628) give Qn = ±IPrQ (629) Tn = T + £ n *·. (630) 160. Scholium. No more natural systeq^of coordi- nates of a curve could probably be devised than its radius of curvature, e, and the angle, *■, which its direction makes with a given direction. A curve is 21*246 DIFFERENTIAL· CALCULUS. [β. II. CH. X. Evolute. readily referred to these coordinates by the equations already given ; and from its equation referred to these coordinates, the corresponding equation of either of its evolutes is readily obtained by means* of (629) and (630). 161. Examples. 1. Find the evolute of the ellipse. Arts. The equation referred to the coordinates of § 160 is _ | A* sin. 2 τ ^ (-d2 cos.2 r -(- B2 sin.2 t) £ ’ 2. Find the evolute of the parabola of equation of § 160. Ans. Its equation is 6 p sin. τ ρ — —----------. cos.4 τ 3. Find the nih evolute of the cycloid. Ans. Its equation is ρ = 4 R COS. T , so that.it is a cycloid precisely equal to the given cycloid. 4. Find the nth evolute of the logarithmic spiral. Ans. It is a logarithmic spiral. 5. Find die evolute of the curve of which the equation results froi^^liminating φ between the two equations y = R sin. φ — R φ cos. φ x z=l R cos. ψ -j- R φ sin. φ.§161.] CURVATURE. 247 Involute of circle. Solution. We have d x — R φ d φ cos. φ d y τ=. R φ d φ sin. φ tang, τ = tang, φ, τ — φ d s = R τ d τ q zr R τ ο ± — D , τ ρ — that is, the radius of curvature of the evolute is constant and the evolute is therefore a circle.248 DIFFERENTIAL· CALCULUS. [b. II. CH. XI. Points of stopping. CHAPTER XI. SINGULAR POINTS. 162. Those points of a curve, which present any peculiarity as to curvature or discontinuity, are called singular points. 163. Whenever a function is discontinuous, the cor- responding curve found, as in $ 5, is also generally discontinuous. Thus if /. x is a function of x, which is imaginary for all values of x less than x — a = AP (fig. 55); for all values contained between x = a' = AP1 and x = a' = AP", between x = au' =z AP,U and x = a™ = AP™, and for all values greater than x = a? = APV; and is continuous for all values of x between x — a = AP and x — a' — AP'y between x = a" = APn and x = ain = APin, and between x = a™ = AP™ and x — av = ;§ 164.] SINGULAR POINTS. 249 Points of stopping. its locus is composed of the different portions MM*, MuMln, and MlVMv. If, for instance, this function were such as to have always the same value b = PM-P'W= P"N", &c. wherever it was not imaginary, the locus of y =f-x would be the portions MN\ N,,NN'^ and NiyNv, of a straight line drawn parallel to the axis of x. 164. Examples. L Construct the locus of the equation y — b= [log. (* — a)]-1 in the vicinty of the point at which it stops; and find its tangent at this point. Solution. The logarithm of a negative number is imagi- nary, and therefore the value of y is imaginary as long as x is less than a; but when x = a, we have y — b = (log. 0)”1 = od-1 = 0 y~~b, so that the point M (fig. 56), for which AP = a, PM = b is the point at which the curve stops. At this point we have, by § 108, tang. * — D . y — — [log. (x — a)]-2 (x — a)-1 = ao T = i so that PM is the tangent to the curve at the point M.250 DIFFERENTIAL CALCULUS. [B. II. CH. XI. Points of stopping. The remainder of the curve near the point M is construct- ed by finding different values of y for different values of x nearly equal to a, and drawing the curve through the points M, M', Mn, dec. thus determined. The figure 56 has been constructed for the case in which a = 2, i = 1, and, for the present example, extends to x = AP > = 2.135, y = 0.5, τ = 118° 26'. 2. Construct the locus of the equation y — b — (x — a) [log. (x — a)]"1 in the vicinity of the point at which it stops ; and find its tangent at this point. Ans. This locus is, for the present example, represented in fig. 57, from x — — qd to a? zzz d —|— 0.5 = AP*. The point where the curve stops corresponds to x = a — AP, where τ = 0, So that ΜT parallel to AX is the tangent. 3. Construct the locus of the equation y — b — (x — a — 1) [log. (x — a)]"1 and find the tangent at the point where it stops. Ans. This locus is represented in fig. 58. The point where it stops corresponds to x — a — AP, where τ = J ?r, so that PM is the tangent. +§ 164] SINGULAR POINTS. 251 Points of stopping. 4. Construct the locus of the equation y — b ~ (x — a) log. (a — a) and find the tangent at the point where it stops. Ans. This locus is represented in fig. 59. The point where it stops corresponds to x — a AP, where r = J sr, so that PM is the tangent. 5. Construct the locus of the equation y — b == (a? — a)2 log. (x — a) in the vicinity of the point where it stops, and find the tan- gent at this point. Ans, This locus is represented in fig. 60, which, for the present example, extends from x = — Qo to x=:AP/—a-\-0.223, where y~b—0.075, τζ=155° 57'. The curve stops at tSe point corresponding to x — AP = a, where T = 0, so that ilJTT, parallel to AX, is the tangent. 6. Construct the locus of the equation y — b = (x — a) [log. (cc — ct)f in the vicinity of the point where it stops, and find its tan- gent at this point. Ans, This locus is represented in fig. 61, which, for the present example, extends from x ~ — qo to x z=z AP' =z a + 0.368 ;252 DIFFERENTIAL· CALCULUS. [b. II. CH. XI. Points of stopping. it stops at the point corresponding to x — AP — a, where so that PM is the tangent at this point. 7. Construct the locus of the equation y — l· = (x — a)2 [log. (x — a)]2 in the vicinity of the point where it stops, and find the tan- gent at this point. Ans. This locus is represented in fig. 62, which, for the present example, extends from x = — qd, to x = AP1 = a -f- 0.38 ; it stops at the point corresponding to x = AP = a, where T = 0, so that MT, parallel to AX, is the tangent. 8. Construct the locus of the equation y = log. («-(-1)4. a^og. x, and find the tangent at the point where it stops. Ans. This locus is represented in fig. 63 ; it stops at the origin where the axis of y is the tangent. 9. Construct the locus of the equation y = mx log. x-f-n (a — x) log. (a — x), and find the tangents at the points where it stops. Ans. This locus stops at the points where x = 0 and x = a ;§ 164.} SINGULAR POINTS. 253 Points of stopping. at which points the values of y are, respectively, y =zn a log. a, and y τ=ζ m a log. a; and the tangents are parallel to the axis of y. Figure 64 represents this locus when a = 1, m — 2, n = 3, and figure 65 represents it when a = 1, m = 2, n = — 3. 10. Construct the locus of the equation y =: mx2 log. x -|- n {a — x) log. (a — a?), and find the tangents at the points where it stops. Ans. This locus stops at the points where x — 0 and x a; at the first of which points tang, τ — — n log. (a — x) — n, and at the second *= i Figure 66 represents this curve when a = 1, m = 2, n == 3; and figure 67 represents it when a = 1, m — 2, it = — 3. 11. Construct the locus of the equation y=f1.x(f.x)"log.f.x, in which/, x is a given function of x; and find the points where it stops. 22254 DIFFERENTIAL CALCULUS. [β. II. CH. XI. Points of stopping. Ans. It stops when f. x becomes imaginary, or when it becomes negative. Figure 68 represents this locus when f1.x = n = 1, f.x = x*—x, in which case it extends from x — — oo to x = 0, where it stops, and extends again from x = 1 = AP to x = oo The tangent at each of the points where it stops is parallel to the axis of y. Figure 69 represents this locus when n = 1, fi .x = x^ f. x — x2 — x, so that the points of stopping are the same as in figure 68. But the tangent at the point A is the axis of x, while that at the point P is parallel to the axis of y. Figure 70 represents this locus when w = 2, f x. x = 1, f. x = x2 £, so that the points of stopping are the same as in figure 68; but the tangent at each point is the axis of x. Figure 71 represents this locus when fx. x = 7i = 1, f. x = x — a?2, so that the points of stopping and the tangents are the same as in figure 68; but the curve extends from one point to the Other. Figure 72 represents this locus when n = 1, fx. x = x, f.x — x — a?2,§ 164.] SINGULAR POINTS. 255 Points of stopping. so that the points of stopping and the tangents are the same as in figure 69 ; but the curve extends from one point to the other. Figure 73 represents this locus when n — 2, fx · X = 1, /. m: X--X2» so that the points of stopping and the tangents are the same as in figure 70; but the curve extends from one point to the other. Figure 74 represents this locus when »=1,/1.* = (10*+1)-4 f.xz=zx(x—1) (a: — 2) (x — 4) (x — 5), in which case it extends from ® = 0 to ® = 1 = AP ^ where it stops, from x = 2 = AP2 to x = 4 = AP±, where it stops, from x =z 5 = APb to·® = od. The tangent at each point where it stops is parallel to the axis of y. Figure 75 represents this locus when n= l,/t .«= (10®+ l)-1 f-x~ — x (® — 1) (x — 2) (® — 4) (a? — 5), in which case it extends from x — — oo to x = 0, where it stops, from ® — 1 — AP x to x == 2 = AP2, where it stops, from x = 4 — AP± to x τ=ζ 5 == APbX where it stops; the points of stopping and the tangents are the same as in figure 74.256 DIFFERENTIAL· CALCULUS. [β. II. Cfl. AT. Points of stopping. Figure 76 represents this locus when n= 1,/j .x = —2) (x—3) (x—4) (2a?-f“^)~2 f.x=zx(x—1) (x — 2) (λ?—4) (x — 5), in which case it extends as in figure 74, and the tangents at the points Pj and P5 are parallel to the axis of y, but the axis of x is the tangent at the points A, P2, and P4. Figure 77 represents this locus when n = l,f1.x = 4(x + l)x*(x—2) (x — 3) (x — 4) (10* +1)"4 f.x——x(x — l) (* —2) (x—4) (x—5), in which case it extends as in figure 75, and the tangents are as in figure 76. Figure 78 represents this locus when »=!,/,.*=(10z+l)-i f.x = x (x— 1) (x — 2) (x—3) (a? — 4) (x — 5), in which case it extends v from x — — ao to x = 0, where it stops, from x — 1 z= APt to x — 2 = AP2, where it stops, from x = 3 = AP3 to x = 4 = AP±, where it stops, from χ — 5 = APb to x = oo ; the tangent at each point where it stops is parallel to the axis of y. Figure 79 represents this locus when η = 1, /i. * = TV (10 x -f 1)—1 /·* =— «(*—'■ 1) (*—2) (x — 3) (λ—4) (x — 5), in which case it extends§ 164] SINGULAR POINTS. 257 Points of stopping. from x z=0 to a? = 1 = -4Pn where it stops, from x = 2 = AP 2 to x = 3 = AP3, where it stops, from x = 4 = -4P4 to λ; = 5 = APb, where it stops; the points of stopping and the tangents are the same as in figure 78. Figure 80 represents this locus when » = 1, = *)**(*—2) (x—3) (x—4) (2*-}-3)-2 fx = x{x—1) (x—2) (x—3) (.r—4) (a?-—5), in which case it extends as in figure 78, and the tangents at the points Pt and P5 are parallel to the axis of y ; but the axis of x is the tangent at the points A, P2, P3, and P4. Figure 81 represents this locus when n = 1,/,. * = 4 (* +1) *2 (#—3) (*—3) (z—4) (lOx+l)-4 f.x = — x {x—1) (x—2) (a?—3) (#—4) (x—5), in which case it extends as in figure 79, and the tangents are as in figure 80. Figure 82 represents this locus when n= 1 = f.x = —6x (x—l)2 (a;—2), in which case it extends from x = 0 to x = 2 = AP2. Figure 83 represents this locus when n = l, fxx — \x {x— l)~l fx = — 6 a? (a? — 1 )2 (a? — 2), in which case it extends as in figure 82. 22*258 DIFFERENTIAL CALCULUS. [B. II. CH. XI. Points of stopping. Figure 84 represents this locus when n = 1, /,·* = 1 f.x— x+ a/x, in which case the portion AM of the curve, which cor- responds to the positive value of the radical, extends from a: = 0 to # = ao ; the portion P x 7kf1, which corresponds to the negative value of the radical, extends from x = 1 = AP1 to x = oo. Figure 85 represents this locus when n = 2, /,.i= 1 f.z = x-\- a/x, in which case the portions extend as in figure 84. Figure 86 represents this locus when n = 0, f1.x= (ia — *) f-x=+*/x, in which case the portions extend as in figure 84. Figure 87 represents this locus when «= h /i-« = log./.a: f.x = x + A/x, in which case the portions extend as in figure 84. Figure 88 represents this locus when » = 1» fx.x = l /·* = (*+ν'*)2»μβί] SINGULAR POINTS. 259 Points of stopping. in which case each portion extends from x = 0 to x — 00 . Figure 89 represents this locus when n=\, f1.x — (x-\-*/x)-' f.x = (x + V*)*, in which case the portions extend as in figure 88. Figure 90 represents this locus when n=l, f1.x = log.f.x f.x= (ζ-Ι-ν'*)3, in which case the portions extend as in figure 88. Figure 91 represents this locus when n — 0, .x — (x2 — x)2 log./.x f.x— (x-\-*/x)2, in which case the portions extend as in figure 88. Figures 92-99 represent this locus when fl.x=n=l, f.x— —*3), in which case it stops at the values «= — APf = — ν' (4 ^— a2) and x == AP" = (4 ■— a*)· The tangents at the points P1 and P" are parallel to the axis of y. - In figure 92, a — - 1.5. In figure 93, a — — 1. In figure 94, a = — 0.5. In figure 95, a =z 0.260 DIFFERENTIAL CALCULUS. [β. II. CH. XI. Points of stopping. In figure 96, a = 0.5. In figure 97, a — 1. In figure 98, a — 1.5. In figure 99, a —2. 12. Construct the locus of the equation y =/·* + mfr. x when m is infinitely small and^ .x a function, which is not infinite while x is finite, but is alternately real and imaginary. Solution. Since m is infinitely small, the part mf1.x may be neglected when^ . x is real, but when^. x is imaginary, the value of y is imaginary, so that if figure 100 is the locus of equation y=f.x; the same figure, with the dotted parts omitted, which corre- spond to the imaginary values of fx. x, represents the locus of the given equation. Thus, the locus of the equation y = \/(R2 — x2) + 0.00000000001 X aT log. z differs insensibly from the semicircle BCB* (fig. 101), of which R is the radius. But it must be remarked, that, when n is unity, the curve is suddenly turned into the form of a hook at the points B and 2?', so ^ to become tangent to the axis of y, assuming a form similar to that of the dotted line, but of indefinitely less extent. 13. Construct the locus of the equation r =/· f + »*/1·φ$ 166.] SINGULAR POINTS. 261 Conjugate points. expressed in polar coordinates, and in-which m is infinitely small, andy^. φ finite when real. Solution. If MM? MuM/n &c. fig. (102) represent the locus of r=f.z=. i?, the curve consists of several successive arcs of the same circle. 165. A conjugate point is one separated entirely from the rest of the curve, but included in the same alge- braic equation. A conjugate point is indicated algebraically by the con- dition that coordinates of this point are real, while the coordinates of no adjacent point are so. 166. Examples. 1. Construct the locus of the equation y =/· * + »n/i · x, in which m is imaginary and/,. x real. Solution. If the curve (fig. 100) is the locus of V —f·262 DIFFERENTIAL CALCULUS. [b. II. CH. XI. Conjugate points. _ and if Μ, M', &c. are the points which correspond to the abscissas, for which Λ · = 0; the locus of the given equation consists of the series of con- jugate points Jf, ΛΡ, &c. without any continuous curve. If f. x = a x -f- by all these points are upon the same straight line. 2. Construct the locus of the equation (y —/· xY + y —fi · xY = °· Solution. The sum of two squares cannot be zero unless each square is zero; so that the^iven equation is equivalent to the two equations y—fx = 0, y—fx.x — 0; that is, the coordinates of all the points of the required locus satisfy these two equations. If, then, APP1 PnP,n Ply PY &,c. (fig. 103) is the locus of the equation and if AP1 P* P ^ Pn P” Pin Piy P1Y &,c. is the locus of the equation y=fi-x> the required locus is the series of conjugate points A, P',P", P1V, &c., in which these curves intersect. Thus the locus of the equation (*-a)2 + (y — ψ = 0§ 166.] SINGULAR POINTS. 263 Conjugate points. is the single conjugate point of which the coordinates are a and b. 3. Find the conjugate points of the locus of the equation in which fx. x and f2 . x are sometimes imaginary. Solution. If fx. x is imaginary for values of x between a and 5, and, if f2-x vanishes for one or more of the values of x contained between a and b; the given equation is reduced for these values of f2 . x to y =/· *» so that the corresponding points of the curve are conjugate points situated upon the locus of the equation y=f.x. In the same way, those points of this locus are conjugate which correspond to values of a?, for which fx. x vanishes, while f2 . x is imaginary. Thus the point P', for which x=z — 1 is a conjugate point upon the axis of x in the curve of figure 76. This locus is represented in figure 104, when /.* = V'(4 — a3), f1.x — \/(l —19)> /2 .* = alog.i—1. It has four conjugate points, M\ M'u M‘\ M”, situated Upon the circle of which the origin is the centre, and of which the radius is AP = — 2. The common abscissa of two of these points is x=z — AP' = — l,264 DIFFERENTIAL CALCULUS. [b. II. CH. XI. Branch. Multiple points. Cusp. and of the other two x = APn = 1.763. 4. Construct the locus of the polar equation of example 13, § 164, when m is imaginary. Ans. It represents a series of conjugate points, upon the curve of which the equation is r =/. ip. These points correspond to the values of φ, which satisfy the equation. Λ·* = o. When /. and extends to X “ QD, y HZ QD. The least value of y in this second branch is found by § 113,to be y = P ’1M[ =2.718, corresponding to x = ΛΡ' =2.718. 4. Find the multiple point of the locus of Example 11, § 164, when n — 0, f1. x — (a:2 — x) f.x = x-\- a/x. Ans. There is a multiple point when x =z l =z AP2, y = 0, at which point the portion corresponding to the negative value of the radical begins, its tangent being P2T2 (fig. 86), drawn parallel to the axis of y, and the portion correspond- ing to the positive value of the radical passes through the same point, its tangent being P2T2, so drawn that T2P2X = 34° 44'. 23*270 DIFFERENTIAL· CALCULUS. [b. II. CH. XI. Branches and multiple points. 5. Find the multiple points of the locus of Example 11, § 164, when » = 1» h-x = log./. * f.x — ^ (*+V4 Ans. There are two multiple points; one is atP1 (fig. 113) where * = 1, y = 0; at which point the branch corresponding to the negative value of the radical begins, while the other branch passes through it; P1T1 is the tangent to the former branch, and the axis of x is tangent to the latter branch. The other multiple point is at M2^ where x — 1.96, y — 0.45 ; the value of τ for the former branch is T = 149° 15', and that for the tatter branch is T = 60° 35 . 6. Find the cusp and the other multiple point of the locus of Example 11, § 164, when n — 1, fi · x - 1, /. x = (x + \Λτ)2· Ans. The origin A (fig. 88) is a cusp of the second kind, and the axis of x is the tangent at this point. The other multiple point Mx corresponds to x = 0.328, y = 0.169. The values of T at this point are r = 69° 29', and τ = 6° 30'.§ 174.] SINGULAR POINTS. 271 Branches and multiple points. 7. Find the branches and the multiple point of the locus of Example 11, § 164, when » = 1, f1.x=(x-\- /.* = (* + V^’)2· Ans. The curve consists of but one branch, for the two portions unite in one branch at the origin A (fig. 89). The multiple point corresponds to x = 0.544, y = 0.634, at which point the two values of τ are t = 76° 35', τ=124°13'. 8. Find the multiple points of the locus of Example 11, § 164, when » = 1> fi-x = log·/· X f.x= (x-\- As/xf. Ans. The origin (fig. 90) is a cusp of the first kind, the tangent at this point being the axis of x. Mx is a multiple point corresponding to x = 0.142, y=z 0.465, at which point the two values of r are τ=114°37;, τζ=2Γ40/. M2 is a multiple point corresponding to x z— 0.544, y — 0.402, at which point the two values of τ are r = 53°17/, τ = 172° 29'. 9. Find the multiple points of the locus of Example 11, § 164, when272 differential calculus, [b. ii. ch. xi. Branches and multiple points. n — 0, x — (xQ — a?)2 log. /. x f.x=(x + */xf. Ans. The origin (fig. 91) is a cusp of the second kind, the tangent at this point being the axis of x. M i is a multiple point corresponding to * = l, y = 0, at which the axis of x is the common tangent to the two branches of the curve, and the contact of the two branches is of the second order. 10. Has the locus of Example 11, § 164, any cusp, when η— 1, frx=i£x (a?— I)-1, f.z =—6a’(i— l)2(a?—2) ? Ans. It has none. 11. Find the multiple point of the locus of Example 11, § 164, when fx.x = n = 1, f.x=a-\-a/(4 — x*). Ans. When a is zero, or negative, there is no multiple point. When a is positive, and less than e"1 = [2.71828]-1 = 0.3679, the curve consists of a single branch without any multiple point. When a — e~l = 0.3679 the 'curve consists of three branches, as in (fig. 114), and has two cusps of the second kind, corresponding to ιη±2, y = — a=z — 0.3679.§ 174.] SINGULAR POINTS. 273 Branches and multiple points. When a is greater than e~l and less than the curve consists of one branch with two multiples, as in figure 115, where a = 0.4, and the two multiple points correspond to x = ± 1.984, y — — 0.275. The values of τ at one point are T zz 102° 38', and τ = 97° 45', at the other T — 77° 22', and rz=82° 15'. When a — ^ — 0.5, the curve (fig. 96) has two multiple points at the beginning and end of its branch, corresponding to x — zh 1.937, y = 0. The values of r are T —90°, T —90°d=45°. When a is greater than ^ and less than 2, the curve con- sists of a single branch, with no multiple points. When a zz 2, the curve (fig. 99) is an oval. 12. Construct the locus of Example 11, § 164, when ft.x — η — 1, /. χ zz a + \/(a2 — ff2). Ans. Where a is greater than the curve is an oval, as in figure 99, where When a -z 2. a — ± — 0.5,274 DIFFERENTIAL· CALCULUS. [b. II. CH. XL Branches and multiple points. the curve (fig. 116) consists of a single continuous branch, which returns into itself; and it has a multiple point at the origin, where the curve has a contact with itself, the common tangent being the axis of x. When a is less than ^ and greater than e-1^: 0.3679, the curve consists of a single continuous branch, which re- turns into itself; and has two multiple points, as in figure 117, where a = 0.4, the two multiple points correspond to « = db 0.31, y — — 0.27. The values of τ at one point are τ — 144° 53', τ—131° 41', and at the other T = 35° 7', T = 45° 17'. When a = e~l = 0.3679, the curve (fig. 118) consists of two branches and two cusps of the first kind, corresponding to x = a, y — — . 16. Construct the locus of the equation y2 = x3 — a;4, and find its cusp. Ans. This locus (fig. 123) consists of a single branch, which has a cusp of the first kind at the origin. 17. Construct the locus of the equation y* = x4 — a?2, and find its branches.976 differential calculus. [b. II. CH. XI. Branches and multiple points. Ans. This locus (fig. 124) consists of two branches, one of which extends from x — — qo to x — — 1, and the other from x = 1 to x — od, and a conjugate point, which is the origin. 18. Construct the locus of the equation — x2 — £4, and find its multiple point. Ans. This locus (fig. 125) consists of one branch, which returns into itself, and has a multiple point at the origin, where the values of τ are t = ± 45°. 19. Construct the locus of the equation = x4 — x6, and find its multiple point. Ans. This locus (fig. 126) consists of a single branch, which returns into itself, and has a multiple point at the origin, where it has a contact with itself. The tangent at the origin is the axis of x. 20. Construct the locus of the equation ?/2 — — x. Ans. This locus (fig. 127) consists of an oval, which extends from§ 174.] SINGULAR POINTS. 277 Multiple points. X — — 1 to X 0, and a branch which extends from x = 1 to x = 00. · 21. Construct the locus of the equation — x5 — x3, and find its cusp. Ans. This locus (fig. 128) consists of a branch* which extends from x = — 1 to x = 0, where there is a cusp and a branch, which extends from x = 1 to x = oo. 22. Construct the locus of the equation (1—a®)8, and find its multiple points. Ans. This locus (fig. 129) consists of two branches, which extend from x — — 1 to x — 1 They cross at the origin, where the values of τ are t = ±45°, and there are two cusps of the first kind, corresponding to x = zh 1. 23. Construct the locus of the equation y* = x4(l_aP)8, and find its multiple points. 24278 differential calculus, [b. II. CB. XL Multiple points. Ans. This locus (fig. 130) consists of two branches, which extend from a? = — 1 to a? — 1. There are two cusps corresponding to the two values of x% and the origin is also a multiple point, where the two branches are in contact. 24. Construct the locus of the equation, x j^=(2 — **)(l—*)(* — *), and find its branches. Ans. This locus (fig. 131) consists of a succession of three ovals, which extend respectively from x =. — \/2 to x = —1 from x — — s/\ to x m */% from x =z 1 to x = \/2. 25. Construct the locus of the polar equation r = a -f- sin. m and find its multiple points and branches. Ans. If m is an integer and a greater than 1, this locus is oval, as in figure 133, where a = 2, m = 3. If m is a fraction and a greater than 1, this locus is a curve, which returns into itself after as many revolutions of the radius vector as there are integers in the denominator of 171. Thus, in (fig. 134), a = 2, m =§ 174] SINGULAR POINTS. 279 Multiple points. there is a multiple point corresponding to φ = 0 or = 360°, r = 2, at which point the values of ε and τ are ε =t = 75° 58', and£ = T- 104° 2'. In (fig. 135) a = 2, m = f, there are three multiple points corresponding to φ = 0° or == 360°, φ= 120° or=480°, = 315° or = 1035°, φ = 495° or = 855”, for each of these points r -= 0.5, ε = 40° 54', and e = 139° 6'. In (fig. 152) a = 1, m — the curve has two contacts at the origin; the two values of r at this point are * '=22° 30', τ= 1123 30'. There are eight other multiple points; four of these points correspond to φ =* 22° 30' or = 382° 30', Φ = 292° 30' or = 652° 30' φ = 562° 30 or = 922° 30', ψ = 112° 30' or = 832° 30', for each of these points r = 1.5, * = 52° 25', and e = 127° 35'; four points correspond to286 DIFFERENTIAL CALCULUS. [b. II. CH. XI. Multiple points. φ = 157° 30' or = 517° 30', φ = 427° 30' or — 787° 30' φ = 697° 30 or 1057J 30', φ = 2473 30' or = 967° 30', for each of these points r = 0.5, ε zn 23° 25', and ε = 156° 35'. In (fig. 153) a 1, m = §, the curve consists of two distinct branches, which are in contact at the origin ; the value of φ at this point is φ = 135°. There are eight other multiple points ; two of these points correspond to φ — 315° or = 1035% φ == 135° or = 1215% for each of these points r = 1.809, s = 82® 36, and ε = 97° 24'; two points correspond to φ =z 45° or = 405% φ = 945° or = 1305°, for each of these points r = 1.309, ε = 73° 48% and ε = 106° 12'; two poipts correspond to φ = 495° or = 855% φ = 1395° or = 1755°, for each of these points r = 0.691, ε — 61° 10% and ε = 118° 50'; two points correspond to φ = 765° or = 1485% φ = 585° or = 1665°,H74] SINGULAR POINTS. 287 Multiple points. for each of these points r = 0.191, = 39° 5', and e = 140° 55 . When a is less than unity, the origin is a multiple point, and several branches of the curve stop at this point, if the nega- tive values of the radius vector are neglected, while they continue through it if these values are retained. This ex- ample, therefore, furnishes an analytic exception to the method of avoiding negative radii vectores given in B. I. § 45. In the following figures the dotted portions correspond to the negative radii vectores. In (fig. 154) a = J, » = 1, the two values of * at the origin are τ = 30°, *=150°. In (fig. 155) a = J, m = 2, the two values of τ at the origin are r ±= 105°, * = 165°. Whether the dotted parts are included or not, the curve is continuous. In (fig. 156) a = J, m = 3, the six values of τ at the origin are T— 10°, T— 50°, T= 70% T — 110% T=130% τ = 170\ In (fig. 157) a = m = 4, the four values of τ at the origin are288 DIFFERENTIAL· CALCULUS. [b. II. CH. XI. Multiple points. T= 52° 30', T= 82° 30', T= 142°30', τ—172° 30'. In (fig. 158) a = }, m = J, the two values of τ at the origin are r = 60°, τ = 120°. There is another multiple point corresponding to Ψ = 0° or = 360°, r = 0.5, τ = 45° and = 135% or <£>=540% r= — 0.5, r=90°. In (fig. 159) a = £, i» = J, the six values of τ at the origin are T= 20% * = 40% *= 80% T = 100°, τ = 140°, τ = 160°. There are three other multiple points, corresponding, res» pectively, to φ = 0° or = 360% φ = 12° or=480% φ = 240° or = 600% for each of which r = 0.5, or to φ = 180% φ = 420% φ = 660% for each of which r = — 0.5; at each of these three points the values of ε are e = 18° 26% £ = 161° 34% and e = 90°. In (fig. 160) «=£, m = the curve has a contact with itself at the origin, the tangent§ 174.] SINGULAR POINTS. 289 Multiple points. at this point is perpendicular to the axis. There are three other multiple points ; one corresponds to φ = 90°= 450°, r = 0.5, £ = 60°, and £ = 120; the two others correspond to φ = 555° 48', φ= 1064° 12% r = 0.408, or to φ = 735° 48', φ = 883" 12', r = — 0.408, and at one of these two points, ε z=z 129° 7', and £ = 71° 8'; at the other, £ = 50° 53', and £ = 108° 52'. In (fig. 161) a = J, m = f. The curve consists of two ovals and a continuous re-entering branch, which has a contact with itself and with each of the ovals at the origin; the value of τ at the origin is t = 45°. There are two other multiple points, corresponding to φ = 225° = 585°, φ = 45° = 765°; at each of these points r = 1, £ = 60°, and £ = 120°. In (fig. 162) a = £, The curve has several contacts with itself at the origin ; the values of t at this point are τ = 67°30', 7= 157*30'. 25290 DIFFERENTIAL CALCULUS. [b. II. CH. XI. Multiple points. There are four other multiple points, corresponding to 0 = 22° 30' or =382° 30', 0=292° 30' or =652° 30' 0= 562° 30' or =922° 30', 0=112° 30' or =832° 30', at each of these points r = 1, e = 40°54', and e = 139<>6'. In (fig. 163) a = m = The two values of τ at the origin are t = 60°, *=120°. There are four other multiple points ; one corresponds to 0 = 0° or =720°, r — 0.5, T = 63°26' and *=116°34'; one point corresponds to 0 = 180“ or = 540”, r = 1.207, t = 81°40' and * = 98»20'; one point corresponds to . 0 = 761” 4', r = 0.322, ε=127°14', or to

=.±n. 180% in which n is any integer. 175. When a curve is continuous at a point, but changes its direction so as to turn its curvature the opposite way at this point, the point is called a point of contrary flexure, or a point of inflexion. Thus M (figs. 166- 199) is such a point. 176. Problem. To find the points of contrary flexure. Solution. It is evident, from the comparison of the two tangents Mf Τ' (figs. 165-169), and M" T,f near ilf, that the value of the angle MTX or τ is either a maximum or a minimum at the point M. The points of contrary flexure correspond, therefore, to the maxima and minima of the angle T. 177. Corollary. When the equation of the curve is given in rectangular coordinates, we have by (549) tang, t = D.y;SINGULAR POINTS. 293 § 179.] Multiple points. so that the maxima and minima of r correspond to those of D. y, except at those points where τ is a right angle. 178. Corollary. It is evident, from (figs. 166-169), that the convexity of a curve is turned towards the axis of x when the angle τ (or its supplement, if the curve is below the axis') increases with the increase of x\ otherwise the convexity is turned from the axis. 179. Examples. 1. Find the point of contrary flexure in the locus of example 1, § 164, and the tangent at this point. Solution. We have, in this case, D . y— — [log. (x — a)]-2 (a; — a)~l D2.y — (x—a)"2 [log. (x—a)]“3 [log. (x — a) + 2]; so that the point of contrary flexure corresponds to the point Μ" (fig. 56) for which log. (x — a) -j- 2 = 0 x = a-f-0.135, y~h — 0.5 T —61° 38'. 2. Find the point of contrary flexure in the locus of example 2, § 164, and the tangent at this point. Ans. It corresponds to a? = a + 7.387, y = J +3.694, r — 26° 33'. 3. Find the point of contrary flexure in the locus of example 3, § 164, and the tangent at this point. 25*294 DIFFERENTIAL CALCULUS. [B. II. CH. XI. Multiple points. Ans. It corresponds to x~a-\- 1, y~b-1- 1, τ — 26°33'. 4. Find the point of contrary flexure in the locus of example 5, § 164, and the tangent at this point. Ans. It corresponds to x=a + 0.223, y = b — 0.335, τ = 155° 57'. 5. Find the point of contrary flexure in the locus of example 6, of § 164, and the tangent at this point. Ans. It corresponds to x = a-\- 0.340, y = b + 0.085, τ 135°. 6. Find the points of contrary flexure in the locus of example 7, § 164. Ans. There are two which correspond to x — a+ 0.683, y = l· + 0.068, r = 162° 8' x = a + 0.073, y — b + 0.035, τ = 31° 6'.§ 180.] APPROXIMATION. 295 Approximate value of an explicit function. CHAPTER XII. APPROXIMATION. Almost all theoretical results, when converted into numbers, are insusceptible of exact expression, and can only be obtained approximatively. Hence, in all its practical applications, ready and rapid means of obtaining approximations are the only object of the exact science of mathematics ; and the great labor, which has been bestowed upon this subject, is the dis- tinguishing characteristic of the modern science. 180. Problem. To obtain by approximation the value of an explicit function. Solution. The only useful method of accomplishing this object is to arrange the function in a series of terms, which are susceptible of easy calculation and decrease as rapidly as possible. I. When the variable is very small, the function is, at once, arranged by means of MacLaurin’s Theorem (447) in a series of terms, which are multiplied by the successive powers of the variable, and are, therefore, usually de- creasing. II. When the values of the function and its differential coefficients are known for a given value of the variable; the function can, for another value of the variable, which differs but little from the given one, be arranged, by means296 DIFFERENTIAL CALCULUS. [β. II. CH. XII. Approximate value of an explicit function. of Taylor’s Theorem (445), according to the successive powers of the difference between the two values of the variable. III. Besides the formulas thus obtained, other formulas can often be found, by processes dependent upon the nature of the functions and the tact of the geometer; and some formulas, often of great use, will be given in the Integral Calculus. Scholium. Formulus (478, 484, 487, 492, 493, 500, 501, 504, 509, 513, 515), are examples of useful developments. 181. Problem. To obtain, by approximation, the val- ues of an implicit function, when its value is known to differ but little from that of a given explicit function. Solution. Let x = the required implicit function t = the given explicit function x — t = e = the excess of x above t. Find the algebraic equation for determining e, and let it be reduced to the form e — F x, where F x is a small function of which we may denote by a z, in which a is any small quantity, and % the function of x obtained by dividing e by a; we have then, e z=z Fx — a z (631) x = t-\-e=zt + Fx=:t + az. (632)§ 181.] APPROXIMATION. 297 Lagrange’s theorem. Now we have by MacLaurin’s Theorem for any function u of x if we develop it according to powers of a, and denote by m0, B.au&c., the values of w, B.au, &,c., when a — O u — u0 —J- B,au0.a + Tfi ^auQ. — -f- D3. auo· 3 &c* Again, if we put (633) BtXuz=zu^B,xu0z=z υ/0, (634) we have by (566), B.tu = u' B tx, (635) B.au = ur B,ax. (636) But the differentiation of (632) gives, by putting zf=B,xz, (637) Β,αχ=ΐζ-\-αΒ,αζζ=ι zaz1 B maXy (638) whence D-I=T^ir <«») In the same way, the differential coefficient of (632), relatively to £, is B'txz=i \ aB tz \ -\- az1 B tx, (640) whence Β·*=Γh? <641> and, therefore, B,ax=iz Bttx B . aUZZZ U* Z B 'tX, (642) (643)298 DIFFERENTIAL· CALCULUS. [β. II. CH. XII. Lagrange’s theorem. The differential coefficient of this last equation, relatively to is D.aD'tu=2D.t(u' zD,tx), (644) or D. a(u' D ,tx) =z D 't(u' z D, tx), (645) in which any function whatever of z may be substituted for u By substituting for u' in (645) the function z11 u' of a?, we have D .a (z”w/ D.t*) = &u*Dmtx). (646) Now the successive differential coefficients of (643), rel- atively to a, are by (646) D2 au=zD a{z ulD tx) = D.t(z* u'D.tx) (647) D*.au = D'tD,a(z* u'D.tx) = D*.t(z* u'D mtx), (648) and in general D\n= DtDa (zn~' uf D. t x) = ΙΡγ1 (zn u> D,t x). (649) Now if in (641 and 646) we take a = o we have D.tx0=l D„u0= D\>u„=, d:^ = d:t'(w0z”), (650) whence by (631) aD ,„u0= au‘0z0 = u'oF.t (651)§ 181.] APPROXIMATION. 299 Lagrange’s theorem. a* &au0=B.t (a2u'0 %\) = D t [u'Q (F. tf\ an Ό:αηυ = Dn~t (an u'e zna) = Dn~l [uf0 (F. i)2], (652) which, substituted in (633), give D,K(F. ψ] u = u0-\-ufoF.t-{- 1.2 ■ &c., which is called Lagrange's Theorem. Corollary. If u =. a?, w' = 1, u0 = t, and (650) becomes D.'(F.t)* D2t{F.ty (653) (654) x = £ + P.i + : 1.2 1.2.3 ·- + &c. (655) Corollary. If instead of (632), a; had been the given function of £ -f- a 2 1 =/· (* + <» *)» (656) we might have put xl — t -j- a z, (65?) and u would have been a function of a?', and that if such η = φ.χ (658) we have u = ..λ. ri Λ- Jr'Vms :vr3Γ VΨiidinys.__PL ,17 Γ.