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Slats 
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 wh os: ' eT i Neer: ‘ 
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 Pia J : : "ps 
 
A TREATISE 
 
 ON 
 
 ANALYTICAL GEOMETRY 
 
 OF 
 
 THREE DIMENSIONS, 
 
 CONTAINING THE 
 
 THEORY OF CURVE SURFACES, 
 
 AND OF 
 
 CURVES OF DOUBLE CURVATURE. 
 
 DESIGNED FOR THE USE OF STUDENTS IN THE 
 UNIVERSITY. 
 
 malin 
 By JOHN HYMERS, ALM. 
 
 FELLOW OF ST. JOHN’s COLLEGE: 
 oa i] >* “ 
 
 CAMBRIDGE: 
 Printed by J. Smitu, Printer to the University ; 
 
 FOR J. & J. J. DEIGHTON, CAMBRIDGE; 
 AND C. J. G. & F. RIVINGTON, LONDON. 
 
 M.DCCC.XXX. 
 
ich abi! 
 Koes eh 
 
 Degen 
 Leh STIR 
 
 ERO 
 
ADVERTISEMENT. 
 
 As there is no separate Treatise upon Geometry of Three 
 Dimensions in the English language, and as in those books 
 where it is introduced as a part of the Differential Calculus, 
 but a small and insufficient space is allotted to it, the Author 
 hopes that the one now offered to the public may be found 
 useful in supplying the deficiency. The Work, being de- 
 signed principally for the use of Students in the University, 
 is divided into Propositions, the most important of which 
 are illustrated by Examples ; but compression has every where 
 been attended to, as far as strictness of demonstration would 
 permit. With a view also to the convenience of the reader, 
 the easier propositions have been made to precede the more 
 difficult; and a very copious Table of Contents has been pre- 
 fixed, which exhibits at once the substance and extent of 
 the Work. Those who wish to prosecute the higher parts 
 of the subject, must have recourse to the works of Euler, 
 Monge, and Dupin, especially the Application de Analyse 
 a la Geometrie of Monge, from which a portion of the matter 
 
 of this publication is extracted. 
 
 St. Jonn’s COLLEGE, 
 March, 1830. 
 
ERRATA. 
 
 Line 
 last line, for axis read axes. 
 5 from bottom, for Z= read z=. 
 last line, for cc’ read 1. 
 15, for mx—y read mx -+-y. 
 6 from bottom, for y= fa read z=fax. 
 c?e? 
 8, for «=—e? read pm ae 
 1 and 3 from bottom, for M read M’. 
 11, for Mread M’. 
 7 from bottom, for —«?(a?+ b?) read — x?(c?-+ 62). 
 last line, for da + By+Cz=D read Aax+ Bby+Ccz=D. 
 5 from bottom, for (4) @-2 read ( Ye z)’. 
 13, for (z,— 2)” read (z,—2z)z”. 
 10, for Art. 43 read Art. 48. 
 | £7 COS? « : 8? COS? x 
 
 THN Sie ara 
 
 6 from bottom, for 
 
SECTION . I. 
 
 On THE PLANE AND STRAIGHT LINE. 
 
 &RT. 
 
 SO. 
 
 . ExpLANATION of three rectangular co-ordinates, and of the 
 
 equation to a SULLAC Oss, Por gis ca ae aihis Pte eae a Ty on iN 
 
 . To find the distance of a point from the origin, both when the 
 
 co-ordinates are rectangular, and oblique; also the distance be- 
 tween two given points. Hence the equation to the surface of a 
 SPUClOa ls wpatat a's tes < a RUSE ee wks amcaadan «0 a Palaeie wae: 
 
 . To find the equation to a plane, either from considering it as the 
 
 locus of a point which is equidistant from two given points, or as 
 cenerated by a straight line which moves parallel to itself, along 
 a straloht: lirte S1VEM, 1D) POSTON. + yw 2 sid qarscomes qiste's HG + oid 
 
 . To find the equations to a straight line in space. Hence the equa- 
 
 tion of condition for the intersection of two given straight lines. . 
 To find the conditions that a straight line and plane may be pa- 
 FHISIOr COMCIGENE tat Grote wea ed pein tosis « ae Be te Bs td ere 
 To find the seatiionee that a straight line and plane may be per- 
 pendicular to one another. Through a given point to draw a 
 straight line perpendicular to a plane. ..........-.0...024. 
 
 . To find the angles which a given straight line makes with the 
 
 axes ; or, which a given plane makes with the co-ordinate planes. 
 
 knowing their equations, or the angles which they respectively 
 
 PAGE 
 
 1b. 
 
 . To find the angle between two given straight lines, either from’ 
 
 make with the axes. Hence the condition that they may be at. 
 
 right angles. Re eS Chee mnae oe care etl dcavestol eaeap bal Ghost hoetnt Dix ie a me cia 
 
 . To find the angle between two given planes, and the condition 
 
 that they may be at right angles. ........ 02.00.30. eeeees 
 
 To find the angle between a straight line and a plane. ........ 
 a 
 
 (oa) 
 
il CONTENTS. 
 
 ART. PAGE 
 11. To find the distance of a point: from a plane... ......-+..006- 10 
 12. To find the distance of a point froma line. .......... Hig in YEA 
 
 13. To find the equations to a straight line which cuts perpendicu- 
 larlypeach ofstwogivempetuGbe dines soe eS... ee oe ee BO 
 
 14, To determine the shortest distance between two given straight 
 lines, and to find an expression for its length. The same geo- 
 
 FILCUTICHUIVORB MMR RL 2's pee cients seta sw wena, 11 
 
 eoeeresneeever ee ee # © 
 
 meee 
 
 SECTION II, - 
 
 ON THE ORTHOGONAL PROJECTION OF AREAS, AND EQUATIONS 
 TO SURFACES. 
 
 15. ‘The projection of an area upon any plane, is equal to the area 
 multiplied by the cosine of its inclination to that plane. ....... 14 
 
 16. The square of any plane figure, is equal to the sum of the squares 
 of its projections on three co-ordinate rectangular planes. The 
 pyramid whose vertex is the origin, and base any plane figure, 
 is equal to the three pyramids whose common vertex is any point 
 in the plane of the figure, and bases its three projections. Hence 
 the equation to the plane which passes through three given points. 
 Given the lengths of three conterminous edges of a triangular 
 pyramid which are mutually at right angles, to find its base, 
 height, and the inclinations of its faces,.............+...0.. tb. 
 
 17. To find the sum of the projections of any number of areas upon 
 RC LVET DING. i, Wik) sean oe A ASI | ahaa ea aN A 16 
 
 18. To determine the plane of a maximum projection. ‘The sum of 
 the projections on any other plane, varies as the cosine of its in- 
 clination to the plane of maximum projection. .............. ib. 
 
 19. To find the equations to the surfaces of an oblique cone, and an 
 oblique cylinder, on circular bases. .., 2.22 ..00seeeeeeeeees 18 
 
 20. To find the equation to any surface of revolution about one of the 
 cb-ordinateiaxes: 40... 020.020.0904 tee Geen ore te tN BER 2b. 
 
 21. To find the equations to the ellipsoid, and elliptic paraboloid. .. 19 
 
 22. Examples of finding equations to surfaces from their geometrical 
 LES IULIOIL ty ay 5s onde | eh CH Re io. | ce oe 20 
 
CONTENTS. ill 
 
 ART. PAGE 
 I. Ifa straight line have three given points in three fixed 
 planes, the locus of any other point in it is a surface of the 
 SOMME SCRT Auld toed uke ret ot? citi a (ks cite adie Bis eas aerate ey 
 
 IT. Two points move uniformly in straight lines, to find the 
 equation to the surface described by the line which joins them. 21 
 III. ‘Two planes are drawn through two given straight lines, 
 so as to be always perpendicular to one, another, to find the 
 equation to the surface described hy their intersection. ..... 22 
 
 IV. ‘To find the locus of a point which is equidistant from 
 two straight lines given in space. .......... Agee apres s. 2b. 
 
 V. To find the equation to the surface described by a 
 
 straight line which always meets three fixed straight lines. 
 
 The same surface might be generated by a straight line con- 
 stantly passing through three different straight lines. ...... 23 
 
 23. To find the equations and forms of the three surfaces of the 
 second order that have a center, ..s..10. -.0000s0 bebe gia), PR: 
 
 24. To find the equations and forms of the two surfaces of the second 
 
 order that have not a center. These are particular ‘cases of the 
 POLICY «evs ete! 9 94 sity atin sii ele inlbla vis Vakekiniierttes eis SYSTEM ss ee 
 
 SECTION III. 
 
 On Tancent Pianes, NorMarts, VOLUMES, AND AREAS OF 
 SURFACES, AND TANGENTS, AND NorMAL PLANES, &c, OF 
 Curves oF DouBLE CURVATURE. 
 
 25. To draw a tangent plane to a curve surface. To find its inclina- 
 tions to the co-ordinate planes, and the volume of the pyramid 
 which it forms with them; also the length of the perpendicular 
 dropped upon it from the origin... 20.55 24,5« omelette. sige OO 
 
 26. Ata given point of a curve surface, to find the line of greatest 
 inclination, and the curve traced on the surface to which it is 
 always a tangent. Ex. To find the curve of greatest inclination 
 Of Cheer aCe, OF CUO IPSOI 7's 0,0 ofa says vi aidie als, ea eg stetde boc 
 
 27. To draw anormal to a curve surface. It is the longest or short- 
 est line that can be drawn from any point in itself to the surface. 34 
 28. To find the length of the normal, intercepted between the surface 
 _ and the co-ordinate planes, and the points where it meets them, 35 
 
iv CONTENTS. 
 
 ART. PAGE 
 29, Examples of drawing tangent planes, and normals to surfaces... 35 
 
 I. ‘To draw a tangent plane to an ellipsoid.........-.-- 20. 
 
 II. To find the length and the locus of the extremity of the 
 perpendicular, dropped from the center of an ellipsoid upon its 
 TANSEML PADS Bir eUNY POWER LL pial, ieee cipete ei plae\s lnaly 6 oe 36 
 
 Ill. If three tangent yee to an ellipsoid are mutually 
 at right angles, their point of intersection will trace out a 
 sphere concentric with the ellipsoid. In the case of parabo- 
 loids, the locus is a plane.......... Bye ine eta stot et aBeLsda. sik 37 
 
 “IV. Jf three planes, eutally at Ae angles, eactly 
 ‘touch the perimeter of a plane curve of the second order, to 
 find the locus of their point of intersection. ........++,+.. 39 
 
 V. The sum of the squares of the reciprocals of any three 
 semi-diameters of an ellipsoid, mutually at right angles, is 
 equal to the sum of the squares of the reciprocals of the 
 BOIMI-B KER. CHG Ste TESS ASS byte ne sian sees Mclelelele wae el setye ) 40 
 
 VI. _ If two concentric surfaces of the second order have the 
 same foci for their principal sections, they will intersect had 
 CTRL TIPNbengies, eet waar ee eth a tc e ee cis Mirtle pee” ) 2Up 
 
 VII. To draw a normal -to an ellipsoid. If normals are 
 drawn from the points in which any plane cuts the ellipsoid, 
 to find the locus of the points in which they meet the plane 
 Be ee a Te Bia Nicos fe he,» eatehos swat dep MEd 
 
 VIII. ‘To draw a tangent plane to the surface whose equa- 
 tion is Va 4+ y+ Vz=Va. The portions of the axes 
 intercepted between the origin and tangent plane have their 
 BUM CONSTANG, cic o's sip sp es vm Ore opie Swe o preje'g ge spt wien) 42 
 
 30. ‘To define a curve of double curvature, and shew how its equa- 
 tiGDS ere Ou. pevelsastte weenie ee aa 
 
 31. To find the equations to the tangent line of a curve of double 
 CUrValUres 2S 0e. S 
 
 32. To draw a normal plane to a curve of double curvature. ...... 76, 
 
 33. To find the equations to the line of intersection of two consecu- 
 tive normal planes, and the equation to the osculating plane.... 45 
 
 34, Examples of finding equations to curves of double curvature, and 
 drawing their tangents, normal planes, &c. ..........0+-24- 46 
 
ART. 
 
 4 
 ox 
 
 ts: 
 
 N 
 
 CONTENTS. 
 
 I. A sphere being pierced by a cylinder, to find the equations 
 to the curve of intersection, and to draw a tangent line, nor- 
 mal plane, and osculating plane to it. : 
 
 II. ‘To find the equations to the helix, .............. 
 
 III. -To find the equations to the path of a point moving 
 uniforrily along a great circle of a sphere which revolves uni- 
 formly about a diameter; and the area included between the 
 path, and a great circle perpendicular to that diameter. .... 
 
 IV. To find the equations to the Screw of Archimedes. 
 
 . To find the differential coefficients of the volume, and curvilinear 
 
 surface of any figure bounded by planes parallel to the co-ordinate 
 planes, and a curve surface whose equation is known. In some 
 cases it is convenient to express the differential of the volume 
 by polar co-ordinates. To find the differential coefficients of the 
 volume and surface of a wedge contained by the plane of za, a 
 plane parallel to’ zy, and a a through the axis of z, and by a 
 given curve surface. ........4.6. 
 
 cee ezeeeefeeeesee epee eeosa eve 
 
 . To find the differential coefficient of the arc of a curve of double 
 
 CUT VAUUEOs. aia ec ote<.x 
 
 eoeeeorvr esc eeee ee ee eee weerewmeseerteseeeesgese 
 
 . Application of the formulz for volumes and surfaces to examples. 
 
 I. ‘To find the volume and surface of any portion of a sphere. 
 
 II. To find the volume of any wedge of an ellipsoid inter- 
 cepted by two planes: passing through the axis.’ 
 
 IIf. To find the surface of an ellipsoid which differs little 
 from a, sphiereeparacense id. duie ik Wek MA. PN SMR APS 
 
 IV. <A sphere being pierced by a cylinder, to find the vo- 
 lume and surface of the portion of the sphere intercepted by 
 the cylinder 
 
 eoeeeceoeeeeeeeet*ereev et Pease eseseeeeaeeeoee @e ee 
 
 _  V.. To find the volume of the cono-cuneus of Wallis 
 
 VI. The axes of two equal cylinders intersect at right 
 angles, to find the volume of the portion which is common 
 to both, and the surface of one cylinder intercepted by the 
 OUDETs axe 6 
 
 VII. ‘To determine the surface of an oblique cone on an 
 elliptical base, in certain cases. .........4.. 
 
 oem ee ee eer eee 
 
 VIII. To determine the curve to be traced on the surface of 
 a sphere, so that its length shall always be equal to (7) times 
 the co-ordinate (x) of the describing point..... 
 
 eoeeo2es 
 
 49 
 
 50 
 
 54: 
 
 ib. 
 
 ab, 
 
 55 
 
 56 
 
 58 
 60 
 
 2b. 
 
 61 
 
 64 
 
vi CONTENTS: 
 
 SECTION IV. 
 
 On TRANSFORMATION OF CO-ORDINATES AND ITS APPLICATIONS. 
 
 ART. PAGE 
 38. To pass from one system. of co-ordinates to another, supposing 
 the first rectangular, and the second oblique. We may obtain 
 
 expressions free from equations of condition...............+- 606 
 
 39. To pass from one system of co-ordinates to another, supposing 
 both rectangular; the six equations of condition in this case may 
 be replaced by six different equations.” ....... 0c ce ee cece ee 67 
 
 40. When both systems are rectangular, to express the formule in 
 terms of the inclination of the axis of 2’ to the plane of xy, and 
 of the inclination of its projection on that plane, to the axis of x. 68 
 
 41. To find the equation to the section of a surface made by any 
 DIST Cui & hs alsin ohnte Min Revita eihiisdectia ae Wii eaiek ee, Rubee. aie tee OG 
 
 Ex.]. ‘To find the equation to the section of an oblique cone, 20, 
 
 IJ. Of all sections of a paraboloid of revolution made by 
 planes passing through a tangent to its base, to’find that 
 MANIC IAS SLL MD TOU CCE LM ABCE Ua tat e's nitty sucnr cas va ay aie ster) CO 
 
 III. To find the equation to a section of a surface of the 
 second order, made by any plane, .............. A Pees eam fa | 
 
 42. To find the co-ordinates of the center of a surface of the second 
 order; and the relation among the coefficients of the general 
 equation, when tlie surface it represents has not a center, .,... 72 
 
 43, To find the equation to a diametral plane of a surface of the 
 second order, that is, to the locus of the middle points of a sys- 
 tem of parallel chords. — If the surface have a center, any diame- 
 tral plane is parallel to the tangent plane applied at the extremity 
 of that diameter to which the bisected chords are parallel. .. :, 73 
 
 Ad, There is an infinite number of systems of oblique co-ordinates, 
 but only one system of rectangular co-ordinates, which will cause 
 the terms involving xy, xz, yz, to disappear from the general 
 equation of the second. order. That equation, except when it 
 represents a cylinder, can always be reduced to one of the forms 
 
 Av+ByY4+C2=D, ByY+4+C242Ar=0. 74 
 
ART. 
 45, 
 
 46, 
 
 47. 
 
 48, 
 
 CONTENTS. 
 
 In a surface of fie second order that has a center, the sum of 
 the squares of any system of conjugate diameters, and the 
 volume, and the sum of the squares of the faces, of the paralle- 
 lopiped described upon them, are invariable. Hence to find the 
 locus of the extremities of the equal conjugate diameters of an 
 ellipsoid. Of all systems of conjugate diameters of an ellipsoid, 
 the principal diameters have their sum a minimum, and the 
 equal diameters their sum a maXimuM...........ee ee eeeee ; 
 
 Every surface of the second order, except the hyperbolic para- 
 boloid, may be generated in two different ways by the motion of 
 a circle of variable radius parallel to itself, the center of the 
 circle moving along a diameter of the surface. For the ellipsoid, 
 hyperboloid of one sheet, and hyperboloid of two sheets, the cir- 
 cular sections are parallel to the mean axis, the greater of the 
 real axes, and the greater of the imaginary axes, respectively ; 
 for the elliptic paraboloid, they are perpendicular to the principal 
 section whose latus rectum is least. ............0. Papeete sas 
 
 To find the relation among the coefficients of the general equa- 
 tion of the second order when it belongs to surfaces of revolution, 
 and the equations to the axis of revolution. ................ 
 
 Problems on surfaces of the second order, illustrating the preced- 
 THEE PFODOSILIONS., 201s wl eles AO eRe stale igs Awe. Atte tees 2 - 
 
 I. Three straight lines mutually at right angles always meet 
 the perimeter of a given plane curve of the second order, to 
 find the locus of their point of intersection. ... ........... 
 
 ‘Tl. If three straight lines mutually at right angles be applied 
 ‘to a surface of the second order, to find the locus of their point 
 Ghantérsectiany. Ve ek oe naan wae a Wena a eer « ° 
 
 III. If a surface of the second order be cut by a plane 
 which always passes through a fixed point, to find the qocus 
 of the centers of all the plane sections. ......... 0... eee 
 
 ae To find the locus of the middle points of chords of a 
 surface of the second order, all passing through a given 
 POWta EME Sse slat ir Gxses cee iersiaiaeite cordate eee Tae 
 
 ) V. ‘The curve of contact of a surface of the second order, 
 . and of a cone, the vertex of which is given, is a plane curve; 
 _ and if the plane of contact always pass through a fixed point or 
 
 17 
 
 80 
 
 82 
 
 84 
 
 2b. 
 
 88 
 
 89 
 
CONTENTS. 
 
 a fixed line, or constantly touch a surface of the second order, 
 the locus of the vertex of the cone will be respectively a 
 plane, a straight line, or a surface of the second order, and 
 Conversely), /TA, = 5ehcthis chev pla eiete Cpateoude seksbu inte Wataeis 5's cin higeed 
 
 VI. ‘The line joining the center of a surface of the second 
 order, and the vertex of the enveloping cone passes through 
 theicentexpiitheicurveiot contacts u. sxtiy nee teks os wk 
 
 Senne: “aunenanE 
 
 SECTION V. 
 
 PAGE 
 
 89 
 
 On THE CURVATURE OF SURFACES AND CuRVEs OF DOUBLE 
 
 49. 
 
 ap 
 
 CURVATURE. 
 
 To find the requisite conditions, in order that two surfaces may 
 have a contact of the first, second, &c., order. Number of dis- 
 posable constants necessary for a contact of any order......... 
 Ex. I. To find the equation to the plane that at a given 
 point of a surface, has with it a contact of the first order.... 
 II. Given the radius, to find the co-ordinates of the center 
 
 -of a sphere, that has a contact of the first order with a surface 
 
 ALAN YAO be atstliste leo ahinie wR oie Hoth ie tis wieasimicie eefeeaeee 
 
 . To find the radius. of curvature of. any.normal section at any 
 
 point of a curve surface. ‘The sum of the curvatures of any two 
 normal sections at right angles to one another, is invariable, at 
 the same point of A surfaehe een sees mn cian te atoms ee 
 Of all normal sections of a surface at any point, to determine 
 
 _those of greatest and least curvature; and to shew that they are 
 
 Or 
 to 
 
 54. 
 
 at right angles to one another............... UY ie Simeden* 
 
 . To express the radius of curvature of any ‘section, in terms of 
 
 the greatest and least radii of;curvatore. ..... 0+. cenwenan 
 To determine a paraboloid, the vertex of which shall have a 
 complete contact of the second order with a given surface at a 
 
 DISPOSED. PONT es oh axel aly colts an Aare m0 i re eaere 
 If a surface be cut by a plane, parallel and indefinitely near to the 
 tangent plane at any point, the section is ultimately a curve of 
 the second order. It is called the indicatrix of the surface, and 
 will be an ellipse or hyperbola, according as the curvatures are 
 in the same direction, or opposite directions, and a parabola 
 
 ‘ 
 
 94 
 
 95 
 
 96 
 
 1b. 
 
 98 
 
 99 
 
ART. 
 
 56. 
 
 57. 
 
 58. 
 59. 
 60. 
 
 61. 
 
 CONTENTS. 
 
 when one of them is evanescent. At points where the indicatrix 
 is a circle, a sphere can have a complete contact of the second 
 order with the surface ; these are called umbilici of the surface ; 
 an ellipsoid has four such points, symmetrically placed in the 
 principal section containing the greatest and least axes. ...... 
 
 . To find the radius of curvature at any point of an obhque sec- 
 
 tioMop aicurversurfacels 2. Fee) ) wi ool ieoneewosie 
 
 Having given a point on a curve surface, to find the directions 
 in which we must pass to consecutive points, in order that the 
 corresponding normals may intersect. .......... 2.0 cece ee 
 
 To find the radius of curvature of any section of a surface of 
 TOVOLION UR. "20st Fh Mab clot dal a UA het ha Fe PAS LY NS 
 Ex. To find the radius of curvature of any normal sec- 
 
 tion, at a given point of an oblate spheroid............ te 
 To determine the lines of curvature of a given surface, ...... 
 To determine the lines of curvature of an ellipsoid. ......... 
 
 To construct the projections of the lines of curvature of an ellip- 
 soid on the plane containing the greatest. and mean axes, . .. 
 
 If the plane containing the greatest and least axes be taken for 
 the plane of projection, the lines of both curvatures are pro- 
 jected into ellipses, and are all inscribed in the same equilateral 
 De VAL GIOCEANY seach: dipea Sold ieee 6 tiene gah ats oe. hae 
 
 . To determine the radii of curvature at any point of a surface, in 
 
 terms of the co-ordinates of that point.. ..........-+.00.05- 
 
 Ex.I. To find the radii of curvature at any point of an 
 ellipsoid. The result shews that at points where the tan- 
 gent planes are equidistant from the center, the product of 
 the radii is constant; and that at points which are them- 
 selves equidistant from the center, the sum of the radii varies 
 inversely as the distance of the tangent plane. Given the 
 radii of curvature at any point, and the distance of the tan- 
 gent plane from the center, to find the volume of the ellipsoid. 
 To find the points at which the radii of curvature are equal 
 to one another. ...... 
 
 ° II. To find the radii of curvature at any point of a para- 
 
 DOR»: «33 whe. a ae aint 8 “aa or Sars Bird. Viti tty ePahanss 
 
 ix 
 
 PAGE 
 
 100 
 
 2b. 
 
 101 
 102 
 103 
 2b. 
 105 
 
 107 
 
 108 
 
 109 
 
 110 
 
ART. 
 
 63. 
 
 64. 
 
 65. 
 
 66. 
 
 67. 
 68. 
 
 09. 
 
 70. 
 
 71. 
 
 72. 
 
 CONTENTS. 
 
 Hence to find the equations of condition in order that the cur- 
 vatures of any surface may be in the same, or opposite directions ; 
 or that they may be equal and in the same direction, or equal 
 ana ansopposite directionsyiyh,< sis Urls Ty Sb... ees 
 
 Properties of the line of intersection of two consecutive normal 
 planes of a curve of double curvature, and of the developable 
 surface generated by it. Centers of absolute and spherical 
 CUPM ALN CRAM REGIE em eT eE Ngo ot ema ase 
 
 To find the equation to the surface generated by the perpetual 
 intersections of the normal planes of a curve of double curvature. 
 
 To find the radius, and co-ordinates of the center of spherical 
 curvature; the results are simplified by making the tangent line 
 UNO FAXIG OL ie sate td 5a sakes Ae aN os AL IL VS eatin 
 
 To find the radius and co-ordinates of the center of the absolute 
 circle of curvature. The expressions are greatly simplified by 
 making s, the length of the arc, the independent variable in- 
 stead of x; or by taking the tangent line for one of the co- 
 Ordilinte mimesy ii! F454 te South Sisk Ut. se aGie, aild. Soo testaig o, 
 
 To find the points of inflexion of a curve of double curvature. 
 
 The first flexion of a curve of double curvature is equal to the 
 second flexion of the edge of regression of the developable sur- 
 face of its normal planes ; and the first flexion of the latter is 
 equal to the seconid flexion of the former... 2.5.2... eee. ce ee 
 
 Every curve, whether plane or of double curvature, has an 
 infinite number of evolutes, all of which le on the developable 
 surface generated by the intersections of the normal planes. ... 
 
 The locus of the absolute center of curvature is not an evolute, 
 except the curve bea platie curve. 700" 00 I eas ae 
 
 To find the equation to any proposed evolute of a curve of 
 MOUDIe CUDVALUT Er AEM. co cate /akie ates hips cing’ s MMU srs) ow os 
 
 Exemplification of the theory in the case of the curve of 
 double curvature, resulting from the intersection of a sphere 
 BEdLerUndets day sice fre dias bmbe hates 4a vitipiia* 2.419 pig ae 
 
 . Examples of finding the involutes of curves of double curvature. 
 
 I. The involute of the curve of equal inclination, that is, 
 one which is traced on any surface in such a manner that its 
 
 PAGE 
 
 113 
 
 114 
 
 115 
 
 116 
 
 117 
 119 
 
 120 
 
 hea 
 
 122. 
 
 123 
 
 123 
 123 
 
ART. 
 
 74. 
 (2: 
 76. 
 
 77. 
 78. 
 
 79. 
 
 CONTENTS. 
 
 tangent line is inclined at a constant angle to the plane of xy, 
 is a plane curve. ‘To find the equations to the curve of equal 
 HCH DES Sas sl ea itale ae os ae meg a PY ie ly al 
 
 II. When the curve of equal inclination lies on a surface 
 of revolution, its equation is readily obtained by polar co-or- 
 dinates. If the surface be a paraboloid, the projection of the 
 curve of equal inclination on the plane of the base, is the 
 involute of a circle; if a cone, the projection is an equi- 
 
 angular spiralyed a ge a aoe ia ween Atte Aaa di ona Be, 
 III. Example in which the involute lies on the surface of 
 @ soherea. xe Seg. ain garph aasiedaints ao Sei alas were te dae oie 
 
 pee Se ee 
 
 SECTION VI. 
 On SURFACES IN GENERAL, 
 
 To find the general equation to cylindrical surfaces. ... .. 
 To find the differential equation to cylindrical surfaces. . 
 Given the direction of the generating line, to find the equation 
 to the cylindrical surface which envelopes a given curve surface. 
 Ex. To find the equation to the cylindrical surface which 
 envelopes a given oblate spheroid. The cylinder meets the 
 tangent plane at the pole of the spheroid, in an ellipse, of 
 which that” pointiis, Ch6, fogusy Wy sw. cae Ma ee laa: 
 To find the general equation to conical surfaces............- 
 Application of the last article to find the projection of a curve 
 updn any plarieR Ce Hv koe i. tA ATE Seis Oat fick: ote 
 Ex. To find the stereographic, gnomonic, and orthogra- 
 phic projections of-any circle of a sphere. .........0006- 
 The gnomonic, orthographic, and stereographic projections of 
 the rhumb line on the surface of a sphere, are three of Cotes’s 
 
 SD ei tate Els gk Biats seule ots ane oats yh 
 
 If an angle on the surface of a sphere have one of the arcs 
 which contain it a meridian, the tangent of its stereographic 
 projection is a mean proportional between the tangents of its 
 orthographic and gnomonic projections. Hence we can deter- 
 mine the orthographic and gnomonic projections of any angle 
 whatever on the surface of a sphere, .......... fe Pe sy ee 
 
 PAGE 
 
 125 
 
 126 
 
 127 
 
 129 
 130 
 
 130 
 
 131 
 132 
 
 133 
 
 134 
 
 135 
 
xil CONTENTS. 
 ART. PAGE 
 80. To find the differential equation to conical surfaces........... 138 
 
 81. Given the position of the vertex, to find the equation to the 
 conical surface which envelopes a given curve surface........ 20. 
 
 Ex. To find the equation to the conical surface which 
 
 envelopes a given surface of the second order. .........+-- 139 
 
 82. To find the general equation to conoidal surfaces............. 140 
 Ex. When the directrix is a helix, to find the equation 
 
 tothe conoidal surtace. 275i dd vre''s Siles U8 sb eee hes 141 
 
 83. To find the differential equation to conoidal surfaces. ......... ab, 
 
 84, To find the general equation to surfaces of revolution.......... 142 
 
 Ex. To find the equation to the surface generated by the 
 revolution of any straight line about the axis of z. The sur- 
 face is an hyperboloid of revolution..............-..0.66- 2b. 
 
 85. To find the differential equation to surfaces of revolution....... 143 
 
 86. A curve surface revolves round an immoveable axis to which 
 it is fixed, to find the equation to the surface which touches and 
 envelopes it AN eVeEry POSILION se ei ss ke awe pe winds so eck eet a's 144 
 
 Ex. When an oblate spheroid revolves about any di- 
 POUT Meas artis a piel ais fom ie pe teil ele lei a ce eens ete, yee 145 
 
 87. To find the equation to a surface which envelopes a series of 
 surfaces produced by giving all possible values to a parameter 
 contained in their common equation; and the equation to the 
 characteristic, or curve in which any given surface in the series, 
 is intersected by the consecutive one. .......-...00eereseees 2b. 
 
 68. To find the equation to the edge of regression of the envelope 
 
 considered. In the. preceding Brticles >. ig ieswasiars'niels di’ »:+ eye ble mie 146 
 $9. Examples:of the two last articles, 9... si5...22 00.0 500.305.. 147 
 I. To find the equation to the envelope of all right cones 
 Of a Constant volume. <.ssc ace eee sees Veen bene 5 OY 2b. 
 II. If the center of a sphere describe a plane curve, to 
 find the equation to the annular surface which envelopes it 
 INVEVETY POSIUON. +. eee. pat eer tee ee ree sa cee 148 
 
 III. Supposing the vertex of a right cone to describe a 
 curve in the plane of ay, and its axis to be always perpen- 
 dicular to that plane, to find the equation to the envelope... 149 
 
ART. 
 
 90. 
 
 Ol. 
 
 92. 
 
 CONTENTS. 
 
 If the differential equation to the envelope be known, the equa- 
 tions to the characteristic may be deduced from it. .......... 
 
 Every partial differential equation of the first order, may be 
 considered as belonging to the envelope of surfaces represented 
 by F(a, ¥,2,@,p~a)=0. Hence the use of the characteristic 
 in the integration of partial differential equations. ............ 
 
 Ex. I. To integrate the general linear equation of the 
 first order 
 
 ceoeeevr eee eee eeeeeeaeeeeveoe eevee ee eee eee eee eee eee 
 
 II. To integrate the equation Pp-+ Qq=0. ......... 
 
 III. To find the equation to the surface which cuts at 
 right angles all surfaces comprised in the same total differ- 
 ential equation. » <a. sty 2. PVE ON, HESS ee waltla 
 
 IV. To find a surface where the portion of the axis of z 
 between the origin and tangent plane, bears a constant ratio 
 to the distance of the point of contact from the origin. .... 
 
 Case in which p and q can each be determined in terms of the 
 co-ordinates, and the problem leads to a total differential equa- 
 LGU tele aban cure, fe en ciel sigh eke Mb diate Shaiya al Seatac Ge « apalaln’s wis ayo 
 
 Ex. I. To find a surface such that the normal at any 
 point passes through the center of gravity of the triangle 
 whose sides are the co-ordinates w and y of that point. .... 
 
 IJ. To find a surface which belongs both to conical sur- 
 faces, and surfaces of revolution about the axis of z 
 
 93. To determine the envelope, when the equation to the series of 
 
 surfaces contains two independent parameters 
 
 Ex. J. To find the equation to the surface touched by a 
 series of planes so drawn that the product of the perpendicu- 
 lars let fall upon them from two given points is invariable. . 
 
 I]. To find the surface which envelopes one whose equa- 
 
 mM m mM 
 
 : seal y z : ss ; 
 tion Is me rein = 1, in all the positions it can. assume, 
 
 subject to the condition a” + b” +c” = k”, or abc = h3, where 
 k is invariable 
 
 IIT. To findthe surface touched by a series of planes 
 which retrench from a given right cone, an oblique cone such 
 that its volume is constant, or such that the transverse axis 
 of its elliptic base is constant.¢..s.cercevsccceccorer 
 
 xu 
 PAGE 
 
 150 
 
 151 
 
 152 
 153 
 
 2b. 
 
 154 
 
 155 
 
 2b. 
 2b. 
 
 156 
 
 157 
 
 158 
 
 159 
 
Xiv CONTENTS. 
 
 ART. 
 94. To find the general equation to developable surfaces. ....... 
 
 95. Upon the assumed hypothesis, they may be made to coincide 
 in all their points with a plane, without tearing or rumpling... 
 
 Ex. Ifa series of planes passing through a fixed point 
 in the axis of z, have their traces on the plane of ry all 
 equal, to find the surface formed by their intersections. ... 
 
 96. 'To find the differential equation to developable surfaces, and 
 to their characteristic. The differential equation may also be 
 obtained by eliminating the arbitrary functions from the equa- 
 ONS £0 Lhe eeneraing, NG. sh betas With tehsil 's'e co bble + =: 
 
 97. To find the equation to the developable surface which touches 
 at the same time two given curve surfaces... .......+.250%- 
 
 9s. A curve being traced on a developable surface, to find its 
 nature when the surface is made plane; and conversely... ... 
 
 Case I. When the surface is cylindrical. Character- 
 istic equation to curves which after developement become 
 BILAL Lines late 9% peat HCO e ta Sie. oes, « sal aeee ule 
 
 Ex. A curve traced on the surface of a cylinder has its 
 projection on a plane through the axis a circle, to find its 
 nature when the cylinder is developed... ............... 
 
 99. Case II. When the surface is conical. Characteristic 
 
 equations to curves which after developement, become equi- 
 angular spirals, or straight lines...............0220000. 
 
 100. A given curve being traced on the surface of a right cone 
 whose base is a circle, to find its equation when the surface is 
 developed scandconversdly! CONDI PS 20. OLE Sa ae ees oe 
 
 Ex. I. Ifasemi-circle be described on the bounding ra- 
 dius of a quadrant: of a circle, to find its projection upon 
 the base, when the quadrant is formed into a cone; also to 
 determine the projection of the chord of the quadrant. . ... 
 
 Ex. II, A right cone being intersected by a sphere 
 whose center is in its surface, to determine the equation to 
 the curve of intersection when the cone is developed, and 
 to trace Itun the several Cases 0. rat. ee a, ono aac ele 
 
 101. A curve of double curvature traced on a surface of revolution 
 always cuts the generating curye at the same angle, to find 
 
 ’ 
 
 161 
 
 166 
 
 167 
 
 169 
 
 2b. 
 
CONTENTS. XV 
 
 ART. PAGE 
 its equations. Examples, when the surface is a sphere, a 
 right cone, or an oblate spheroid. 0.65.02. .u. sco pececcaes 173 
 
 102. ‘To draw the shortest line between two given points on a sur- 
 face of revolution. Application to the cases of a paraboloid, 
 sphere, and oblate spheroid... ............ ola’ ode! SR eee cuales « 175 
 
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ANALYTICAL GEOMETRY 
 
 OF 
 
 a SE 
 
 THREE DIMENSIONS. | 
 
 ¢ 
 5 fy 
 
 SECTION I. 
 
 ON THE PLANE, AND STRAIGHT LINE: 
 
 1. Iw order to determine the position of a point in space, 
 some fixed point is taken for the origin of the co-ordinates, and 
 through it are drawn three fixed planes at right angles to one 
 another, which are called the co-ordinate planes; and their in- 
 tersections are called the axes of the co-ordinates. ‘Then the 
 distance of a point from each of the co-ordinate planes being 
 given, it is manifest that its position is completely determined. 
 Thus the pomt A (fig. 1.) being the origin of the co-ordinates, 
 AB is called the axis of x, AC the axis of y, AD the axis of 
 z3 also the plane CAB is called the plane of xy, BAD the 
 plane of xz, and CAD the plane of yz. The co-ordinates of 
 the point M are ABor EM=2, BN or FM=y, and MN=z; 
 and if a given equation exist between these three quantities, and 
 their values be always taken so as to satisfy it, the different 
 positions of M thus determined will all lie in a curve surface, 
 whose nature is defined by the equation, and which may be traced 
 out*. 
 
 * Since the co-ordinates x, y, z, are measured from the point 4 
 along the lines 4x, Ay, Az, if any co-ordinate be measured in the 
 contrary direction, it will be negative. Thus, if a point be situated 
 below the plane of xy, its co-ordinate parallel to Az will be —z; and 
 
 A the 
 
2 
 
 If in the equation to a surface we make x constant and 
 =ON=c, (fig. 3.) we obtain the equation to the plane curve 
 PQ, which is the intersection of a plane, drawn parallel to that 
 of yz at a distance (c) from it, with the surface; and by making 
 r, Y, 2, separately =Q, we find the equations to the plane curves 
 AC, AB, BC, formed by the intersections of the surface with 
 the co-ordinate planes, which are called its traces on those 
 planes. As many real values as z has for given values of 2 
 and y, so many sheets will the surface have; if z is impossible, 
 there is no point in the surface corresponding to those values. 
 
 2. Since (fig. 1.) AN*= AB’ + BN’, 
 and AM*=AN*°+ NM’, 
 
 ye Ree Ai +y°+2° the distance of the point M from the 
 origin. 
 
 Also if AM=r, and a, $B, y, be the angles which AM 
 makes with the axes of x, y, z, respectively, then « =r cosa, 
 y=rcosP, z=rcosy, 
 
 *, r =r’ cos*a+r cos B+r° cos ¥, 
 or cos’a-+cos’ B+ cos’ y=]. 
 
 Suppose wv’, 7’, z’ the co-ordinates of a second point. Then 
 if the origin be removed to the first point, the co-ordinates of the 
 second point will be v’'—x, y'—y, z —2z; and therefore its 
 distance from the new origin, that is, the distance between the 
 
 two points = 4/ (2° — 2)’ +(y —y)? +(2 — 2)”. 
 
 Hence also, since all its points are equidistant from the 
 center, the equation to a spherical surface whose radius is (7) 
 
 the co-ordinates of a point situated in the compartment 2’y'z’, will all 
 be negative. In some cases it is requisite to take the co-ordinate 
 planes, not rectangular, but inclined at given angles to one another. 
 Thus, (fig. 2.) if «Ay, zAx, zAy, be three planes intersecting in 
 the lines Ax, Ay, Az, and MN, NB be drawn respectively parallel 
 to Az, dy; AB, BN, NM are the three co-ordinates of the point 
 M referred to the oblique axes Az, Ay, Az. 
 
3 
 
 and the co-ordinates of its center x’, y’, 2’, is(w— a’) +(y—y')? 
 +(z—- z')? =r’, which becomes x” + y + z° =r, when the center 
 is taken for the origin. 
 
 If the co-ordinates are oblique (fig. 2.) let 2rAy=y, 
 ZvAz=a, LyAz=f, and let AB=z, BN (parallel to 
 Ay)=y, NM (parallel to Az) =z, be co-ordinates of the point M; 
 
 “. P =AN?+NM’?+2AN. NM coszAN; 
 but AN’=2°+y'?+22ry cosy, 
 
 and AN cos zAN=An=Am+mn=zx cosa+y cos f, hak 
 Bm be perpendicular to Az; 
 
 *. P=ae ty? +2°4+ ry cosy +2xz cosa +2yz cos B, 
 
 which gives the distance of a point from the origin, and also the 
 equation to a sphere whose radius is (7), 
 
 3. To find the equation to a plane. 
 
 Der. A plane is a series of points, of which each is equi- 
 distant from two given points. 
 
 Let one of the points A (fig. 4.) be the origin, and let 2’, y’, 2’ 
 be the co-ordinates of the other point M; and let P be a point 
 in the plane whose co-ordinates are w, y, 25 
 
 * APP =x? +y? +27, 
 and PM*=(2' —2)*+(y' -— yy) +(2’— 2); 
 and these by the definition are equal, 
 pyr ter=(al—axyt+(y—yy +(e 2) 
 or by reduction 2(ra! + yy +22))=2"? +y? +2", 
 the equation to the plane, which is of the form z=Ax+By+C. 
 
 Hence every equation of the Ist order involving three variables 
 is that to a plane. 
 
 Cor. 1. AM is manifestly perpendicular to the plane, and 
 is bisected in the point Q in which it meets the plane. Let 
 therefore AQ=6, and let the angles which it makes with the 
 axis of x, y, z be a, B, y, 
 
4 
 
 then x’ =2d cosa, y'=205 cos B, z'=20 cos ¥; 
 , 40 (4 cosa+y cos B +2 cos y)=40, 
 
 or x cos a+y cos B+ cos y=, another form of the equation 
 to a plane. 
 
 Cor. 2. Ifa, b, ¢ be the parts of the axes of x, y, Z, in- 
 tercepted between the origin and the plane, a= sec a, b=6 sec B, 
 c=6 sec ry, and therefore the equation to the plane may also be 
 
 Ty 4 Bz 
 put under this form, —- +> + -=1. 
 ST ATS vars 
 
 Cor. 3. We may also define a plane as generated by a 
 ‘straight line, which moves in a direction parallel to itself, along 
 a straight line given in position. 
 
 Let AB (fig. 5.) in the plane of xz be the line to which the 
 generating line PQ is always parallel, and AC in yz the line 
 along which it moves. 
 
 Let z=Ax+C be the equation to AB, 
 and z= By+C be the equation to AC, where C=O4A; 
 
 then PQ being parallel to 4B, we have PN=A,MN + QM, 
 but QM=B.MO+ OA, .. PN=A,.MN+B,MO+0OA; 
 
 or z=Ax+ By +C, the equation to the plane. 
 
 AB, AC, are called the traces of the plane on the co-ordinate 
 planes of rz, yz. Hence we perceive the meaning of the con- 
 stants in the equation z= Ax+ By+C, for C is OA the part 
 of the axis of z intercepted between the plane and the origin, 
 and A and B are the tangents of the angles of inclination of the 
 traces AB, AC, to the axes of x and y. If AC be parallel to 
 OM, B=0; in this case the plane is perpendicular to that of rz, 
 and its equation is z=Axr+C, the same as that to its trace 4B. 
 If the axes are oblique, let 4zOr=a, 4zOy= 8; and let 
 the inclinations of AB, AC to the axis of z be respectively d 
 and @; then in the same manner as above, it may be shewn that 
 sin (a—@) sik sin(B—9) | 
 sin @ sin 0 
 
 Hence the equation to the plane always preserves the same form. 
 
 the equation to the plane is z =a 
 
5 
 
 4, To find the equations to a straight line in space. 
 
 A straight line is the intersection of two planes, and con- 
 sequently is given, when the equations to any two planes which 
 contain it are given. 
 
 For the sake of simplicity, to represent a straight line we 
 employ the planes drawn through it perpendicular to the planes 
 of xz, yz; and the equations to these planes are called the 
 equations to the straight line. 
 
 Thus let NC, BQ, (fig. 6,) be the intersections with the 
 co-ordinate planes xz, yz, of planes drawn through any line MP 
 respectively perpendicular to them. Then NC, QB, are 
 called the projections of the line MP; and the equations to 
 the planes, MN, MQ, or, which is the same thing, to the pro- 
 jections NC, QB, are the equations to the line MP, which are 
 consequently of the form r=az--a, y=bz+f. 
 
 It is easily seen that a and # are the co-ordinates AN, NP, 
 of the point P, in which the straight line meets the plane of xy, 
 and a, 0b, are the tangents of the angles which its projections 
 
 NC and QB make with the axis of z. 
 
 By eliminating z we find ay— bx =a8—ba, which is the 
 equation to the projection on the plane of ry. When the 
 co-ordinates are oblique, the equations to a straight line will 
 manifestly retain the same form. 
 
 Cor. Ife=azta r=adzt+a 
 
 , , 
 
 y=bz+PB y=0z2+h 
 
 two lines which intersect, then these four equations must be 
 
 satisfied by the same values of x, y, z, namely, by the co-ordinates 
 
 of the point of their intersection. Hence eliminating 2, y, z, 
 
 we find the equation of condition for the intersection of two 
 straight lines 
 
 (a -a)(O- b= (B' — B)(a’ — a) =0. 
 When this is satisfied, the co-ordinates of the point of intersec-— 
 tion are the values of x, y, z, given by the above equations. 
 
 be the equations to 
 
 5. Given the equations to a straight line and plane, to 
 find the conditions, that they may coincide, or be parallel to. 
 one another. 
 
6 
 
 Let z= Ar+By+C be the equation to the plane, 
 a=az+a, y=bz +f the equations to the straight line ; 
 
 Substituting az+a for 2, and bz+ for y, in the first, 
 
 | 7(Aa+ Bb—1)+ Aa+ BB+C=0, 
 
 and since this equation must be true for all values of z, 
 “,4a+Bb=1, Aat+BB+C=0, 
 
 are the required equations of condition. 
 
 If the plane and straight line are parallel, then a plane and 
 straight line drawn through the origin respectively parallel to 
 them, coincide; therefore the above equations must be satisfied 
 when we suppose, a, (9, C, to vanish. Hence the remaining 
 equation of condition is Aa+ Bb—1=0. 
 
 6. Given the equations to a straight line and plane, to find 
 the conditions that they may be at right angles to one another. 
 
 Let the straight line OP, (fig. 7,) be a perpendicular from the 
 origin on the plane ABC; from P draw PN perpendicular to 
 AOB, and join ON, then ON is the projection of OP. Also 
 OC, PN, being parallel, are in the same plane, and therefore 
 the lines CP, ON being produced will meet in Q; then the 
 plane COQ is perpendicular to both the planes AOB, CAB, 
 and therefore AB, the intersection of the two latter planes, is 
 perpendicular to COQ (Euclid XI. 19); and AQis at right angles 
 both to OQ and CQ. Hence when a straight line is perpen- 
 dicular to a plane, the trace of the plane and the projection of 
 the line on any one of the co-ordinate planes, are at right angles. 
 Also the Z CQO measures the inclination of the plane to that of 
 xy, and is eqnal to 4 COP which the perpendicular makes with 
 the axis of z. Now let 2=Ax+ By +C be the equation to the 
 plane, r=az-+a, y=bz +, the equations to the line. Then 
 the trace of the plane on rz has for its equation z=Ar+C, 
 and this is perpendicular to the projection of the line on rz, 
 
 Sate oe te A ee 
 whose equation is z =~ — ~;therefore— +1=0. Similarly the 
 a a a 
 
7 
 
 straight lines whose equations are z= By+C and y=bz +, 
 
 B 
 are at right angles, and therefore b +1=0. 
 
 Hence d+a=0, B+b=0, are the required conditions. 
 
 Cor. Hence we may draw a straight line perpendicular to 
 a plane through a given point. 
 
 Let 2’, y’, 2’ be the co-ordinates of the given point, r=az+ta, 
 y=bz+ 3, the equations to the line; therefore 2’ =az’+a, 
 y =bz' +B, and therefore eliminating a and B, x — «’ =a(z — 2’) 
 y—y =b(z—2'). But the line being perpendicular to the 
 plane whose equation is 
 
 z=Ar+By+C, a=—A, b=—B, 
 
 therefore x —a' + A(z—2/)=0, y—y'+ B(z—2z)=0, are the 
 required equations. 
 
 7- Given the equations to a straight line, to find the 
 angles which it makes with the axes; also to find the angles of 
 inclination of a plane, whose equation is given, to the co-ordinate 
 planes. 
 
 Let r=az+a, y=bz+f be the equations to the given 
 line. Through the origin draw a line CA, (fig. 8,) parallel to it, 
 then the equations to CA are r=az, y=bz. 
 
 Let a, 8, yy, be the angles which CA makes with the axes 
 of x, y, 2; also let CA=1, and x, y, z be co-ordinates to the 
 point A, 
 
 then r=cosa, y=cosP, z=cosy. But AC°=2°+y' +2’, 
 
 1 
 or 1 =(g" + b° + 1)2*, ere Le 
 J/1+a?+o°’ 
 1 B Or G 
 e COS SY SS —— cos b=) ———_— ,—, 
 - J ita +e?’ Jitaet 
 
 a 
 
 Jitete 
 
 cosa=>= 
 
5 
 
 Next let z= Axr+ By+C be the equation to a plane pér- 
 pendicular to CA. Then y, (, a are the inclinations of this 
 plane to the planes of ry, xz, yz, respectively; alscoa= — A, 
 b= — B, Att: Os 
 
 iy + 
 cosa= 5.0089 = — 
 nf 1+ +B? J 1+ A? + B 
 
 1 
 
 iy Wiha death Be: 
 
 8. Given the angles which two straight lines make with 
 the axes, to find the angle which lines parallel to them through 
 the origin include. 
 
 Let CA (fig. 8.) make with the axes of x, y, z angles a, 8, y;, 
 and CB make with them angles a’, (3, y’; take CA=1, and let 
 z, y, z be co-ordinates of A, then r=cosa, y=cos[, 
 z=cos yy; similarly let CB=1, and 2’, y’, z’ be co-ordinates of 
 B, then «’ =cosa’, y' =cosP', 2’=cosy. 
 
 Join A Beand lett 2 ACR= C, AB ec. 
 
 then 2 cos C=1°+ 1 —°?=2 — f(x—a'P +(y—y' 
 +(z—z)}=Qaer'+ yy +227); 
 *, cos C=cos a cos a’+cos 3 cos BP +cos ¥y cos 4’. 
 
 But if c=az+a, r=az+d 
 
 y=bz+B, y=Uzt+Ph 
 lines to which CA, CB, are respectively parallel, 
 
 be the equations to the 
 
 then r=uaz, y=bz are the equations to CA, 
 and w’=a’z’, y'=b'z' are those to CB; 
 1 
 
 * l= ty4t2=2114+0+0), &«. 2= —, 
 saa E | Tee 
 
 ae ; 1 
 sitiilarly, 2. 
 /1 +a? +02 
 aa +bb'+1 
 
 COS C= aa +bb +cc)z27 = a ee 
 . Sata? +b +a? +b") 
 
9 
 
 Cor. Hence the equation of condition, in order that the 
 
 lines may be at right angles, is 
 
 aa +bb'+1=0; 
 at which we may arrive immediately, by observing that the 
 equation to a plane through the origin perpendicular to the first 
 line is z+ax+6y=0, which must necessarily contain a line 
 through the same point parallel to the second, whose equations 
 are r=a2z, y=Uz, ! 
 
 *, aa +bb'+1=0. 
 
 9. To find the angle of inclination of two planes whose 
 equations are given. 
 
 Let z=Ar+By+C, z=A' r+ By +C’, be the equa- 
 tions to the two planes. Then the equations to two lines, re- 
 spectively perpendicular to them through the origin, are 
 
 = — Az y= — Bz; r='— Az, y= — Baz (Art. 6.); 
 and the Z (0) of inclination of the two planes is manifestly equal 
 to the angle between these two perpendiculars; therefore (Art. 8.) 
 
 1+AA’+ BB’ 
 
 J (1 + A+B) i+ A? + BY”) 
 
 Hence, in order that two planes may be perpendicular to one 
 another, we must have 1+AA’+ BB’=0. Also, if a, B, y3 
 a’, 2’, y’, be the angles at which the planes are inclined to the 
 co-ordinate planes of yz, rz, xy; these are also the angles 
 which the perpendiculars on them from the origin respectively 
 make with the axes of x, y, z, and therefore (Art. 8.) 
 
 cos 0= 
 
 cos 9=cosa cosa’ + cos PB cos 8’ + cos y cos ¥’. 
 
 10. ‘To find the angle between a straight line and a plane. 
 Let z=Ax-+ By+C be the equation to the plane; then 
 the equations to a straight line, through the origin perpendicular 
 to it, are r= —Az, y= — Bz. Let r=az+a, y=bz+f 
 be the equations to the given line. ‘Then, if @= Z which the 
 straight line makes with the plane, 90 — @ is the angle which it 
 makes with the perpendicular to the plane, and therefore 
 
 1— Aa— Bb 
 J + a?+ 8?) (14+ A?+ B’) 
 B 
 
 cos (90 —9) or sinO= 
 
10 
 
 11. To find the distance of a point from a plane. 
 
 Let x’, y's 2’ be the co-ordinates of the givell point P, (fig. 9.) 
 z=4Ax+ By+C the equation to the given plane ABC. 
 Through P draw a plane GK perpendicular to the line AB; 
 then this plane is also perpendicular to the given plane ABC. 
 (Euc. x1. 19.) Let GH be the intersection of the two planes, 
 to which draw PR perpendicular. Then P& is perpendicular 
 to the plane ABC; and PR=PQ sin PQR= PQ cos GHK. 
 
 But PQ=PN-QN=2 —-(Ar'+ By'+Q), 
 I 
 
 and cos GHK = ——————., 
 / 1+ A’ + B? 
 
 (Atta); 
 
 z' — Ax'— By’—C 
 
 .. PR the required distance = a 
 /i+A4+ B 
 
 12. ‘To find the distance of a point from a line. 
 
 Let P (fig. 10.) be the point, whose co-ordinates are 2’, y, z; 
 AB the line, whose equations are v=az+a, y=bz+P; PB 
 a perpendicular from P on AB=0. 
 
 Then z—2'= —a(x—2')—b(y—y’) is the equation to a 
 plane through P, perpendicular to AB; therefore (Art. 11.) 
 AB a tet SBE + ax EERE 
 
 JSita+b? | 
 and AP=+/ (x —a)y? +(y'— BY + gin’ 
 128 Phos? ae Ar 
 
 {2 +4 (2 —a) + b(y'’— B)\? 
 
 ee eee Ne pe over 78 ou 
 (x —ay+(y Py +2 pate? 
 
 B 
 
 or se @ az — a? + (y'—b2'—BY + (6@'—a) —a(y'—B)}* 
 l+a° +b" | ° 
 
 13. To find the equations to a straight line, which cuts 
 perpendicularly each of two given straight lines. 
 
11 
 
 Let r=az+a, v=a'z+a’ 
 
 y=be+B, y=l24+ 8 
 
 given lines; then, if z=>Ax+By+C be the equation to a 
 plane which is parallel to them both, 
 
 Aa+ Bb=1, Aa’ + Bb’=1, (Art. 5.); 
 b' —b a—a 
 
 B= 
 
 “eD—ap. 
 
 be the equations to the 
 
 °* Ac 
 
 Ae AE 
 
 Let z= A’x + B’y + C’ be the equation to a plane, which 
 contains the required line, and the first given line ; and which 
 must, therefore, cut the plane parallel to both the given lines 
 at right angles; 
 
 -. AA’+ BB’+1=0, A'a+ Bb—1i=0, A’a+B'B+C=0; 
 hence C’=—A’a—BfB, 
 and ACA Ge Rae eB) B Oba ond Bye (ase): 
 or, substituting the values of A and B, 
 ,_ @—atb(a'b—ab) , ¥—b+aab'—a’'b) 
 a(a’ —a)+b6(b' — 6)’ ~ a(a —a)+b(b —b)’ 
 *. z= A’(x—a)+ B(y-§), 
 or z{a(a —a)+b(b'— b)} =(ex—a) {a —a+b(a'b— ab)} 
 + (y—B){b —b+a(ab'—a'b)}. 
 
 Similarly, the equation to the plane which contains the 
 required line, and the second given line, by interchanging 
 
 a b'a’ 3’ and abaf, is 
 z{d(a—a)+0(b—0)\ =(w—a)fa-a +0 (ab —ab’)y} 
 +(y—B){b-b 4+ a (ab—ab)}. 
 These two equations determine the line required, since they are 
 
 the equations to two planes, each of which contains it. 
 
 14. ‘To determine the shortest distance between two 
 straight lines, whose equations are given. 
 
12 
 
 Let x=az-4-a c=azta 
 
 fenton Gh y=b24+f' 
 
 two lines, and let x, y, z be co-ordinates of a point in the 
 
 first, 2’, y’, 2’ those of a point in the second; therefore 
 (dist.)” of these points 
 
 I be the equations to the 
 
 / 
 . 
 
 =u=(r—-x) + (y-y) t+ (7-2) 
 
 which, by virtue of the above equations, is a function of z 
 7 
 and z; 
 
 du ; dx dy 
 ot. —— o£ _ W%2r—z)= sig eke Semone ome — =0:; 
 i Wr—wx )at+WMy—y )b+2Az—z )=0, aE. ae b; 
 du | dx’, dy’: 
 Ss ae CN ae INS ae es id aha 2 eke , SOD 
 CE a a im ae het RL rk tet 2 el mart ait 
 w. a(r— ar) t+ b(y—y) +2—-2' =0....00 (1), 
 a (x—2)t+b (y—y/)+2—2 =0....%....(2), 
 
 or aE ALES EOE LAR ipinadbe 
 a (az—a2z+a—a)4+0(bz—0'2' + B—B)+2-—2/=0 
 
 from which equations, z, z’, and thence (uw) may be determined. 
 
 Cor. We observe that equation (1) is that to a plane which, 
 passing through the point xyz in the first line, and 2’y’z’ in the 
 second, cuts the first line at right angles. Similarly, equation (2) 
 is that to a plane which, passing through the same points, cuts 
 the second line at right angles. Hence, the shortest distance, 
 which resuits from the intersection of these planes, cuts both the 
 given lines at right angles; its equations are found in the last 
 Article, and tle expression for its length may be obtained as 
 follows. If planes be drawn through each of the lines parallel 
 to the other, the perpendicular distance of these planes will be 
 the shortest distance. of the lines. As in the last Article, the 
 equations to the planes are 
 
 z=A(r-~a)+ Biy-fB), z2z=Al(r—a’)+ Bey 8B), 
 
 b'—b a—a 
 where A Diane 7) te 
 
 i 2b. awe 
 
13 
 
 and the lengths of the perpendiculars dropped upon these from 
 the origin, are respectively 
 Aa+ BB Aa + BB 
 af te Be x/ + AS Re 
 A (a - a) + B(B'—B) | 
 R/ in Aes Be 
 
 therefore, substituting for A and B their values, the shortest 
 distance between the two lines 
 
 Baar eal lay bry 9 me BY Cara @) 
 J (a — ay + (b — by + (ab’—a' by 
 If this distance =0, (a’ —a) (6 — b)=(3' — B) (a —a), which is 
 
 the equation of condition, in order that two lines may intersect, 
 found in Art. 4,* 
 
 (Art. 11.); 
 
 the difference of which = 
 
 * This Proposition may be proved geometrically, as follows. 
 Take one of the given lines for the axis of «x, (fig. 40.) and let 
 the other meet the axis of z in the point A, and the plane of «ry 
 in B. Draw By parallel to Ox, join Ay, and draw OM perpen- 
 dicular to it; draw MN parallel to By, make Ow equal to MN, and 
 join Nx. Then, since MN is parallel to Ox, OM is perpendicular 
 to each of the lines Ay, MN, and therefore to the plane 47B; 
 also MN being equal to Ox, the figure NO is a rectangle, and Nx 
 is equal and parallel to MO; therefore Nx is perpendicular to Ox, and 
 also to the plane Ay, and therefore to 4B; hence it meets each of 
 the given straight lines at right angles. It is also their shortest 
 distance, for take any point K in AB, draw KL parallel to By, 
 and join OL; then the distance of the point K from Oz is OL, 
 because 2 LOw is a right angle; and OL is manifestly greater 
 than OM. 
 
SECTION ILI. 
 
 ON THE ORTHOGONAL PROJECTION OF AREAS, AND 
 EQUATIONS TO SURFACES. 
 
 —@-—— 
 
 15. Der. A system of points is- said to be orthogonally 
 projected upon any plane, when each point is projected by 
 means of a perpendicular dropped upon that plane; and if 
 the sides of any plane figure be projected orthogonally upon 
 a plane, the figure formed by these projections is called 
 the orthogonal projection of the given figure upon that plane. 
 
 The projection of an area upon any plane, is equal to 
 the area multiplied by the cosine of its inclination to that plane. 
 
 Let ABC (fig. 11.) be a triangular area traced in a 
 plane which intersects the plane of the paper in GK, and from 
 the angular points draw AG, CH, BK, perpendicular to GK ; 
 Let abc be the orthogonal projection of ABC on the plane 
 of the paper. ‘Then because, Cc, Dd are parallel, 
 
 a@c=DC "cos Did, or dc= DC tos 4; 
 
 or triangular area abc = triangular area A BC. cos i. 
 
 Now any plane figure whatever may be divided into 
 triangles, and hence the Proposition is manifest. 
 
 16. The square of any plane figure, is equal to the sum 
 of the squares of its projections on three co-ordinate rectan- 
 gular planes. 
 
 Let the given area be denoted by A, and its projections 
 on the planes of ry, xz, yz by A,, A,, 4,, respectively. Also 
 
15 
 
 let a, 8, y be the angles which a perpendicular to the given 
 area from the origin makes with the axes of x, y, z. Then 
 (Art. 6.) -y is the inclination of the given area to the plane of 
 xy, and therefore A,=A cos y; similarly 
 
 A, = A'cos 3, °A,=A cosa; 
 v AS +4/7+4,7 = A’ (cos*y + cos* B+ cos* a) = A’. 
 
 Cor. 1. The equation to the plane in which A lies 1s 
 x cosa+y cos +z cosy =6, (Cor. 1. Art. 3.), 
 if 0 be the perpendicular upon it from the origin. Hence, 
 
 winds A 
 multiplying by 2 
 
 We ( pia BY 
 Acosa.= +A cos B.~ + A cos Perales? 
 
 © y Zeorptey 
 or A, Tee 3 +A,-3 Sripie 
 which shews that the pyramid whose vertex is the origin of 
 the co-ordinates, and base any plane figure, is equal to the 
 three pyramids whose common vertex is any point in the 
 plane of the figure, and bases its three projections on three 
 rectangular co-ordinate planes. Hence also the equation to 
 a plane passing through three given points is 
 
 tA,+yA,+z24, =A; 
 
 A being the area of the triangle formed by joining the points, 
 and 6 the perpendicular from the origin upon its plane. 
 
 Cor. 2. Let ABC, (fig. 7.) be a plane, 
 OB=a, OA=b, OC=c; 
 
 (8 aBor= (2) (+(e 
 
 or A ABC=4,/ (aby + (a0? + (bc)?: 
 
16 
 
 ab 
 } 3 = irnrnr>\—°=“—a[TehnDanDnDaaeaoooeaee Ze 
 also COS Yy J (aby + (ac) + (bey 
 and 0=c cosy = ————$—$— Pen wnen OA 2/1 
 OT Jab Fac + (60 
 
 17. To find the sum of the projections of any number 
 of areas upon a given plane. 
 
 Let D’, D’, D”, &c, designate the areas, the planes of 
 which make with the co-ordinate planes the angles a’, b, c’s 
 ae Ee a. b”, ec’; &c; and let the given plane make 
 with the co-ordinate planes the angles a, b, c. Then each of 
 the areas must be multiplied respectively by the expressions 
 
 c t , 
 cosa cosa +cos b cos b +cosc cosc, 
 ” ” Ws 
 cosa cosa +cosb cosb +cosccosc , &c, 
 
 which are the cosines of the angles between the plane of pro- 
 jection, and the planes in which the areas he (Art. 9.). The 
 sum of these products will be 
 
 S=cos a (D’ cos a’ + D" cos a” + D” cos a” + &c.) 
 + cos 6(D’ cos b’ + D” cos b” + D” cos b” + &c.) 
 +cosc(D’ cosc + D” cosc’ + D” cose” +&c); 
 
 or, if we denote the sums of the projections on the co-ordinate 
 planes by A, B, C, we shall have 
 
 S=Acosa+Bcos 6+€ cos c. 
 
 18. ‘To find the position of the plane, on which the 
 sum of the projections of any number of areas Is a maximum. 
 
 Let a, b, c, be the unknown angles which the required 
 plane makes with the co-ordinate planes; and A, B, C, the 
 sums of the projections of the areas upon the co-ordinate 
 planes, which are given; then S=A cosa+Bcosb+Ccosc, 
 which is to be a maximum; now by virtue of the equation 
 cos’ a+cos’ 6+cos’ c= 1, we may regard c as a function of 
 a and b; 
 
17 
 
 dS dc 
 
 hence, — =—Asina—Csinc— =0, 
 
 da da 
 d§ dc 
 — = — Bsinb — Csinc— =0;3 
 db “db 
 Ay aa sin a cosa dc ___sinbccos b 
 Ua Sse Srey el SET OE We oe y 
 da sinc cosc niles sin ¢ COs Cc 
 
 *, Acos¢c=C cosa, Bcosc=C eos 6; 
 ° : s C 
 he! (Anta C3) cos c=C’; or'cosc= Sea BAC?” 
 B A 
 cos b 2) Ola ee, 
 ” SATB +O J A? + B+C 
 
 These angles a, b, c, determine the direction in which the 
 plane is to be drawn; and 
 4 eB: Ge ee 
 ye Se Soa peepee ALB 2'C 
 /A+B+C Vv 
 is the value of the sum of the projections, when it is the 
 greatest possible. 
 
 Cor. If any plane make with the plane on which the sum 
 of the projections is a maximum, an Z =8@, then the sum of 
 the projections upon it 
 
 = cos 0,/ A*+ Be + Cr 
 
 Let a, 8, y be the angles which it makes with the co- 
 ordinate planes, and S’ be the sum of the projections upon it. 
 Then cos 9=cosu cosa+cos b cos 3 -+ cosc cosy; or, sub- 
 stituting for cos a, cos 6, cos c, their values, 
 
 A cosa+ B cos B+ C cosy S’ 
 COD OSS erg pe ag ee | 
 fA? +i B? + C? / AP BEC 
 
 nS 2 cos O SAPP B+ Ce. 
 
 Hence, also, the sum of the projections is 0, on all planes 
 at right angles to the plane on which that sum is a maximum, 
 
 C 
 
18 
 
 19. To find the equations to an oblique cone, and an 
 oblique cylinder; the bases of both being circular. 
 
 Let a, hb, c be the co-ordinates OR, RQ, VQ (fig. 12.) 
 of the vertex of the cone, and 2, y, z the co-ordinates of any 
 point M in its surface. Jom VM, and produce it to meet the 
 circle which forms the base of the cone in P. Also, let 
 
 r—a=a (z—0c), y—b=B(z—o), 
 
 be the equations to the line YP; therefore, when z=0O, we 
 have the co-ordinates of P, or 
 
 ON=a—ac, NP=b—Bc; 
 
 and from the equation to the circular base, supposing its radius 
 (r), and the axis of x a diameter, 
 
 (a—-ac) + (6— BcyY = 2r(a—ao); 
 
 therefore, eliminating a and £, 
 
 G2 Owe he) 
 
 or (az—cx) +(b2z-—cy) =2r(z—c) (az—cx), 
 
 the equation required. 
 
 Again, to find the equation ‘to an oblique cylinder. Let 
 x=az, y=b6z, be the equations to a line through the origin, to 
 which the generating line is always parallel; and let 
 
 t=az+a, y=bz+8, 
 be the equations to the generating line in one of its positions 
 PQ (fig. 13.); therefore, making z=0, we have the co-ordinates 
 
 of P, or r=a, y=; and these values of the co-ordinates 
 of P must satisfy the equation to the circular base; 
 
 . a? + 8? = 2ar; 
 or, eliminating a and #8, the equation required is 
 (w—az)y+ (y— bz) = 2r(e@— az). 
 
 20. To find the equation to any surface of revolution, the 
 axis of the figure coinciding with one of the co-ordinate axes. 
 
19 
 
 Let OC (tig. 14.) the axis of z, be the axis of revolution; 
 CPQ the generating curve, and PR = F(OR) its equation, and 
 let it revolve about OC into the position C P’Q’; 
 
 ON=x, NM=y, MP’ =z, 
 
 co-ordinates of the point P’. Then, by the equation to the 
 curve, 
 
 MP’ = F(OM), or z= F (,/2° +y"), 
 which is the equation to the surface. 
 
 Ex. 1. Ina cone whose altitude =a, and semi-vertical 
 angle =a, the equation to the generating line is z=a~— wx cota, 
 taking the origin in the center of the base; therefore the equa- 
 tion to the surface is 
 
 z=a—cotan/2*+y’. 
 
 Ex. 2. In the paraboloid whose altitude =a, the equation 
 g 
 
 to the generating curve is z=a ep) if 7 equals its latus rectum; 
 
 ty? 
 therefore, the equation to the surface 1s Be i a If 
 the origin be in the vertex of the paraboloid, its equation is 
 g 2 
 z 
 i TY ; 
 l 
 
 21. ‘To find the equation to the ellipsoid, and elliptic 
 paraboloid. 
 
 The equation to a sphere is r +y°+ 2°=a’, where (a) is 
 the radius, and the center 1s made the origin of the co-ordinates, 
 
 2 y 2 
 t y i] z D . - 8 . 
 
 or = ++3 + == 1. Suppose all the z’s diminished in the 
 a rarer e 
 
 ratio of a : c; then it is manifest that all the sections parallel to 
 the plane of yz become ellipses, whilst those parallel to the plane 
 
 of xy continue circles. The solid then still continues to be one of 
 oO 
 x = Meee A 
 revolution, and its equation becomes — + — + y= 1, whicn 
 am a” G 
 
 is the equation to the common spheroid, where (a) and (c) are 
 the semi-axes. But, suppose next, that we diminish the y's 
 
& 
 
 20 
 
 in the ratio of a: 6, then it is manifest that all sections parallel 
 to the plane of ry will also become ellipses, and the solid will 
 cease to be one of revolution. In such case it is denominated 
 an ellipsoid, and the equation becomes 
 
 GI 3) in aiee 
 
 Again, to find the equation to an elliptic paraboloid. 
 
 If the axis of a paraboloid of revolution coincide with the 
 axis of z, and the origin be in the vertex, the equation to the 
 surface is 2° + y* =lz. 
 
 Let all the x’s be diminished in the ratio of te lat 
 and all the y’s in the ratio of ALE Jb; the surface is then 
 called an elliptic paraboloid, and its equation is 
 
 2 
 
 i y 
 esis aE i 
 
 22. The followmg Problems are examples of finding equa- 
 tions to surfaces from their geometrical description, and serve to 
 illustrate the preceding Propositions. 
 
 Pros. 1. If a straight Ime have three given points in 
 three fixed planes, to find the locus of any other point in it. 
 
 Let the planes be at right angles to one another, and let 
 them be taken for the co-ordinate planes. Let the straight line 
 meet them in the pomts A, B, C, (fig. 15.) and let P be the 
 describing pomt, the co-ordinates of which are 
 
 ON=27, NQ=y, PQ=z. 
 Also let 
 
 |e baa PB=b; PSE tACa—v, £QbN=¢; 
 . 2=C¢ sin 0, r=a cos 8 cos @, y=b cos @ sin @ 5 
 
 re ee Laie q a 
 : () + (3) + () = cos 6+ sin? d=], 
 a OD c 
 
 the equation to an ellipsoid. 
 
21 
 
 Pros. 2. Two points move in straight lines with uniform 
 velocities, to find the equation to the surface generated by the 
 straight line which joins them. 
 
 Let AA’, BB’ (tig. 16.) be the two given lines, AB their 
 shortest distance, which make the axis of z; also take the plane 
 bisecting AB at right angles for that of xy, and the line bisect- 
 ing the angle between the projections Ob, Oa of the lines, for 
 the axis of x Then, if OA =OB=a, and tan rOa=m, the 
 equations to the lines A.A’, BB’, are manifestly 
 
 yYrme, z2—A3 yYour—-me, o=— —a, 
 
 Let 6, b’ be the initial values of x for the points moving in 
 AA’, BB’ respectively, c, c’ any other corresponding values, 
 2 : 1 the ratio of their velocities; therefore e—b=n(c —0’). 
 But if c=Az+a, y= Bz+, be the equations to the line 
 joining the points, they must be satisfied by the co-ordinates of 
 the points ; 
 
 “. c=Aata, c= —Aata, 
 mce=Ba+B, -—mc = —Bat 8, 
 
 which give 
 
 / , , 
 ac 3: BS jaca = Yeisen : 
 2 2g Qa Qa 
 / ee 
 OC OG Cre c—C 
 r= Star eed 5 y=m Z2+m 3 
 Qa 2 2a 
 
 which equations give 
 a(me+y) , al(mr—y) 
 : = ——___— ¢ = ————. 
 m(a+z) ” m(a— 2) 
 
 Hence, by substitution, the equation to the surface, is 
 
 Mi ied a oY) as b=n ‘a Sitar — 0) : 
 . m(a+ z) m (a— Zz) 
 
 or, by reduction, 
 
 (n+ 1) (a°y—mazx)+(n—1) (azy—ma'e) = m(b—nb’) (a?—2°). 
 
22 
 
 Pros. 3. Two planes are drawn through two straight 
 lines given in space, so as always to be perpendicular to one 
 another; to find the locus of their intersection. 
 
 Taking the same axes as in the last Problem, the equations 
 to the lines are 
 
 Y=MI, 2=a; Yr-mMY, 2=—4. 
 
 Let . 
 
 z=Avt+ By+C, z=A'rt+Byt+C, 
 
 be the equations to the two planes. Then, because the plane 
 z=Ax+By+C, contains the line y=mx, z=a, 
 
 a=Ar+Bme+C; oe MI, A et Mee 0) 5 
 
 similarly, a= —C’, A’ — Bm=o. 
 Hence 
 
 z-a=A (2-#), one eee B= +wuiea_a 
 
 b] 
 mx—y ” me—y 
 
 _ m(z +a) Be z+a 
 
 mMa—Y d ML—~—Y 
 
 zta=A'(si7), or A’ 
 
 But, because the planes are at right angles to one another, 
 AA'+ BB'+1=0; 
 m(2* — a®) 2° —a” 
 Se en ea ene ol har 
 (mz) —y° (may —y 
 or (m?— 1) (22—a’) + (mry — ¥°=0, 
 is the required equation to the surface. 
 If the lines are parallel, m=0, and y” + 2° =a’, the equation 
 to a cylinder, as it ought to be. 
 Pros. 4. To find the locus of a poiwt which is equi- 
 
 distant from two lines given in space. 
 
 Retaining the same axes and notation as m the last question, 
 it will be found that the equation to the surface is 
 
 may + (1 +m’) az=0. 
 
23 
 
 Pros. 5. To find the equation to the surface generated 
 by a straight line which constantly passes through three fixed 
 straight lines. 
 
 Let the axes of the co-ordinates be parallel té’ the thre 
 
 given lines, then the equations to the latter will be ay “Y 
 
 ra pe o pb ce 
 y= os? Pe sep ae 
 Let the equations to the generating line in any positiqn be 
 r=mzetm, y=nzt+n, oe 
 and consequently 
 
 7 
 
 , a 
 mn — mn — 
 y= —2 +p, (where p =. ————+——" J". 
 m 
 
 m \ 
 
 Then, since it meets each of the three given lines, we have 
 three equations of condition, viz. 
 
 , 
 
 a=metm, V=ne+7, b=—atp; 
 m 
 
 . . . - , , 
 and it remains to eliminate m, m, n, n, between these, and 
 the equations to the generating line. By subtraction we find 
 
 r-ad=m(z- 0), y= F=n(z—c), y-b= Gaba), 
 m 
 
 therefore, eliminating m and 7, 
 (x —- a) (y— 0) @—c)=(4@—a) (y — 6) @-C) 
 
 the required equation, which is of the second degree, because 
 the term xyz disappears. 
 
 By interchanging the quantities abc, a’'b’c’, the final equa- 
 tion is not altered; therefore the same surface would be gene~ 
 rated by a straight line constantly passing through three different 
 lines, whose equations are 
 
 , 
 
 $e r= a) ty 
 
 ? 
 qf 03 ey Qiiwa OC 
 
 23. ‘To find the equations and forms of the three surfaces 
 of the second order that have a center. 
 
24 
 
 Surfaces, in the same manner as lines, are divided into 
 orders according to the degree of their equations. The plane 
 is the surface of the first order, because its equation is of the 
 first degree; the sphere, cone, &c, are surfaces of the second 
 order, because their equations are of the second degree. The 
 general equation to surfaces of the second order is 
 
 ax® + by" +cz°+ 2a'yz+ W'rz4+ 2cxy t+ 2a,x+ 2b yt 2,z+d=0 
 
 it will be shewn. hereafter, that this equation can always be 
 reduced to one of the forms, 
 
 Ax? + By’ + C2,-—1=0, or By + C2+r=0; 
 
 at present, we shall examine the figures and properties of the 
 different surfaces, which these equations represent. 
 
 The equation Ax + By’ + C2? — 1 =0, offers three 
 varieties; all the coefficients may be positive, or one may 
 be negative, or two may be negative; in all cases the surface 
 is symmetrically situated with respect to the axes, and the 
 origin is the center of the surface, that is, all lines drawn 
 through it, and terminated by the surface, are bisected in it; 
 as is manifest from the form of the equation. 
 
 I. Let all the coefficients be positive, and let the 
 equation be | 
 2 fe 4 
 
 i y z 
  @ me oa he ~V~ = ] 5 
 a b° Cc : 
 
 the surface is the ellipsoid, and may be generated in the 
 following manner. 
 
 In the planes of xz, yz, describe ellipses AD, DB, 
 (fig, 17.) whose. 4 axes are AC =a, BC=b, DC =c; and 
 let a variable ellipse move parallel to the plane of ry, having 
 the ordinates NR, NQ, for its axes... Let x, y, z, be co- 
 ordinates of the point P, 
 
 N b° b° b° 
 (ty: .(RN*® — x°) = QN*— rat = a(c’ — 2°) — se", 
 RN a a’ 
 
 *O+Q+O- 
 
25 
 
 The figure represents an eighth part of the surface, which, 
 as appears from its equation, is limited in all directions. The 
 ellipses BD, AD, AB, are its principal sections, the equa- 
 tions to which result from the equation to the surface, by 
 making v, y, 2, separately =O. Hence the quantities a, b, c, 
 represent the 4 axes of the principal sections, and also the 
 Z% axes of the surface. ‘The points A, B, D, are called the 
 vertices of the surface ; it has six on the whole, one at each of 
 
 the extremities of each axis. 
 
 II. Let one of the coefficients, viz. C, be negative; and 
 let the equation be 
 
 The surface is the hyperboloid of one sheet, and may be 
 generated as follows. In the planes of xz, yz, describe hy- 
 perbolas AQ, BR, (fig. 18,) having } axes AC=a, BC=6, 
 and the same minor or conjugate axis DC=c; and let a 
 variable ellipse move parallel to the plane of xy, having the 
 ordinates NQ, NR, for 4 axes; then x, y, z, being co-ordinates 
 of the point P, 
 
 RN* b° b* b° 
 
 PMe ty" No? (NQ’ — x*)= RN’ — fa ale + c’)— aa" 
 
 od y” 7° La 
 A a came 
 The principal sections are the ellipse A.B, which forms the 
 interior limit of the surface, and the hyperbolas BR, AQ. 
 If we make w constant, and =a, the equation to the section 
 parallel to yz, 1s 
 
 bee) G a 
 which, as long as a < a, represents an hyperbola with its vertex 
 in AB, and conjugate axis parallel to CD; when a=a, it 
 represents two straight lines; when a>a, the equation to 
 the section is 
 eye 
 2 b? 
 
 sor 
 
26 
 
 this also represents an hyperbola, but in a new position, viz. 
 with its vertex in AQ, and conjugate axis parallel to CB. 
 If a=b, all sections parallel to ry become circles; hence the 
 form of the surface resembles that generated by the revolution of 
 an hyperbola about its minor axis; and since it 1s continuous, 
 it is called the hyperboloid of one sheet. ‘This surface has four 
 vertices, two of which are A and B. 
 
 Since = : 
 ince “ 2B 1; 
 ail A aca Se Gaby i 
 cu ab (ay) + (62) 
 MPS ACI CL alias eel 
 ab > J (ayy + (ba 
 ad x y 
 Hence, amy + re 
 
 is the equation to the asymptotic surface, which is a right cone, 
 (whose vertex is C, and base an ellipse, center D and 4 axes 
 parallel and equal to AC, CB,) and is a particular case of the 
 
 general equation. 
 
 III. Let two coefficients, viz. B and C, be negative, and 
 let the equation be 
 
 the surface is called the hyperboloid of two sheets, and may be 
 generated by describing hyperbolas AQ, AR, (fig. 19.) in the 
 
 planes of xz, vy, having f axes 
 AG Sway, DC=ec; 
 and making a variable ellipse move parallel to yz, having the 
 
 ordinates QN, RN, for semi-axes. Then, as before, it will 
 be found that the equation to the locus of P is 
 
27 
 
 In this case, the principal sections are the hyperbolas 
 AQ, AR; the principal section in the plane of yz being 
 imaginary. This surface has only two vertices; its form re- 
 sembles that generated by an hyperbola revolving about its 
 major axis, and consists of two distinct portions; it is therefore 
 called the hyperboloid of two sheets. As before, it may be 
 shewn that the equation to the asymptote is 
 
 2 2 g 
 eta) z 
 
 a Hee LM eo A 
 which represents a right cone on an elliptic base, and is a parti- 
 
 cular case of the general equation to surfaces of the second 
 order. 
 
 The above are the three varieties of the first class of surfaces 
 of the second order, viz. those which have a center. If any 
 result be obtained for the ellipsoid, the corresponding result for 
 the hyperboloids may be deduced, by writing 
 
 baie) for 6; or bA/ —1, an / al) for 6 and c. 
 
 24. The second class of surfaces of the second order, 
 that is, those which have not a center, represented by the 
 equation By* + Cz? +2 =0, affords but two varieties, viz. 
 when B and C are of the same sign, and when of different. 
 
 I. Inthe former case, let x be measured so as to make 
 
 6 
 
 4 
 ~ 
 
 1 z ae ee 
 the equation of the form ye this is the elliptic 
 
 by ainhar 
 paraboloid, which may be generated by describing a parabola 
 Ak, (fig. 20.) in the plane of ry, and making another parabola 
 RP, whose equation is z*=/'x, move parallel to the plane of zz, 
 with its vertex in AR; for x, y, z, beimg co-ordinates of any 
 point P, 
 
 fal RM=U (2~ "4 has Ns tt Plan 
 
 ‘Hence /, I’, are the latera recta of the principal sections AR, 
 AQ, which are parabolas; the principal section in the plane of yz 
 
28 
 
 is imaginary, but all sections parallel to that plane are ellipses 
 with their centers in the axis of v, and vertices in the parabolas 
 AQ, AR; and hence the name of the surface. It has but one 
 vertex A, and resembles in form the paraboloid of revolution, to 
 which it is reduced when /=1’. 
 
 II. Let C be negative, and let the equation be 
 2 2 
 
 N 
 
 this is the hyperbolic paraboloid, which may be generated by 
 making a parabola PR, (tig. 21.) whose equation is 2° = —/2, 
 move parallel to the plane of vz, with its vertex in a parabola 
 AR traced in the plane of vy 3 for x, y, z, being co-ordinates 
 of any point P, 
 
 Mi, hy A bind ie “celyty | tava 
 fal MRal (7 v), SP cig 
 
 The principal sections are, the parabola AA, which forms the 
 
 interior limit of the figure, the parabola AQ, and two straight 
 
 lines, represented by the equation /z*=/'y’, in the plane of yz. 
 
 If we make r=a, the equation to the corresponding section 
 / 
 
 parallel to yz is 2 = 7(y la), which represents an hyperbola 
 whose vertex is in AR, and conjugate axis parallelto Az. If a be 
 negative, the equation is z= F (y° +a), which also represents 
 
 an hyperbola, but in a new position, viz. with its vertex 
 in AQ, and conjugate axis parallel to Ay. 
 
 ng Sai ACY lax 
 Since 2 = a(1- —), 
 
 WR A? ‘Aaelahs 
 ae VEC ee —&e.). 
 
 x 
 y 
 Ny 
 Hence, the equation to the asymptote is s=yV 5 » which 
 
 represents two planes perpendicular to the plane of xz. 
 
29 
 
 s 
 
 Cor. ‘The elliptic and hyperbolic paraboloids are parti- 
 cular cases of the ellipsoid, and hyperboloid of one sheet, viz. 
 when the center is removed to an infinite distance. In the 
 equation, 
 
 write x — a for x, then the equation, 
 
 Ys 2 2 4 
 C “) +4455, 
 a b c 
 
 is reckoned from the vertex of the surface. Call p and p 
 the distances of the foci of the principal sections in vy, and xz, 
 from the vertex ; 
 
 aS b =a —(a—p)y =2ap—p’, c= 2ap +p ; 
 
 hence, by substitution, 
 
 5 2 9 
 r AG i Cas 
 gt rar: Gaeta 4 1— 129 = 0, 
 a a 2ap —p 2ap +p 
 
 2 2 2 
 or ssn eh ty a J gust 12 =0; 
 Pp pon pu 
 2 Dre anit 
 
 therefore, making (a) infinite, 
 
 7 47-9 
 2p 2p 
 
 Hence, if any result be obtained for the ellipsoid, or hyperboloid, 
 
 it will be adapted to the paraboloids, by the modification just 
 
 indicated ; viz. by transferring the origin to the extremity of an 
 
 axis, and making that axis infinite. 
 
SECTION III. 
 
 ON TANGENT PLANES AND NORMALS- TO SURFACES, AND TANGENTS 
 AND NORMAL PLANES TO CURVES OF DOUBLE CURVATURE. 
 
 25. To draw a tangent plane to a curve surface. 
 
 Let z=f(x, y) be the equation to a curve surface ; 
 2 = Ar’ + By'+C 
 the equation to a plane touching the surface at the poimt P, 
 (fig. 22.); then, if x, y, 2 be the co-ordinates of P, we must 
 have z= Axv+ By+C, and therefore, eliminating C, the equa- 
 tion becomes.2/—z=A (2° —2) + Biy' -y)- 
 
 Let there be drawn through the point P a plane parallel to 
 that of «z, cutting the surface in the curve PC, and the tangent 
 plane in the line PT. Then the line PT’ must be a tangent 
 to the curve PC, otherwise the tangent plane would cut the 
 surface. 
 
 Now, for every point in the line PT, y'=y, and therefore 
 its equation, deduced from the equation to the plane, is 
 
 2—z=A(x —2z). 
 
 Aliso, the equation to PC is deduced from z=/(a, y), by 
 making y constant in it; and in order that PJ’ may touch PC, 
 
 dz 
 we must have A=, (y) being supposed constant in the 
 
 - 
 
 differentiation. Again, through P draw a plane parallel to that 
 of zy, cutting the surface in the curve PD, and the tangent 
 plane in the line PR; then, as before, the equation to PR is 
 
 Zm—z2=By'—y), 
 
Sl 
 » 
 gid the equation to PD is obtained from z= Ss y) by 
 regarding x as constant; and since PR touches PD, we must 
 dz , ge 
 have B= ay’ (x) being supposed constant in the differentiation. 
 Hence, substituting for A and B their values, the equation to 
 
 the tangent plane at a point epee” Ts 
 
 wh 
 
 —— 2 
 
 7 ON dZ \2.; 
 — (x —x) + — —y);3 
 ras ney vay) 
 or, as it is usually written, 
 z—z=p(c’—2)+9q(y'—y), 
 
 when p and q denote the partial differential coefficients of z, 
 derived from the equation to the surface. 
 
 Cor. 1. If vy be the angle of inclination of the tangent 
 
 plane to that of ry, then (Art. 7.) 
 
 1 
 UDO Ae ney so eee ee 
 . A (gg fee eit ey 
 GH dy 
 and similarly its inclinations to the other co-ordinate planes may 
 be found; also (Art. 11.) the length of the perpendicular from 
 the origin on the tangent plane 
 By Bilas tod 
 eLepit ve 
 Cor. 2. Let AB, BC, AC (fig. 7.) be the traces of the 
 
 tangent plane on the co-ordinate planes; then, by making 7’ 
 and z’ vanish in its equation, we find 
 
 Fy aN Ae 
 
 similarly, J 
 
 Po 40s mr ak = CO=z2—pr-qy; 
 
32 
 
 therefore, (Art. 16.) area of tangent plane 
 
 BoA =o 
 
 Qpq J! +p +q> 
 : i OAS 
 and volume of pyramid ABCO= ia oe 
 iG] 
 
 26. To find the line of greatest inclination ; that is, among 
 all straight lines which touch the surface in a given point, 
 to determine that whose inclination to the plane of ry is the 
 greatest. 
 
 As all these lines must lie in the tangent plane, the line 
 required is obviously that which is perpendicular to the inter- 
 section of the two planes. Now, the equation to the intersection 
 of the tangent plane at a point xyz, with the plane of ay, is 
 
 ’ Pir z 
 —y= —-(t —27)—-3 
 i bem A q q 
 
 t q- , 
 ae — — “(a — 7) 
 ier p 
 
 is the equation to a line perpendicular to the preceding, and is, 
 therefore, the equation to the projection, on the plane of xy, 
 of the line required. 
 
 Again, let 2 ~v=a(z'—2z) be the equation to its projec- 
 : a ‘ 
 tion on the plane of xz, then y —y= Ee -2) is the equa- 
 me 
 
 tion to its projection on the plane of yz; and, since the line in 
 question coincides with the tangent plane, 
 
 qa L 
 pa+—q=1, (Art.5.); 0. a= 
 Pee ip pP+q 
 Hence, 
 , P , , q , 
 v—r= -(z)—z), and y —y==(t —2), 
 pt ae TP at p 
 
 are the equations to the line of greatest inclination; and its 
 length 
 
33 
 
 Cor. 1. If a curve be traced in the plane of ry, the 
 equation to whose tangent is 
 
 / Dep 
 errs Y p 
 d q 
 
 or whose differential equation 1S = — = 
 x 
 
 B 
 and aright cylinder be erected upon it; the curve of double 
 curvature formed by the intersection of the cylinder and surface 
 
 will have the property, that its tangent is always inclined at the 
 ereatest angle to the plane of ry. 
 
 3 
 
 If, therefore, we derive the values of p and g from the 
 equation to the surface, independent of z, and substitute them 
 in the equation 
 
 PY aie 
 dus; p. 
 the integral of this, together with the given equation to the 
 surface, will be the equations to the curve of greatest inclination. 
 
 2 Q 2 
 5 x Mee Zz 
 Ex. In the ellipsoid — + laps >z=1, 
 Gs b c 
 2 2 we 
 Cat RG, q a 
 oe PS me me; ! Sooner te el and puke 5) Zi 
 QZ pie Pp 2 
 dy a’ y 
 dz b? x’ 
 
 ° . 2 2 . : 
 and therefore, integrating, y’ =Cx2, is the equation to the 
 
 projection, on the plane of xy, of the curve of greatest 
 inclination. 
 
 Cor. 2. Also, if the projection on the plane of xy be 
 given, we may find the surface. Thus, let «+ y°= a’; 
 
 q x 
 - .. == —-~, or pr-+qy=0; 
 dx sy Bhd eas ini 
 
 hence, integrating, z = f (=) is the equation to the surface. 
 s 
 
 ry E 
 
34 
 
 27. To draw a normal to a curve surface. 
 
 Let x =az'+a, y’ =b2'+ 3, be the equations to the 
 normal. Then, since it passes through a point whose co-ordi- 
 nates are xv, y, z, we must have r=az-+a, y=bz+ 8; 
 therefore, eliminating a and #, | 
 
 v—x=a(z—2z), y'—y=b(z —2). 
 
 Now this line is perpendicular to the tangent plane, whose 
 equation is 
 
 é 
 
 eee HAE AES 
 Zz —2= 5 ( —«)+ ay eriv Als 
 
 dz d 
 ote (Art. 6.) a “ —_ = 0, b+ ae = O. 
 - ar dy 
 
 Hence, substituting these values of a and 6, the equations 
 to the normal, at a pomt ryz, are 
 
 ; ie i ee hen / sina Be eyiyy 
 L—-T eat ter 280, pie ay z) =0. 
 
 Cor. 1. The equations to the normal may also be found 
 from the consideration, that it is the longest or shortest line 
 which can be drawn from any point in itself to the surface. 
 Let 2’, y', 2’, be co-ordinates of a fixed point in the normal, 
 and x, y, z, of any point in the surface; then, if 6 = distance 
 of these points, 
 
 o = (a — al +(y — yy + — 2°), 
 a function of two variables x and y; for z is given in terms 
 of x and y, by the equation to the surface. ‘Therefore, in order 
 that 0? may be a maximum or minimum, 
 
 x — 4 By =O : i a 2) =O 
 £ ag z) =0, Upy qua =0. 
 
 These are the equations which determine the position of the 
 shortest line that can be drawn from the fixed point a y'z’, 
 to the surface, and therefore those to the normal. 
 
35 
 
 28. To find the length of the portion of the normal, 
 intercepted between the surface and the plane of xy, and the 
 co-ordinates of the point where it meets that plane. The 
 co-ordinates of any point in the normal being 2’, 9’, 2, and 
 vt, y, z, those of the point where it meets the surface, the 
 distance of these points 1s, by the last Art., 
 
 SE a ON + (J) + (=) : 
 
 But at the point where the normal meets the plane of ry, z’=03 
 therefore the required length 
 
 ‘ / dz\2 dz\2 
 sie NA Ge) i 
 Also by making 2’ =0, in the equations to the normal, we find 
 r=rt aa Wade ine 
 dx dy 
 the co-ordinates of the point where it meets the plane of ry; 
 
 and similarly we may find the points, where the normal meets 
 the other co-ordinate planes. 
 
 29. The followmg Problems will exemplify the method 
 of drawing tangent planes, and normals to surfaces. 
 
 Pros. 1. To draw a tangent plane to an ellipsoid. 
 
 Q 
 x 2” 
 
 ve 
 The equation is — ZF +—= ] 
 q a? + b? c 5) 
 
 £L 
 te gemieharg aca a yO 
 ie teh GeALLT ; 
 
 ax 
 
 Hy i a2. Vas i 
 Hence, substituting for —, a? their values in 
 Y 
 
 Z—z=p(e- 2) +qiy'—y) 
 
36 
 
 the equation to the plane touching an ellipsoid at the point ryz, 1s 
 
 kN, ae 
 z—-zt>54-@-arvnt+at(y—-y=0 
 az ba J ’ 
 zz oer L ” ’ 
 Opie ey a owt ap! 
 Cc Cc a a b b 
 4 
 rx Z2 
 0 ee te re | 
 a b 
 
 If A, B, C, be the poimts in which the tangent plane cuts the 
 2 
 
 axes, (fig. 7.) by making y’ = z’ = 0, we find OB = x = ~ 4 
 
 2 
 C 
 
 he b 
 
 similarly OA = 7; wo a hence the pyramid OA BC 
 will be a minimum, when vyz =a maximum, that is, when the 
 co-ordinates of the point touched by the plane are 
 
 a b C 
 0 ee eS ee 
 So ob a Js 
 Pros. 2. ‘To find the length of the perpendicular dropped 
 
 from the center of an ellipsoid on the tangent plane, and the 
 locus of its extremity. 
 
 The equations to a perpendicular from the center on the 
 tangent plane, are 
 Cobie be pen y a 
 Vili tiaeree A) Be Er ae ua 
 3 y bz ? 
 which, combined with the equation to the tangent plane, give 
 by eliminating 2, 7, 
 
 $o7n° af" 
 “ , 
 A ade = 3s 
 cee Zz 
 c* 2 ar by 
 
 TOR? 8a. Ou ae eD: Liat LOT PD Of et eg en ee 
 12 12 123 19 2 79 9 72 tc / 
 Vie comet ae OLE Se Ie i aa sa oe) i 
 
 2 2 2 
 a 
 
 ie V4 ta 
 But {tet at) 
 
 oc i= 
 
 we (ary + (by') + (2)? = (ve? + y? + 2%, 
 
37 
 
 the required equation. Next let é be the length of the perpen- 
 dicular, 
 
 2 h 2 12 ; 
 OS e+ y+ 2 = —; 
 z 
 . . . , , Bs 
 but by Er ee x,y, between the same equations, 
 2 
 
 ,fO8 rey 2” 
 z 4 = weat4t+4. 
 
 a b+ C 
 
 Bo 
 
 Pros. 3. If three tangent planes to an ellipsoid are 
 mutually at right angles, their point of intersection will trace 
 out a sphere concentric with the ellipsoid. 
 
 Let é, the perpendicular from the center on a tangent plane, 
 make angles a, (, yy, with the axes of xr, y, 23 then the 
 equation to that plane is 
 
 , ¢ , 
 © cosa | Vases Oi a COS Ue: 
 ) 6 , : 
 which must be identical with the equation to the tangent 
 plane, found in Prob. 1, viz. 
 
 — +22 44521, 
 a: b2 2 
 x0 yo 20 
 cosa=-—>, cosB= grr cosy = 5 
 a 
 2 
 Py UH 
 hence, since RE ae P ++5= 1, 
 
 a 
 se) 4 er @) = a’ cosa +b’ cos’ 3 +c cosy, 
 
 which gives the length of the perpendicular in terms of its 
 inclinations to the axes. Now let there be two other per- 
 pendiculars 0’, 6,, which make with the axes the angles a, By; 
 a, B,> y,» respectively; and suppose the three Abs pecdiculaie 
 to be mutually at right angles ; 
 
38 
 
 0? = a’cos’a’ + 6’cos® 3’ + ¢* cos" y’, 
 
 6° = a’cos a,+ bcos’ B, + cos’ ¥y,. 
 But a, a, a,, being the angles which a line (viz. the axis 
 of x) makes with three rectangular axes (viz. the three per- 
 pendiculars,) 
 
 cosa + cos’a’ + cos*a, = 1; 
 similarly 
 cos’ 3 + cos? SB’ + cos*B, = 1, cosy + cos’ + cosy, = 1; 
 hence, by addition, 
 O+0°4+ S02? = 84+ 0 +c’. 
 
 But if g be the distance of the point of intersection of the three 
 planes from the center, then 
 
 oi= & 4 84 88, p Bafa +0 Eo 
 
 the radius of the sphere described by the point of inter- 
 section. 
 
 In the case of hyperboloids, one at least of the quan- 
 tities a, b°, c’, is negative; and hence their sum imay be 
 negative, or nothing; in the former case, there is no point in 
 space, through which three rectangular planes touching the 
 hyperboloid can be drawn; and in the latter, the center is the 
 only point which has that property. 
 
 Again, if we transfer the origin to the extremity of the 
 Z-axis (a), and then make (a) infinite, as in Cor. Art. 24, 
 the equation to the surface will become 
 
 2 ae 
 tale t+ —, = 22, 
 of Minctinl 
 
 and that to the sphere 
 —2=p+p'; 
 
 therefore the locus, in the case of paraboloids, is a plane per- 
 
 ; 
 pendicular to the axis of the surface, at a dist. = ~ PP 
 from the vertex. 
 
39 
 
 Pros. 4. If three planes, mutually at right angles, con- 
 stantly touch the perimeter of a plane curve of the second 
 order, to find the focus of their poimt of intersection. 
 
 xv 2 , 
 Let Soitiea 1, be the equation to the curve, 
 is 2 
 
 x cosa+y’ cos [3+2 cos y=0, 
 
 the equation to one of the rectangular planes; then, making 
 z =0, the equation to its trace on the plane of ry is 
 
 U / 
 «a COS a cos 
 tse hee y' cos =a) 
 
 r) é a 
 Lx yy 
 
 which must coincide with — + b2 
 - 
 
 = 1, 
 the equation to the tangent to the curve; 
 
 oy 
 / cosas, cos p=503 
 
 dr ,2 ou y? i : 
 hence &=(—) + (*) = a°cos a’ + b° cos’ 3’; 
 
 similarly, 
 
 ° c C 6 
 0? =a’cos’a” +b? cos’ B”, 0,’ =a? cos*a,* + 6 cos’ B’. 
 
 Hence, by addition, 
 O+0°+0%=a +0", or p=r /at+b, 
 
 if p=distance of the point of intersection from the center. 
 Hence, the locus is a sphere concentric with the curve. 
 
 For the hyperbola, p=./ a’ — 6°; therefore we must have 
 a> 6, otherwise the Problem is impossible, if a=6, as in 
 the rectangular hyperbola, the locus is a point, viz. the center. 
 If the curve be a parabola, it may be shewn, as above, that 
 the locus is a plane, perpendicular to its own, through the 
 directrix. 
 
40 
 
 Pros. 5. Ifr, 7’, r,, be three semi-diameters of an ellipsoid, 
 
 mutually at right angles, then 
 1 1 1 1 
 
 1 
 arte eer Q° 
 a” 2 
 
 | 
 a 
 - 
 Il 
 
 This property, analogous to that for the perpendiculars, 
 may be proved in a similar manner. Let a, 6, y be the 
 angles which (r) makes with the axes of v, y, 2; then the 
 co-ordinates of its extremity are 
 
 r=rcosa, y=rcosB, z=rcosy; 
 hence, by substitution in the equation to the ellipsoid, 
 
 cos a , cos BY Coney 
 
 a’ b? Se r 
 
 A sa) , 
 In like manner, for the semi-diameters r and 7,, we have 
 the equations 
 
 cos'a cos? 3 4 cos” ry’ 1 
 c = ys 3 
 oP b? ce yr? 
 cos a cos” cos” same 1 
 v7] / i 
 9° + —_— a ug 
 
 hence, by addition, as in the last Problem, 
 
 Pros. 6. If two concentric surfaces of the second order 
 have the same foci for their principal sections, they will cut one 
 another every where at right angles. 
 
 g 2 y” ° 
 
 Tn Yy Z r fe 
 
 Let —+54+5=1 ee me Ft 
 nip eee rye ea i OM Bet 
 
 be their equations; then the equations to the tangent planes are 
 
 , , U / / / 
 mini) ya gee Ey | aly ee 
 Ae aeryeae. ergs)! + as + >= =1; 
 a b c B Y 
 
 2 
 a 
 
Al 
 
 and in order that these may be at right angles, we must have 
 
 (Art. 9.) 
 2 2 . 9 
 (=) 1 Ga) . (a AT eed SVE 
 
 which gives the relation among the co-ordinates of the points, 
 in which the surfaces may intersect at right angles. But, by 
 subtracting the two equations, we find 
 
 (= =) “4. 3) *+(; 1 er 
 cee bee 0 ae vk ear § RT! at rs = 
 a” a’ b° p° J Cc y ; 
 
 the relation among the co-ordinates of the actual points of 
 intersection ; which must be identical with equation (1), if the 
 surfaces cut one another every where at right angles; therefore, 
 equating the ratios of corresponding coefficients, 
 
 2 2 2 
 ae—-vw=b—-S=c—-y; 
 on a? —b =a’ — 8°, Av eaa =a? — +7, be — ce =B’—+'; 
 which three equations shew that the principal sections have the 
 same foci. 
 
 In like manner, if the surfaces have not a center, and we 
 represent their equations by 
 
 2 2 2 2 
 z y 2x z YO UG 
 <q CIET ee he RS een ee 
 Gein a DP fuded 6 ama: 
 they will intersect every where at right angles, provided 
 Qo 9 9) 29 Oo (94 
 ao-C ay a—b a—p 
 EN es ee? Bi Ln Sane 
 
 that is, the foci of the principal sections in the planes of xz 
 and xy, must be coincident. 
 
 Pros.. 7. To draw a normal to an ellipsoid. 
 
 The equations are, since 
 
 dx oe el die ch bees. 
 2 2 
 ; Bey oye ae, 
 rors —-(*—-2), y-y=y ~(% —2). 
 Gi 4 " iws 
 
42 
 
 By making z’=0, we find the co-ordinates of the point 
 where the normal meets the plane of vy, 
 
 v=r(i-S), yay (1-4) 0s 
 
 and (Art. 28.) the length of the normal 
 
 Pee Lee er wee Ce aA g 
 = IER a a VAs . 
 Vit S45 Gad Stata=5; 
 
 ~ 
 ~ 
 
 oe 
 
 by Prob. 2. Art. 29. 
 
 Cor. Suppose normals are drawn from points where a 
 plane, whose equation 1s 
 
 z=Ar+ By+C, 
 
 meets the ellipsoid; combining this with the equation to the 
 ellipsoid, we have 
 
 x f y i (Ax+By+Cy ee 
 
 2 b2 C 
 
 Therefore, substituting for x and y their values, derived from 
 equations (1), 
 
 (2). Goel a Gast ee) = 
 
 which is the equation to the curve of the second order, traced 
 out by the extremities of the normals in the plane of vy. 
 
 Pros. 8. To draw a tangent plane to the surface whose 
 equation is 
 
 Se eee: -/s, 
 
 dz Ravhes dz z 
 Here dz_ ft ce AWE 
 dx Ne. rp: 
 
 ¥Y 
 
AB 
 
 therefore the equation to the tangent plane is 
 
 4 \/* eee \/2 / 
 Sat i! (2 v)+ PAC — y)=0, 
 
 ¢ y — — ol <a 
 Tat fet pen Vat ly tena 
 
 Make y=2=0; .*. (fig. 7.) 2 =BO= J ax; 
 similarly, AO=/ay, CO=,/az; 
 
 AO + BO+ CO=n/a (Jt + /y +/2) = 
 
 or their sum is constant. 
 
 30. ‘To define a curve of double curvature, and shew how 
 its equations are found. 
 
 A curve of double curvature may be considered as generated 
 by a point, which changes continually, not only the direction of 
 its motion, but also the Riker in which it moves. The nature of 
 a curve of double curvature depends on the relation of its three 
 rectangular co-ordinates ; and this relation is expressed by two 
 equations between these quantities. 
 
 Let PQ (fig. 23.) be a curve of double curvature, pq its 
 orthogonal projection on the plane of xy. ‘Then the equation, 
 y=x, to the plane curve pq, is one of the equations to the 
 
 curve PQ. 
 
 Similarly, let P’Q’ be the projection of PQ on the plane of 
 xz, then the equation, y= fx, to the plane curve P’Q’, is the 
 other equution to PQ. And these two equations are sufficient 
 ta determine all the points in the curve; since, for any assumed 
 value of x, we find values for y and z. ‘The equation to the 
 projection of PQ on the plane of yz, is deduced from the 
 other two, by elimimating 2. 
 
44 
 
 From this, it appears that every curve of double curvature 
 may be considered as formed by the intersection of two cy- 
 lindrical surfaces, erected on its projections pg and P’Q’ as 
 their bases, perpendicular to the co-ordinate planes. 
 
 31. To draw a tangent line to a curve of double curvature. 
 
 Let PQ (fig. 23.) be a curve of double curvature, PT’ a 
 tangent at P; P’Q’, P’T”, the projections of the curve and its 
 tangent on the plane of xz; pq and pT’ their projections on the 
 plane of wy. Then it is manifest that P’T” is a tangent to Pie 
 and p!'a tangent to pg. Let z=o2, y= fx, be the equa- 
 tions to the projections P’Q’, pq; and let x,, y,, z, denote 
 the co-ordinates of any point in the line PT’. Then 
 
 are the equations to the lines P’T", pT, which are the pro- 
 jections of PT’; and consequently they are the equations to PT, 
 a line touching the curve of double curvature in the point ryz. 
 dz dy 
 
 and — by 2 and y’, the equations to the 
 dx dx y J? ‘d 
 
 If we denote 
 tangent are 
 ! Z,—2=2(2,-2), y,—y=y (1 —2). 
 32. To find the equation to the normal plane, that is, to- 
 
 the plane which is drawn perpendicular to the tangent line of a 
 curve of double curvature, through the point of contact. 
 
 Let z =Ar,+ By,+C be its equation; then, since it 
 passes through a point ryz, 
 
 z=Ax+ By+C; 
 therefore, eliminating C, 
 z—z=A(r,—r) + Bly, —y). 
 Also, it 1s perpendicular to a line whose equations are 
 
 G 
 Z,—-Z2=2(0,-%),: ¥ ~y=y(@ 2), 
 
45 
 
 or (taking the equations to the projections on the planes of 
 wz and yz), 
 1 y 
 Tp BG ty 2), ¥, ~Y =~ 2)3 
 
 therefore, (Art. 6.) 
 
 1 
 A+-—=0, B+ 5=0. 
 zz z 
 
 Hence, substituting for A and B their values, the equation to 
 the normal plane at a point ryz, is 
 
 (2,2) + (yy) y¥ +, — 2) 2s 0. 
 
 35. To find the equations to the line of intersection of 
 two consecutive normal planes to a curve of double curvature ; 
 and the equation to the osculating plane. 
 
 The equation to the normal plane at a point xyz, Is 
 2 ceca (y,-yy + (2,— 2) pad, moma aE 
 
 and if we consider the normal plane to the consecutive point, it 
 will manifestly cut the former in a straight line, whose co- 
 ordinates have not varied, whilst x has changed to x+Ah. 
 Hence, if we join to equation (1), its differential with respect 
 to 1, viz. 
 
 (yy) f+, — 22" — 2 —y* — 1=0, 
 
 we have the equations to the line of intersection of two consecu- 
 tive normal planes; or, taking the projections on the planes of 
 vz and yz, the equations are 
 
 U 
 
 re a, Pa PF 7 : 
 2, a= (2) — (be? +42, 
 y y 
 
 Ze 7 £2 49 1 
 Y, SY ac ct Gist 2 ok y=. 
 a ue 
 
 Next, let 
 Op Te ee A (t,— ©) + Biy,- y) 
 
AG 
 
 be the equation to a plane, drawn through the point xyz of the 
 curve of double curvature, perpendicular to the line of intersec- 
 tion of two consecutive normal planes through the same point; 
 
 os , a 2’ 4? z 
 ._ A= at a en =i. (Arta) 
 if a 
 
 Hence, substituting these values of A and B, the equation to 
 
 the plane is 
 2y 4) 
 
 yz z 
 sal gl Ala Ay ans 
 y tf miay 
 
 This plane is called the osculating plane, or the plane in 
 which the curvature lies; it manifestly contains two consecutive 
 tangents to the curve, and passes through three consecutive 
 points of the curve. 
 
 34. The following are examples of finding equations te 
 curves of double curvature, and drawing their tangents, normal 
 planes, &Xc. 
 
 Pros. 1. A sphere is pierced by a cylinder, the diameter 
 of whose base is equal to the radius of the sphere; and their 
 surfaces are in contact. ‘To find the equations to the curve of 
 double curvature formed by their intersection. 
 
 Take a plane drawn through the center of the sphere per- 
 pendicular to the side of the cylinder, for the plane of ry, 
 (fig. 24.); and let AN=zx, NM=y, MP=z, be the co- 
 ordinates of the point P in the curve, AB=2a. Then, 
 because P is a point in the sphere, 2° + y*-+ 2°= 4a’; and 
 because M is a point in the semicircle, 2 + y*=2aza; 
 
 “yy =8ar—x, and 2*°=4a’—2ax, 
 are the equations required. 
 
 Therefore, differentiating these equations, 
 
 dy da VS 2 
 y a » (2+: Fes —/l\s 
 dy _ a—xX diy yf +(a-ay ay 
 
 a Tie : 
 "dx y dx” y y 
 
dz dz\* d*z 
 ” = aR a, ary 2 = Ons 
 
 dx ax dx 
 de a d* a 
 dx a ee Bae Dg 
 
 Hence, the equations to the tangent are 
 a—x a 
 Y,—Y= (v,—«), and z4,—z2=— -(1,—2); 
 
 in which, by making z,=0, we find the co-ordinates of the 
 point 7’, where the tangent meets the plane of ry, 
 
 = 
 
 3 
 Aas te 9% : -, (a, 2a)? 
 & =4a—2, BO pay Or ae Yo 
 
 x AQ x 
 
 is the equation to the locus of T. 
 
 The equation to the normal plane 1s 
 
 (6 Ve 
 
 ( ) 
 bia e 
 y Y, ¥ 
 
 =(z,—2) =27—-x-+ 
 z 
 And the equation to the -osculating plane is 
 a’ a® a’ (a—2x) a’ 
 ee ar cS (z,— 2) & a je saamt) 7 (Y,-Yae 
 o (yy, —y)y— (2-2) P =(@,—-2). far +(a—x)y'*}. 
 
 Pros. 2. To find the equations to the curve of double 
 curvature called the helix. 
 
 While the rectangle ABCM, (fig. 25.) revolves uniformly 
 about its side AB, the point: P moves uniformly along the 
 parallel side MC, and thus generates a curve called the helix. 
 When the rectangle is in the plane of xz, let P be at D, and 
 let the uniform velocity of the poimt M be to the velocity 
 Siete 3: L's eran lena =a, INI =o) % EVN: be.co- 
 ordinates of P, and AM=a. Then the projection of the curve 
 on the plane of xy is the circle MD, whose equation is 
 
 bs 
 r? +y’=a?; and PM=n.arc DM=na cos” * - ; 
 
48 
 
 -iI 
 i: y= fa? — x, and z=nacos” -, 
 ; a 
 
 are the equations required; the latter’ of which may also 
 
 1¥ 
 
 be written z=nasin~ =. 
 a 
 
 Cor. Since the corresponding increments of DM and PM 
 are always in a given ratio, the curve DP always cuts the 
 generating line CM, of the cylinder on which it is traced, 
 at the same angle; and the line touching DP is always 
 inclined to the plane of xy, at a constant angle (a) whose tan~ 
 gent=n; therefore also the length of the helix 
 
 : 7 £ 
 DP =arc DM ,/1+n?=asecacos~'~. 
 a 
 
 Pros. 3. Let A (fig. 26.) be the pole of the great circle 
 «By, of a sphere whose center is C3; and as the quadrant 
 AB revolves uniformly about AC, let the point P move 
 uniformly along AB; to find the curve which is the locus 
 
 Of. 
 
 Let CN=x, NM=y, MP=z, be co-ordinates of P, and 
 AC=a; and suppose that when the plane ACB coincided 
 
 ; ; 1 
 with that of wz, that P was at A; and let P’s velocity = - of 
 n 
 the velocity of the pomt B. Therefore are Br=n.arc AP, 
 re 
 
 1 
 
 or tan~ =ncos 
 
 RB Te 
 
 9 
 Hence, a? -+y°+2° =a", and tan’ 4 
 : x 
 
 are the required equations. 
 
 Cor. To find the area of the spherical surface included 
 between the plane of vy, and the path of P. 
 
 Let AB’ be a position of the quadrant very near to AB, 
 meeting the curve in P’; draw Pp parallel to BB’. Then the 
 
49 
 
 zone BPp B’ is the element of the required surface, and_ its 
 
 area= BB’. PM=n. P’p. PM=nad@. a cos8, if ACP=86; 
 
 Therefore area APBr=na’ sin@ =na’, from-0 =0, to @= = 
 
 Bo = 
 
 If n = 2, the curve is the spiral of Pappus, and the area=2a’. 
 
 If n=1, or the velocities of P and B are equal, the 
 equations to the curve are 
 
 2 g 
 QZ 
 
 z Aotetag! 
 Yhay n/a 2, and y = 
 a 
 
 Pros. 4. To find the equations to the Screw of 
 Archimedes. 
 
 This curve is the helix, described in Problem (2); but 
 the axis of the cylinder on which it is traced, is inclined to 
 the plane of xy, instead of being perpendicular to it. 
 
 Let the axis of the cylinder lie in the plane of xz, (fig. 27.) 
 then the base ACO of the cylinder is perpendicular to that 
 plane. From P, a point in the curve, draw PC parallel to the 
 axis of the cylinder, and CD, Dd, respectively parallel to the 
 axes Oy, Ox. Let AC=90; therefore CP =@ tana; if asthe 
 constant angle at which every element of AP is inclined to 
 
 ACOA -Let.4 A072; 
 
 then dist. of P from the plane of zz =CD=sin 0; 
 dist, of C from the plane of zy = Dd=(1-+ cos 9) cos e, 
 and therefore, dist. of P from that plane 
 = CP sin e— Dd=60 tan a sin e—(1+c0s 6) COs @; 
 dist. of D from the plane of ry = Od=(1-+ cos 8) sin e; 
 and therefore, dist. of P from that plane | 
 
 = CP cos e+Od=8 tan a cos e-+(1-- cos 8) sin e. 
 G 
 
50 
 
 If therefore, a, y, z, be co-ordinates of P measured from O, 
 parallel to the axes Ox, Oy, Oz, 
 
 then + =@ tan a sin e—(1+ cos 0) cos e, y=sin 9, 
 
 z=6 tan a cos e+(1+cos 8) sin e......(1); 
 
 and the equations are 
 x hin ad tana sin e —(1 ne Ae —y") Cos é. 
 ze=sin~*y tanacose+(1 + xf 1 —y’) sin é. 
 Cor. The value of 6, corresponding to that point in the 
 
 screw whose height above the plane of ry is a maximum, Is 
 tana 
 
 given by the equation sin 0= ane? 28 appears by making 
 an e 
 
 dz ; j 
 
 do =0, in equation (1). 
 
 35. To find the differentials of the volume and surface 
 of a solid, bounded by the co-ordinate planes, and a curve 
 surface whose equation is given. 
 
 Let z= @ (x, y) be the equation to the surface, and through 
 a point M of it, (fig. 28.) draw planes parallel to the co-or- 
 dinate planes; then the volume V of the figure NMEF is a 
 function of x and y; and if x and y receive increments h/ and k, 
 the point M will be brought to C, and the increment of the 
 solid, contained by the parallel planes ME, SD, SB, FM, 
 will be 
 
 dV ai; fa OY a°V ave ik 
 —h+ —k+ — — Lk; ae & 
 dx yt aes 1.8 dxdy a NE 
 
 But if we had made x alone vary, the increment would 
 have been 
 
 dV avi ke 
 MBRF= ar eh HF jlo +&c3 
 
51 
 
 similarly, if we had made y alone vary, the increment would 
 have been 
 GV) + hyd? Var? 
 
 MDEQ= Bet dy? 1,2 + &c} 
 
 therefore, by subtraction, we have the volume 
 7 
 
 dxdy 
 
 MCRQ= hk + &c, 
 
 which is ultimately equal to the prism whose base 1s QR and 
 altitude MP, and whose volume therefore = zhk; therefore 
 
 9 
 
 uh Ak+&c. 
 limit of dep ronyss 9h} <a or atv 
 zhk dudy 
 Also if S=surface NM, by a similar process, we may shew 
 that the surface MC = saan &c, which «is ultimately 
 
 equal to the portion of the tangent plane at M, itelizepted 
 by the prism erected on the base QR. But if c=angle of 
 inclination of tangent AG at M to the plane of ry, 
 
 Ter 
 
 therefore intercepted portion of tangent plane =Ak. sec 7 
 
 as SSS» (Art. 25, Cor. 1.) 5 
 
 ee lad 
 therefore limit of he Nata Hat tas as Ls 3d = |e 
 
 Ak 1+ Gye ey 
 
 eo Wis (2) +4). 
 
52 
 
 Cor. 1. In some cases, it is convenient to express the 
 differential of the volume by polar co-ordinates. Let Qp, qr, 
 (fig. 29.) be portions of concentric spherical surfaces, intercepted 
 between two planes passing through the axis of z, and two 
 planes perpendicular to that axis, all indefinitely near to one 
 another; OP=r, rOM=¢, COP=86; therefore Pp=rd0, 
 Qq=dr, PA=PN.dp=rsin Odd (PN being perpendicular 
 to OC); now the elementary figure gp is ultimately a rectan- 
 gular parallelopiped, and its volume=r'sinOdrd@dq@; there- 
 fore the volume of the body =ff[fr° sin@drdOdq@. If 2, y, 2, 
 be co-ordinates of the point P supposed to be situated at the 
 surface of the body, it is manifest that 
 
 z=rcos@, r=r sinO cosd, y=r sin® sing; 
 hence, by substitution, we shall have the equation to the surface in 
 polar co-ordinates; and thence a value of (7), in terms of @ and 0, 
 to be substituted in the above expression for the volume. 
 
 For instance, in a sphere, 
 
 3 
 
 a° ae oT a oy Qa” 
 = 5 ff sino d0dp = : [sind dd = (1 — cosa), 
 
 =O, r=a the radius. 
 between the limits «@=0, P= 27. 
 0=0,'0—a: 
 
 which is the expression for the volume of a spherical sector, whose 
 vertical angle = 2a. , 
 
 Cor. 2. There is also another way of expressing the 
 differentials of the volume and surface of a solid, which is 
 often convenient in practice, 
 
 Let DMQ, (fig. 30.) be a curve surface, QD being its trace 
 in the plane of xz. Draw planes MP, Np, parallel to ZY 3 
 and MA, NA, planes passing through Az. Let a, y, z, be 
 co-ordinates of the point J, and uw=tanZ HAP, therefore 
 y=uxr; also let rth, (vx +h) (u+@) , be co-ordinates of the 
 pomt K. Then A AHP =42"u, 
 
 z 0 oO 
 and area KH =(a-+h)’ fo ho («+ & 
 
53 
 
 Now the volume DMP may be considered as a function 
 of x and zw; and its increment, contained between the planes 
 DK, Dk, Pm, pN, 1s 
 
 dV dV a’V ih’ dV 
 
 Ve 
 —0s ee dx’ 1.2 + ton dué 1.2 
 
 + &Xc, 
 
 Also MpQk, arising from the variation of x only, 
 
 ave. EV he 
 
 D de dat) 2 
 
 and Am4H, arising from the variation of wu only, 
 
 dV d’V @ 
 
 + &¢3 
 
 wl 
 *. by subtraction, M Khn= aa hO + &ce 
 MKhn 
 But limit of Se AWGE MH, 
 d°V 
 hO+&c. 
 ae x du ; av 
 .. limit of —————————- =z, and therefore ——— = xz. 
 hO (« +2) dx du 
 2 
 
 Similarly, if S = area of the surface DMQ, 
 
 d’S 
 ee ea a 
 MmNn Tdythh Cc; 
 
 N 
 area KH 
 
 but limit of = secant of inclination of the tangent plane 
 
 at M, to.the plane of ry; 
 
 ra) 
 
 ~ 
 
 *. limit’ of ————_—_—_—_ dvdu 2/14 (8) (BY 
 Baas CAG) 
 
 Vin (G) + G@) 
 old Th} dx sae 
 
54 
 
 36. To find the differential of the arc of a curve of 
 double curvature, 
 
 Let z=@2, y=fx, be the equations to the curve; and let 
 kK and / be the respective increments of y and z, when z be- 
 
 comes «+h; 
 
 ey 4 ei § + &c, 
 dx dx 1.2 i 
 y dy h’ 
 k= — 4 & 
 detate & 
 
 Let R=length of the chord joining- the two consecutive 
 
 points in the curve ; 
 
 RaR+P+Pe aR jit (2) + +5 ~) } + AR + &; 
 - (Gz) = tinitot Feat Gj LY +(Z =) ; 
 o 7 14 (4) + us ih 
 
 37. Application of the formule for volumes and: surfaces 
 to examples. 
 
 I. To find the volume and surface of any portion of a 
 
 sphere (fig. 31.). Here 2* + y’? +2° =a’; 
 
 aVv C 2 Q 
 
 ay —o a? x ae y 3 
 Vides doy ne m 2 — Gj ar ; ES vie 
 j —=f rr»fa —x —y =circular area (rad. = cara) 
 
 and abscissa=.2) from t=O, to r= CN= Ja — y"; 
 
 = “dys Py Ai 3 («'y-*) , 
 
 no correction being added, for V and y vanish together. 
 
55 
 
 3 
 -. volume PQC =" (vy - =) ; and when y =a, 
 
 Tv 
 
 volume ABCD= - pik ene 
 nae ak 0 
 
 ee 
 
 therefore volume of whole sphere = 8 . et = = 
 eae S OGD Ga. N/a as 
 A —_ = ot ES ue. 
 gain ade Gs 4) (; ean | I+ at a 
 
 . 7h aoe 
 : rie a sain 
 59 —Te Ta 
 (from 2 =O} toe =, [a —y’) => : 
 
 therefore S = oe = surface PA BQ; 
 
 and when y=a, surface DAB = ste : 
 
 al 
 
 2 
 
 Ta : 
 *, surface of whole sphere ap m8 = 47a. 
 
 II. ‘To find the volume of any portion of an ellipsoid. 
 
 2 2 
 
 a tae Ft a 
 The equation is 3 ae = “— 2 =1, (fig. 31.) 
 
 a°Vv j x y 
 .—— HS T2Z=—CLr —-— 4% 
 dxrdu an 6 
 
 age ve : ( me =) 
 — as Ta _ za 
 a” bay 
 
dV c 1 
 .~—= C -—- —— it (+ \ 
 du ae 4 a 
 3 (aa) 
 1 
 from + =0, mA r= ON= ARES 
 u 
 : 2 2 
 ; +4 y Ten! Nae 1 
 in the plane U8 Rie +- Re SRS A 2 ar Rn ~-) 
 C 
 .”. substituting its value for Cc, — = —>-. 
 du 1 u 
 3 (es + 
 * b? 
 au 
 sings abe eal) pis Obey = age 
 
 1) 
 
 the expression for the volume of the wedge ABCQ; and by 
 making w infinite, 
 abc 
 
 volume of ABCD= er 
 
 is 
 
 4mrabe 
 
 therefore volume of the whole ellipsoid = 
 
 III. To find the surface of an ellipsoid which differs little 
 
 from a sphere. 
 
 xr y ‘ 
 2? 
 
 y+ ey ee 
 
 a 
 
 therefore 
 
ied 
 making =M, —>=-p. 
 Let y=b sin@ cos d, r=a sinO sing, and therefore z =c cos6; 
 differentiate the equation « =a sin@ sing, regarding 6 as constant, 
 
 2 
 
 then dv=a sin®@ cosh dd; and = + - = sin’, 
 z 2 
 
 therefore ydy = b° sin® cos0 dé. 
 
 Hence dxdy=abd@dé sin 8 cos 6; 
 4 as b e fa) & -* Oo ¢) ) in’@ 
 oo = ; See sin” cos” F 
 ibd ab sin (u sin’ + vcos @)s 
 
 =absin 0/1 —p Yh making #2 soaitin =). 
 
 = ab sin 0 (1—+ p sin*O— =P" 5 5 sin°O—&c.). 
 Each term to be integrated is of the form 
 JS sin” 640, which= Se (n odd) from 0=0, to0=-. 
 dS§ ] 1 1 
 gtr  acd Ry ee ie SEY Bdartee 
 (MA has PI suk a 
 
 Now let f pdp= ae J pdgo = z P,; &c.,. between 
 
 the limits @ =0, paar therefore surface of $th of ellipsoid 
 
 mab 1 1 
 = —— }]—1P, —-—P,-— — P,—&c.!. 
 2 TOI We a ete 
 uty one Bs 1.1 aE PSs 
 where P, = Thee 1 aia orlin tt aia +p mare eect 
 
58 
 
 and in general, P,, is the coefficient of z”, in the development of 
 
 (1 —pz)74. (1 —vz)73. 
 
 Cor. In the case of a spheroid a= 6, and therefore «=v; 
 eth 
 '" dodé 
 
 =a’ sin 0,/1—p sin’ @; 
 
 Hence for an oblate spheroid, 4 =e’, and whole surface 
 ‘tana Ji 
 = on Sai-+ < log ea : 
 e 1 +.¢ 
 
 for a prolate, ¢ is greater than a, 
 
 a. 
 “. w= —e’, and whole surface =27c Sco+- - sin~ "et. 
 e 
 
 IV. A sphere being pierced by a cylinder, as in Prob. 1. 
 Art. 34, to find the volume and surface of that part of the 
 sphere, which is intercepted by the cylinder. 
 
 Let 4BAM=60, and u= : = tan 0; (fig. 24.) 
 therefore, (Art. $5. Cor, 2) 
 
 ; Ba oe Pe 
 Sy ey pee) eT Se at: 
 
 dxrdu 
 dV Lasik 
 dig ala atest}, 
 : a 
 from xr =0, to w=AN=acos’6@= ———;; 
 1+u 
 therefore, substituting for C, 
 dV “f 1 a> r 
 ae Sof 
 du 3l1i+u a42)t5 
 ga Sr et A x 
 ges ty wear ane eS 
 Cl + wu") (l+u 
 
: ] 
 VY= Sf tan= tu + airs =t+C, 
 from w=0, tou= @ 5 
 3 
 eeardad Se, 2} = me _ 2a 
 3 le 3) 9 
 1 Ada?® Ta” 
 Now the volume of ABCD=-. alr az i a . Therefore 
 
 3 
 
 part of ABCD not comprised in the cylinder = aa . Conse- 
 quently, if the whole sphere be pierced by two equal cylinders, 
 
 the part not comprized in them = = (diameter) 
 Again, to find the Aa ie 
 
 eaystV “@ +@)a=V +E) +@ 
 
 =-oOoCr OE OO * 
 
 JJ +u?)’ 
 
 FE a oS Se —x(l+), 
 
 a UE OE - 
 from t=0, tor= Rie. ; therefore, substituting for C, 
 ub 
 
 ayn 
 du =0'(e wo sre ae 
 
 Sa’ (tan-'w+ aaa) +6 from u=0, to u= © ; 
 u 
 
 or S=a’ G -). 
 
 a 
 Now the surface BCD =~ 4nd? = 
 
 g 
 part of BCD not intercepted=a*. And if the whole sphere 
 
 Therefore the 
 
60 
 
 be penetrated by two equal cylinders, the part not comprized in 
 them = Sa* = 2 (diameter). 
 
 V. To find the equation to the Cono-cuneus of Wallis, 
 and its volume, Let BOG, fig. 32, be a quadrant of a circle, 
 whose plane is parallel to that-of +z; and let a straight line 
 DN move parallel to the plane of yz, with one end in the 
 circumference of the quadrant, and the other in the axis of 73 
 the surface generated is the Cono-cuneus of Wallis. Let 
 
 ING ar iis Uae ate ha 
 
 ZC : 
 and .. DC =—; hence #” + G) =f, 
 J 
 LY Spare ae : goo 
 or z= or: raf a — x, 15 the required equation. Hence 
 
 =! alae, ek ay 
 ae ~ O¢ Ja mad Sa 
 
 dV 
 
 c 
 from y = 0, to y =¢3 or gel A ee 
 
 *, volume ABOD = 5+ (circular area BOCD); 
 
 and whole volume = 
 
 twola 
 AN 
 (oe) 
 
 VI. The axes of two equal cylinders intersect at right 
 angles, to find the volume of the portion which is common to 
 both. Let OC be the axis of one cylinder, (fig. 33,) AO that 
 of the other. Then the equation to AB, or to the cylinder 
 
 whose base it forms, is z = Ka — ys 
 a2V 
 
 dV 
 asap iN ee re dy ~=N4 —y +C, 
 
 one garg dV 
 from «= 0, to r= ON = fa — x: or = a? — y's 
 y 
 
 y 
 * V = a'y ~~ (from y = 0, lo y= 4) = a 
 
61 
 
 16a° 
 
 Hence, multiplying by 8, the whole volume = 3 
 
 We may also determine the surface of one cylinder which 
 which is intercepted by the other. 
 
 zs RN eer ike Bea RN St) a3 
 
 and .. S = ay =a’; and .*. whole surface = 8a’. 
 
 } $ 
 Cor. =i=nafau- yy =“ =a sw's, if BM=s; 
 on. «. RIM = 2 y rE $3 
 hence the curve of intersection BD lies all in one plane, and is 
 therefore an ellipse; and if the cylinder were developed, it 
 
 would become the line of sines. 
 
 VII. If a plane be drawn through the axis major, 
 perpendicular to the plane of an ellipse; and on it an hyper- 
 bola be described, whose vertex is in the focus of the ellipse, 
 and focus in the vertex of the ellipse; then the surfaces of all 
 cones whose base is the ellipse, and whose summits are on the 
 hyperbola, or its asymptote, can be squared respectively by 
 circular or elliptic arcs. 
 
 Case I. C the center of the ellipse, AC, CB, its semi- 
 axes; V the vertex of the cone, (fig. 34,) on the hyperbola ; 
 
 CN = x = asing, and, therefore NP = y = bcos@, | 
 
 co-ordinates of a point P in the ellipse; arc BP =s; 
 
 aas =// du +dy* =dd/ a*— c’ sin’ d, (where c* =a’ — b”). 
 
 CM =/f, VM = h, co-ordinates of V; draw MQ perpendicular 
 
 on thy tangent to the ellipse at P; .. CT = — = ——_; 
 
62 
 
 at b(a — fsing@) | 
 c —/) binge Ages: = rain 
 
 ae ante J ie + 6° ae ar Li 
 — c* sin Ribu 
 
 If therefore § = ese of the cone, 
 
 a8 = ds = Ldpr/h'(a* — c’ sin?) + 0? (a — f sing) 
 
 di dd sf a*h? + a7b* — ch’ sin® p—2ab’fsingd + f° sin” Pp. 
 But the equation to the locus of V is 6? f* = c*(h* + 6°); 
 os te td J a’ h? + ab — 2ab’ f sing + 6c? sin’ p 
 
 = 1d (2) _ sto! eb sin@ + (bc sin )* 
 
 b 
 = ido (4 ~ besing) 
 Cc 
 
 ih + be cosh) +, C; 
 
 s=1(p~ 
 
 suppose the generating line at first in the position VM, or that 
 
 9 
 
 10139 
 
 S=0, when x = — a, and therefore @ = — 
 
 .*. substituting for C, s=4} & - >) of + be cos} ; 
 
 b 
 Hence surface MV B = 5 (= Te + be) , when d=0, 
 
 € 
 
 Tv 
 
 b 
 surface MV A = 4 (x cae ) , when O= -. 
 2 Cc 9) 
 
 b 
 *. by subtraction, surface BVA = 4 E% ais bc Ni 
 
63 
 
 Hence the difference of surfaces MVB and BV A = bc, and is 
 
 the same for all cones, that have their vertices on the hyperbola. 
 
 abf wf 
 
 Whole surface of cone = 7 -—~ = 
 
 a een 
 sai a8 
 
 Cor. 1. Since the difference of the surfaces MV B and 
 BVA=be, we have the following problem. 
 
 If a plane be drawn through the minor axis of the base, and 
 the vertex of any cone described as above ; the difference of the 
 portions, into which its surface is thereby divided, is equal to the 
 rectangle contained by the minor axis, and eccentricity of its base. 
 
 Cor. 2. Suppose the base circular, and let ACP=¢; 
 therefore MQ=a—/f cos @, since MQ is parallel to CP ; 
 
 ». dS=Lado J +(a—f cos $y, 
 
 which can only be integrated by elliptic arcs. 
 
 Case II. When the vertex is on the asymptote to the hy- 
 perbola, and therefore bf = ch, 
 
 then dS=3dq s/f (ah) + (aby — 2ab? fsin dp ; 
 let p=2y— =, “. sng = — cos2y; 
 “ dS=dwW/ (ah) +(aby + 2abch cos2y 
 hn? = ach 
 = bd SiGe) ay (1-42 sin” 
 WV a+ (=) + ; (1 — 2 sin’ W) 
 ah” 5 le ee oy Ct 
 =bdw (c+ a +6 ‘a oid Wy, sincea =O +c’. 
 Let (c+ NED a eo = VM 
 ; : 2 
 
 ahye A i ; ya 
 (c-=) +6 =(a—fyY+h=a =VA 
 
dS =bdw ns br a (a”* —a’) sin? 1p; 
 
 +, S=b. fan elliptic arc, axes a, a’, and abscissa r=a’ sin}. 
 
 e . T 
 Supposing S to vanish, when Y =0, and therefore P= — 5\3 
 that is, when the generating line coincides with VM. 
 
 . . Tv Tv 
 Hence making @ = oe and therefore ~ = nos 
 
 entire surface=2 (elliptic quadrant, whose 5 axes are VM, VA). 
 
 VIII. To determine the curve to be traced on the surface 
 of a sphere, so that its length shall always be equal to () times 
 the co-ordinate (x) of the describing point. 
 
 Let the radius =1, then the equation to the sphere is equiva- 
 lent to the equations. 
 
 r=cosd, y=singsnO, z= sing cosO; 
 +, ds’ =dx’+dy’'+dz2° = (-sing dp)’ 
 + (cos @ sin@ dod + sing cos @ dé)’ 
 + (cos f cos6 dg — sing sin@ de)’ 
 = (dp) + (sing dé)’: But s=nxr=n cos@, 
 *, ds’ = (—nsing dd)’; 
 . n® sin’ ia = (doy + sin’ p (dO, 
 
 or dO = PENG e nin? hl sin pb — 1 
 
 ae 
 
 Sain OS eam. \ eae 
 
 iy: Jn? sin? pb — 1 sing ./n* sin? p—1 
 Sd (cosp yy Ll EoD 
 
 Hi ties Oy (eee 
 
65 
 
 2 COS , ot 
 te 6.=3 1.008 + al + sin yeas 
 
 Jn a | nm —1 
 
 he ay, OB: 
 
 or sin 5 =n cos 
 /l—x J n?—1 
 ab x 
 ++ sin 
 
 / (nv? = 1) (127) pie 
 
 the equation to the projection of the curve, on the plane of ry. 
 
SECTION IV. 
 
 ON THE TRANSFORMATION OF CO-ORDINATES, AND ITS 
 APPLICATIONS. 
 
 —g-—— 
 
 38. To pass from one system of co-ordinates to another, 
 supposing the first rectangular, and the second oblique. 
 
 First, to change the origin of the co-ordinates without altering 
 the directions of the axes, we must substitute 2 +a, y’+8, 
 z'-+ry, for x, y, 2, respectively; a, 3, yy, being the co-ordinates 
 of the new origin. 
 
 Next, suppose the directions of the axes are to be changed, 
 retaining the same origin. Let x, y,z, be rectangular co-ordinates 
 of the pot P, (fig. 35.) parallel to the axes Ox, Oy, Oz; 
 
 ON=7, NM=y, MP= 2, 
 its co-ordinates parallel to the oblique axes Ox’, Oy’, Oz’. 
 Through the points P, M, N draw planes parallel to zy, 
 cutting Ow in the points p, m, n; then, because the distance be- 
 tween two parallel planes is equal to the intercepted portion of 
 
 any line, multiplied by the cosine of its inclination to the line 
 which is perpendicular to them, 
 
 , « 
 . mp = MP cosz’Oxr=z' cosz'z, denoting the Z 2’ Ox by z'2; 
 . . / 
 similarly 2m =y' cos y'a, On=2x' cosx’x; andr= Op; 
 “. c=2 cosxxrt+y cosy’ r+2' cosz’7r; 
 
 that is, each primitive co-ordinate is equal to the sum of the pro- 
 jections of the three new co-ordinates upon its axis. 
 
 Hence 
 y=xr cosxyt+y' cosy y+z' cosz'y. 
 
 z=2' cosx2z-+y cosy z+2' cosz’z. 
 
67 
 
 Of the nine angles involved in these formule, six only are 
 independent; there being three equations of condition; for, since 
 wax, x'y, x 2, are the angles which a straight line, namely the axis 
 of x, makes with the three rectangular axes of x, y, z, 
 
 “. cos’ 2 x+cos a’y+cos az = 1 
 similarly cos’ yx +cos’ y’y-+cos’ y'z = 1} «.(1). 
 cos’ z’°x+ cos’ 2’ y + cos* 2’z = 1 
 
 Cor: 1. We may however obtain expressions free from 
 
 equations of condition, thus ; 
 
 L= MZ) LSmMz) c= m,z 
 Let ; be respectively the equa- 
 
 y=nz) y=n2) y=n2 
 
 tions to the axes of a, y’, z. 
 1 i 
 Also let 0 = —jp—————— , PO ee 
 /1 +m +n nL tm + n° 
 1 
 L 
 
 nea AG 
 Then cos 2’ x = Im, cosry =In, cosxz =, &c; 
 / , sds , 
 m= lngitit my + bm,z, 
 y=lIne +l n'y + ln,z, 
 , Te Ws , 
 z= lz tL york Jz. 
 
 39. When the new system is rectangular, as well as the 
 primitive, we may arrive at the expressions for x, y, 2, briefly 
 thus. Let (7) be the distance of any point P from the origin ; 
 .cos Pxr=cos P2’ cosrx’+cos Py’ cos ry’ + cos Pz’ cos x2’; 
 hence, multiplying by (7), and observing that 
 
 rcos Pr =z, rcosPx =z, &e, 
 we have # = a’ cosa’x + y'cosy’x + 2'cosz x; and so on, 
 
 for the others. 
 
 In this case, there are only three independent angles. For, 
 since the axes of x and 7’ are at right angles, 
 
 A , , , , , ) 
 ~. cosy xcosy % + cosxycosy y + cosxzcosyz =O. 
 
 7 
 
68 
 
 similarly, cos x’x cos zx + cos2’y cosz’y + cosa’z cosz'z = 0. 
 , , , / , / 
 cosy «cosz xz + cosy ycosz y+ cosy zcoszz=0. 
 which three equations of condition, must be joined to the three 
 equations (1) of last article. 
 
 Cor. It is also to be particularly observed, that these six 
 equations may be replaced by the following, which are equivalent 
 to them, and which are obtained by referring the axes of x, y, 2, 
 to those of x, y’, 2’; 
 
 cos’rar + cos*ry’ + cos rz = 1 
 cos’ yx + cos yy + cosyz = 1 
 cos’ zx’ + cos’ zy’ + cos’ zz’ = | 
 because x2’, xy’, xz’, are the angles which a straight line (viz. 
 
 the axis of x) makes with the three rectangular axes of 2’, y's Zz 
 and so on for the others. Again, 
 
 , / , 4 , , 
 cosrx cosyx -- cosry cosyy + cosxz cosyz =O 
 , / , / , / 
 cosrx coszx + cos xy cos zy -+ cos xz coszz =O 
 , 
 cosyx coszu + cosyy’ cos zy’ + cosyz coszz’ = O 
 because the axes of x and y are at right angles; and so on for 
 the others. 
 
 40. When both systems are rectangular, the expressions 
 for the transformation of co-ordinates may be reduced to others 
 more convenient in practice, in the following manner. (Fig. 36). 
 About O the origin, describe a sphere radius (1); and let Gzz’ 
 be the plane of 22’, Oy’ perpendicular to it the axis of 7, 
 which will le in the planeyOr; 2 GOr = ¢, 4 GOzd' = 8. 
 
 ‘Then, by Napier’s rules, observing that 2 z’zx = 180 — @, 
 i z2y = 90 + dQ, and that 2’ z2, Zz 2%, are quadrantal triangles ; 
 
 cosra = cosOcosd, cosyx = cosOsing, cosza’ = sin O; 
 cosry = — sing, cos yy = cos d, cos zy' = 0; 
 cosrz = —sinOcosd, cosyz = — sinfsing, coszz’ = cos8; 
 
 hence, by substitution, 
 x =(x' cos 0— 2’ sin 0) cos bp —y’' sing, 
 y= (2 cos 9—2’ sin 8) sin pty’ cos d, 
 2=2 sinO+z' cos 0. 
 
69 
 
 41. To find the equation to the section of a surface, made 
 by a plane perpendicular to that of xz, and inclined at anZ =0 
 to the plane of ry. 
 
 This is effected by transforming the co-ordinates so that the 
 cutting plane shall be that of xy, and then making 2’ = O in the 
 resulting equation. Since each original co-ordinate is equal to 
 the sum of the projections of the three new co-ordinates upon 
 its axis, 
 
 “.r=2 cosx x+y’ cosy’x+ 2’ cos? r=x' cos0—2’ sinO, 
 because Za r=0, Ly'x=90, 42/x=90+0; 
 similarly y=.2' cosa y+y' cosy’y+2' cosz y=y’, 
 z=2 cosaz+y' cos yz+2' cosz’z=2" sinO+2’ cos6. 
 
 Since 2’ is to vanish in the final result, we are at liberty to 
 make it =O, in the values of x, y, 2, before we substitute them ; 
 and therefore it will be sufficient to write 2 cos@ for x, y’ for y, 
 and x’ sin@ for z; and the resulting equation between 2’ and 9’, 
 will be that to the section. 
 
 Cor. We may obtain more general formule, by making 
 z'=0 in the results of (Art. 40.); this gives, 
 x=2x cosO cosh — y' sing. 
 y=wx cos sind + y' cosd. 
 
 z=a sin@. 
 
 By substituting these values, we shall obtam the equation to 
 the section made by the plane of xy’; that is, by a plane whose 
 inclination to that of ry is 2 @, and whose trace on that plane is 
 the axis of y’; @ being the Z which the projection ofthe axis 
 of x’, on the plane of xy, makes with the axis of x. 
 
 This of course includes all planes. 
 
 Examp.e 1. To find the equation to the section of an 
 oblique cone. 
 
 Let the axis be in the plane of rz, then 6 =0, and the equa- 
 tion to the surface is 
 
 (cy)? = (cr —az) [2cr + (a—Qr)z—cx} (Art. 19.); 
 
70 
 therefore the equation to the section is 
 (cy’)* =(c cosO ~ a sin) {2cra’ + x? (a—@r. sin@—ccos8)}. 
 
 This will be an ellipse, hyperbola, or parabola, according 
 as the coefficient of 2” is —, +, or O; that is, according as 
 
 which = tan SEB (fig. 37.); or, 
 
 tan @: <i > yor 
 a—Qr 
 
 under the same circumstances as in the right cone. 
 The section will be a circle, when 
 c’? =(c cos0—a sin®@) (c cos0—e sin@), if e=a—2r; 
 or c’ +c? tan?O = c*—(ce + ac) tan@ + ae tan’@; 
 c(a + e) 
 
 ? 
 
 “? tan 0 —=.0, a tan Os 
 ae—c 
 
 the first value of @ gives the base of the cone; the second value gives 
 
 Gurnee 
 _— + —_ 
 re G4 
 tan? = > = tan(SEB + SAB), 
 c 
 Le 
 ae 
 
 . 02—-SAB=SEB, or SAD=SEB; 
 that is, the cutting plane makes the same angle with one side of 
 the cone, that the base does with the other. This section is called 
 the sub-contrary section. 
 
 If ¢ be the eccentricity of the elliptic section, it appears 
 é >»  (c¢ —ae)sin’@ +c (a + e) cosO sinO 
 from its equation that « = Saeaahy air ya urd Toma oat 
 Ex. 2. Of all sections of a paraboloid of revolution, made 
 by planes passing through a tangent to its base, to find that 
 whose area is the greatest. 
 
 If the base of the paraboloid be the plane of wy, anda 
 tangent to the base the axis of y, and consequently one of: its 
 diameters the axis of #, the equation to the surface will be 
 
 2 
 
 b 
 (b—# +y = SAG = 2) 
 
 where a = altitude, and 6 = radius of base, 
 
71 
 
 Hence, if a plane passing through the axis of y, make an angle 
 0 with the plane of xy, the equation to the section is 
 
 Qg 
 2bhxcos@ — x cos?@ — y = —-r sin@; 
 ) a 
 
 one 
 or y* =2 (2b cos0 — — sin8)— 2° cos’ 
 a 
 
 b 
 — 2 US die pie lind, 
 = cos 0 )2b2 secO (1 <a tan@ ) a My 
 
 which represents an ellipse, whose 4 major axis 
 
 = b secO (1 — x tan), and $ minor axis = b(1 —n tan8), 
 where n= — 3 
 Qa 
 
 .. the area of the section will be a maximum, 
 when sec 0 (1—n tan 9)’ is so; 
 
 Mite Nil CATs Oe a’ — 6b" 
 
 this gives tan@ = = 
 x On 3b ; 
 
 hence (a) must be greater than 6 r/ 6: 
 
 the upper sign corresponds to a maximum, the lower to a mini- 
 mum. 
 
 Ex. 3. To find the equation to the curve formed by the 
 intersection of any plane with a surface of the second order. 
 
 Let ax? + by’ + cz? + 2ex =O be the equation to the 
 surface, which may represent all surfaces of the second order ; 
 therefore, substituting for x, y, z, the values found in(Cor. Art. 41.) 
 
 z= x’ cosO cosd —y' sing, 
 
 y= x cos@ sin p +4 cos, 
 
 z=2 sinO, 
 the result, developed and arranged, will be 
 x’? \cos’0 (a cos’ +6 sin’d) +¢ sin’ 8} + y* (a sin’ @ -+ b cos’) 
 +22’ y' (b—a) cosO cosh sin@+2ex' cosP cosp—Zey’sinp=0; 
 
72 
 
 the equation to a curve of the second order, which will be an 
 ellipse, parabola, or hyperbola, according as the quantity 
 (b—a)’ (cos@ cos@ sing) — (a sind + b cos’) x 
 fc sin’ O + (a cos + 6 sin’) cos’@} , 
 or —abcos’@0 —ac (sin@ sin@)*— bc (cos@ sin 6)’, 
 
 is negative, nothing, or positive. 
 
 Hence, every section of an ellipsoid is an ellipse, because 
 all the quantities a, b, c, are positive; for an hyperboloid of one, 
 or two sheets, since any two of them may be negative, or positive, 
 the section may be an ellipse, hyperbola, or parabola. For 
 paraboloids a=0; therefore for the elliptic, in which case 6 and c 
 have the same sign, the section is an ellipse, except when 0 =0, 
 
 ti, Pure ; 
 co 37 in which case it 1s a parabola; for the hyperbolic 
 
 paraboloid, since 6 and ¢ are of contrary signs, the section is an 
 hyperbola, except when 
 
 OfE210, or o = 
 
 > 
 
 214 
 
 when it is a parabola. 
 
 42. ‘To find the co-ordinates of the center of a surface of 
 the second order; and the relation among the coefficients of the 
 general equation, when it represents surfaces that have not 
 a center. 
 
 The general equation of the second order is 
 ax + by’ +cz + 2ay2 + 20 az + 2c ry 
 + Qaxer + 26y + 2c2 +d=0;3 
 if, by assuming another origin, we can make the terms multiplied 
 by x, y, 2, disappear, that point is manifestly the center; for 
 then the equation will not be altered by changing z, y, z, into 
 — a, —y, — 2, and therefore all lines drawn through the 
 origin, and terminated by the surface, will be bisected in it. 
 Let therefore r =e +a,y=y +68, z=27+y, 4,8, ¥ 
 
 being the co-ordinates of the new origin; then we must have 
 
 coefficient of 2, aat+b'y+c’ B+a,=0, (1); 
 
73. 
 
 coefficient of y, b8 + ay + ca+tb =0...(Q), 
 coefficient of 2, cyt a B+ ba+c,=0...(3). 
 Eliminate 3 from the first and second equations, 
 “(ab — ce’) a + (60 — a’) ytab = b,c’ =, 
 Again eliminate 8 from the second and third, 
 vac — bb)a + (a? — boy +ab — be, = 0. 
 Hence a will be infinite, when 
 (ab—c”) (a? — bc) + (a'c — bY =0, 
 or abc — aa’ — bb? — cc”? + 2a'b'c = 0; 
 which is the relation, when the center of the surface is at an 
 infinite distance. 
 
 When this relation does not hold, equations (1), (2), (3), 
 will give the values of a, 8, yy, the co-ordinates of the center 
 
 of the surface ; and the origin being in that point, the equation 
 | alae tet 
 
 becomes a2” + by? + c27 + Qa'y' 2 + Aba’ 2’ + Qoyx 
 taa+bB+ey +d=0. 
 
 A3. ‘To find the equation to a diametral plane of a surface 
 of the second order; 1. e. to the locus of the middle points of a 
 system of parallel chords. 
 
 Let r= mz, y= nz, be the equations to a Ime drawn 
 through the origin, to which the chords are parallel. ‘Take the 
 middle point (a-+y) of any chord for the new origin, that is, 
 maker = 2x +a, y=y' + BP, z= 2’ + ¥;3 then the equa- 
 tions to the chord from the new origin being x =mz2’, y =n2z’, 
 to get the co-ordinates of the points where it meets the surface, 
 we must write mz’ + a, nz + B, 2 ++, for x, y, 2, respec- 
 tively, in the general equation to the surface given in the last Art. 
 *. a(mz +a) + b(n2’ +B) +e (2) +y) +2 (nz’ + B) (2 +) 
 +26 (mz +a)(2’ +r) $2 (mz' + a)(n2’ + B)+24,(m2’ + a) 
 
 + 26,(nz’ + B) + 2c,(2 +y) +d =0;3 
 but since the origin bisects the chord, the two values of 2’ in 
 this equation are equal, therefore the first power of 2° must 
 disappear; hence 
 
 Q2ama + 2bnB + Qty + Qa! (ny + 3) + 2b' (my + a) 
 + 2c (mB + na) + 2am + 2bn + @, = O, 
 K 
 
74 
 or (am+ecn+b)at+(cm+bn+a)pB 
 +(Um+an+eo)y + anm+bn+¢,=0; 
 
 the relation among the co-ordinates a, (3, yy. of the middle point 
 of a chord, which we see is the equation to a plane. 
 
 Cor. If the surface have a center, and the origin be in 
 that point, we must make a,, 6, c, = 0. The equation to the 
 diametral plane then becomes 
 
 (am+cn+b)at+(om+bn+a)B+(Um+ta'n+c) y=03 
 and it is parallel to the tangent plane applied at the extremity 
 of the diameter x = mz, y = nz. 
 
 For the equation to a tangent plane is 
 Z2—z=p(e—a+qy-y 
 ax+bztcy,, bytazter, 
 re dy hea ) cztaytbe Ys 
 or z(czt+aytba +e (art h24+ cy) 
 +y(byta@z+¢rx)+d=0;> 
 
 and if it be applied at the extremity of the diameter x = mz, 
 y = NZ, its equation becomes 
 
 , , , , , , ‘ , ’ d 
 (am+en+b)x +(cm+tbn+a)y+Om+an+c)z thy =0, 
 which represents a plane parallel to the diametral plane. 
 
 4A, ‘There is an infinite number of systems of oblique co- 
 ordinates, but only one system of rectangular co-ordinates, which 
 will cause the terms involving ry, xz, yz, to disappear from 
 the general equation of the second order. 
 
 Suppose the original axes of the co-ordinates rectangular, 
 and the new axes oblique; and let «x = mz, y = nz be the 
 equations to the axis of 2’, r=m'z, y=n'z those to the axis 
 of y’, c=m,2, y =n,z those to the axis of 2’. Then (Art.38.Cor.} 
 
 c=lmie tl my +hm2, yslna +l ry +1n7, 
 zg=lz’+ Ly’ +12’; 
 substitute these values in the general equation 
 
 ax? +by? +e 4+2a'yz+ 2b rz+Qcxry 
 
: 75 
 +2aa+2by+2c,z2+d=0, and we find 
 
 coefficient of ay! 
 (am tent’) m +(om+bnta)n +b m+an+c=0... (1), 
 coefficient of 2’ z’ 
 (am+en+b')m,+(cm+bnt+a)n, +0m+an+c=0... (2), 
 coefficient of y'z’ 
 (am, +n, +0) m +(c'm,+bn,+a’) n' +0 m,+a'n, +¢=0...(3). 
 If moreover we suppose the new axes to have given inclina- 
 tions, viz. ay’ =, y'2 =, 22’ =a, which gives the equations 
 
 cos a = ll (mm, ai nn, + 1), 
 
 cos B=1'l (m'm, + n’n, + 1), 
 
 cos y= ll (mm + nn’ + 0), | 
 we have six equations for determining the six unknown quan- 
 
 tities, and therefore the positions of the new axes. 
 
 Hence, since the angles a, 9, y are arbitrary, there is an 
 infinite number of systems of oblique co-ordinates which fulfil 
 the three required conditions. But if the new axes are to be 
 rectangular, since equations (1) and (2) shew that the axes of 
 y and 2’ lie in a plane whose equation is 
 
 (am+c'ntb') r+(cmt+bn+a) y+ mt+a n+c) z=0...(4), 
 the axis of 2’, whose equations are r=mz, y=nz, must be 
 perpendicular to that plane, 
 
 * amacn+b =M(U m+ a NC) ....00200000-(5), 
 
 omtbnt+a =n(U mtd NC) ..cccececesesee-(6) 3 
 whence, by eliminating (m), we find for determining (7) an 
 equation of the form 
 
 An? + Bn? + Crn+ D=O0O,* 
 
 * The cubic equation is {(a—d) a’b'+(a?—6*) c'\n8 
 +{(a—b) (c—b)b'+-(2c?—a?— 6?) b' + (@e—a—b) ae in? 
 + | (c—a) (c—b) c' + (26?—a?—c?) c+(2b--a—c) ab’ in 
 +{(a—c) a'c'+-(a?—c?) b'\ =0. 
 
76 
 
 which must have one real root; and then m is known 
 from equation (6). Also the axes of y' and z are at right 
 angles, therefore m’ m,+n'n,+1=0, which, joined to equations 
 (1), (2), (3), determines the four quantities m’,n’,m,n, But 
 the axis of y’ might have been determined so that it should be 
 perpendicular to the plane of «’ 2’; in this case also we should 
 have arrived at equations (5) and (6); consequently n’ is a root 
 of the cubic; the same is true of the axis of z 3; therefore the 
 three roots of the cubic equation are real, and are the values of 
 n,n',n,; and m,m',m,, are known from equation (6). Hence 
 there is only one system of rectangular axes which frees the 
 general equation from the terms involving xvy, rz, yz. This 
 system exists in all cases, and the above is the process for 
 finding it. By both transformations, whether to oblique or 
 rectangular co-ordinates, the equation is reduced to the form 
 
 Ac®+ By*®+C2"°4+2A2+2By4+2C2+D=0...(7). 
 
 Hence, any one of the co-ordinate planes is parallel to a 
 diametral plane, that 1s, to a plane bisecting the chords parallel 
 to the intersection of the two others. This also appears 
 by comparing equation (4) to the plane of y'z’, with the 
 equation to the diametral plane in the last Art. 
 
 Cor. 1. If none of the squares of the variables are 
 wanting, it is manifest that equation (7), by simply changing 
 the origin, may be further reduced to the form 
 
 Ag OB Varta C sate tO cae neces scree te 
 
 In this case the origin is the center, and any one of the 
 co-ordinate planes bisects all chords parallel to the intersection 
 of the two others, and is itself parallel to the tangent plane 
 at the extremity of that intersection. Diametral planes thus 
 related are said to be conjugate to one another, and their 
 intersections are called conjugate diameters. If the lengths 
 of a system of conjugate diameters be 2a’, 2b’, 2c’, the 
 equation to the surface referred to them, is 
 
 $8 a 
 
 ioe fe Ti 
 
 » 
 t 
 
 bas 
 
 = Les 
 
 ~ 
 ae) 
 
 ar 
 
77 
 
 when they are mutually at right angles, (which is possible for 
 only one system), they are called principal diameters of the 
 surface; and the tangent plane at the extremity of any one 
 of them, is perpendicular to it. 
 
 Cor. 2. If one of the squares as x” is wanting, it is 
 impossible to reduce the equation to the form (8), therefore 
 the surface has not a center; by writing 2 +a, y+, 2 +y, 
 for 2, y, 2 we may always determine a, B, y so as to reduce 
 the equation to the form By?+Cz*+2A,2’=0. If two 
 squares 2”, y’” are wanting, by writing 2+, 2’ +a, for 2’ and 2” 
 the equation may be reduced to the form 
 
 Cz7+2By +2A¢ =O; 
 
 which, since all sections parallel to the plane of x’ y’ are parallel 
 straight lines, represents a cylinder on a parabolic base. If the 
 three squares are wanting, the equation represents a plane. 
 
 Cor. 3. Hence the general equation to surfaces of the 
 second order can always be reduced to one of the forms 
 
 Ax’ +By?+C2°+D=0, or By?+Cz°+2 A,r=0, 
 both of which are included in the equation 
 Ax’? + By +C2+2A,7=0. 
 
 The figures and properties, of the different surfaces represented 
 by these equations, have already been discussed. 
 
 45. In a surface of the second order that has a center, 
 the sum of the squares of any system of conjugate diameters 
 is equal to the sum of the squares of the principal diameters ; 
 also the volume, and the sum of the squares of the faces, of the 
 parallelopiped described on any system of conjugate diameters, 
 are respectively equal to the volume, and the sum of the squares 
 of the faces, of the rectangular parallelopiped described on the 
 principal diameters. 
 
 - 
 
 2 2 2 
 
 x 2 2 : 
 Let — + J + >= 1 be the equation to the surface 
 a~ * Cw 
 
78 
 
 referred to a system of conjugate diameters, the inclinations 
 of which are Zvy=y, 4v2=a, Lyz=fP. 
 
 Let 2’, y’, z’ be co-ordinates of the extremity of a principal 
 diameter whose length is 2r; and let r=mz, y=nz be its 
 equations. Then the equation to the tangent plane at its 
 extremity 1s | 
 
 , , , 
 
 e, of gy we Mm n 1 1 
 The equation to a concentric sphere, radius (7), referred to 
 the same axes is (Art. 2.) 
 
 rar py? +2°4+2xry cosy +2rzcosat 2 yz cos PB; 
 therefore the equation to the plane touching it at the point 2’ yz’ is 
 
 ; / / , 
 @ +y cosy+2Z cosa 
 eae s Y 
 
 7 
 24 eee 
 z+’ cosaty/’ ares 
 
 i y +2’ cosy + 2’ cos B 
 
 2’ +2’ cos aay Baya 
 
 > ° . , 
 which, by reduction, becomes (since'w =mz’, y’ =nz’) 
 
 x 
 (m+ ncosty + cos a) = + (n+ mcosy + cos B) = 
 &4 1 
 + (1+ mcosa+ncos B)= = Fireeseereecee(2). 
 r 
 
 But since a principal diameter is perpendicular to the tangent 
 plane at its extremity, this plane must coincide with a plane 
 applied at the same point to a concentric sphere. Hence 
 equations (1) and (2) are identical, 
 
 J mr nr® 
 AP “a =m+ncos y + cosa, re =n +mcos ry + cos 2, 
 
 ~ 
 
 ? 
 a= 1 + mcosa +n cos f. 
 
19 
 
 Whence, eliminating m and », we obtain the equation 
 ro = 9* (a? +b + 02) +r? §(a’b’ sinry)* +(a'c' sina)* + (b'c' sin B)"$ 
 ~ (a'b'c’)? {1—2 cosa cos3 cosy — cos*a — cos’ (3 — cos*y} =0. 
 
 But if 2a, 26, 2c, be the three principal diameters, then Gen Bt’. 
 are the three values of 7; therefore, by the theory of equations, 
 
 a? tb? t+ CHP +P +c? 
 (a’b’ sin)’ + (a'c' sina)’ +(U'c' sin)” 
 = (ab)? + (ac)? + (BC) .0c2s cee cee cee cee(3) 
 (a' b'c')? {1 —2 cosa cos B cosy — cos’ a — cos” B — cos*y} 
 =i(a Doyiass'sdensscdd ene eve) 
 
 which are the three required results. 
 
 Cor. 1. If the conjugate diameters 2a’, 26’, 2c’, are each 
 
 z 2 2 
 , a+t+b+e < 
 equal to 2R, then R = yt eG, also, since there are 
 
 only two equations viz. (3) and (4) to determine the angles of 
 inclination a, 9, ‘y, of the conjugate diameters, therefore there 
 may be an infinite number of systems of equal conjugate diameters ; 
 and their extremities all lie in the intersection of the surface, and 
 a concentric sphere, whose equation 1s 
 
 aT a te 
 
 v+y+2= ; 
 
 Cor. 2. Of all systems of conjugate diameters of an ellipsoid, 
 the principal diameters have their sum a minimum, and the equal 
 diameters their sum a maximum. 
 
 If possible, let there be a system, not mutually at right an- 
 gles, whose sum is a minimum. ‘Through any two, which are 
 not at right angles, draw a plane; this will of course cut the 
 surface in an ellipse, of which these two will be conjugate dia- 
 meters, and their sum will be greater than that of the axes of the 
 section; if therefore we join the two axes of the section to the 
 remaining diameter of the proposed system, we shall have a sys- 
 tem of conjugate diameters, whose sum is less than that of the 
 proposed system, contrary to the supposition. In like manner, 
 by recollecting that of all systems of conjugate diameters of an 
 
80 
 
 ellipse, those which are equal have their sum a maximum, we 
 may shew that no system of unequal conjugate diameters of an 
 ellipsoid can have their sum a maximum. 
 
 46. Every surface of the second order may be generated 
 by the motion of a circle of variable radius parallel to itself, the 
 center of the circle moving along a diameter of the surface. 
 
 Case 1. Let the surface have a center, and let its equation be 
 Ax* + By’ + Cz? — 1=0. 
 For x, write x +a, then if CA =a (fig. 38.), the origin will 
 
 be removed from the center C, to A, the directions of the axes 
 remaining the same ; 
 
 - Atiwtay+ By? + Cz* — i=0. 
 Now make r=2° cos0, y=y'’, z=2' sinO, Art. 41; 
 2” (A cos*@ + C sin’0) + By? + 2Aa cosOx’ + Aa®—1=0, 
 
 is the equation to the section BAD, where 2 BAxr=8@, and 
 AB, AD, are the axes of 2’ and y. The section is therefore a 
 circle, . 
 if A cos*°O+C sn°O=B. or A+C tan?0= B (1 +tan’8), 
 A-B 
 
 or tan?@ = —— . 
 
 B-C 
 
 The equation then becomes 
 Aad —!I 
 B 
 
 or (2'+ ; a cos) +y°= (= a cos) — Agias we 
 
 Gow | <a). 
 abe delle OMB). 10 Ba 
 
 A 
 w+ y? +95 a cosd a+ = (0. 
 
 Make AO — - a cos@, then O is the center of the circle, 
 
 and 3 = . cos@.a, is the equation to the locus of O, which is 
 
 manifestly a straight line, and a diameter since it passes through C; 
 
81 
 
 (B—A)AC 1 
 
 also the radius of the section = \/ i oma san Teves 
 : (A—C)B 
 
 Cor. 1. Hence for the ellipsoid 
 
 1 ] 1 cc, J7—f 
 A=-3, B=; Go's; tan d= + @ 8 
 z b re a fh? —i¢* 
 
 a C 
 
 therefore a>b>c, or the axis to which the cutting plane is 
 parallel, is the mean axis. For the hyperboloid of one sheet, 
 
 el C= 
 
 et 
 
 supposing the axis of z the imaginary axis of the surface ; 
 
 “. tand= +=/ b*— a" ? 
 4b" 
 
 a C 
 
 therefore b >a, or the cutting plane is parallel to the greater of 
 the real axes. Similarly if we suppose A negative, or the axis 
 of x the imaginary axis, we arrive at the same conclusion; but 
 if we suppose the axis of y imaginary, that is, B negative, tan @ 
 is impossible; in fact we know that all sections through the ima- 
 ginary axis are hyperbolas. For the hyperboloid of two sheets, 
 
 1 ] 1 
 Somme) a ee b= -—- = Gs +25 
 
 Cc 
 
 therefore b>c, or the cutting plane is parallel to the greater of 
 the imaginary axes. Similarly if we suppose the axis of z the 
 real axis, we arrive at the same conclusion; but if the axis of y 
 be the real axis, that is, A and C negative, tan@ is impossible; 
 in fact we know that all sections through the real axis are hyper- 
 bolas. Since tan@ has two values, the cutting plane may be 
 inclined at an 20 or 90+ to the plane of ry, and hence the 
 surface may be generated by a variable circle in two different 
 ways. What we here assume, viz. that every circular section 
 must be perpendicular to a principal plane, appears from Ex. 3, 
 Art. 41; for in order that the section may be a circle, we must 
 
 L 
 
82 
 
 : ° Tv 
 have the co-efficient of vy! = 0, and therefore either 0 = - or 
 
 P= 57 or =O. 
 
 Cor. 2. By putting the radius of the section =0, we 
 find the value of a, when the plane of the section coincides with 
 the tangent plane; the point of contact lies in the plane of rz, 
 
 < 
 
 i - a i : : : 
 and its co-ordinate x= —; which for the ellipsoid becomes 
 
 Casr 2. If the surface have not a center, let By’ +C2°=2a 
 be its equation. Make dC =a (fig. 39.), therefore the equation 
 reckoning from C, is By’ + Cz° —2(a +a)=0; and the equation 
 to the section is 
 
 By* + C2” sin’ 9 — 2 (x cos +a) =O, 
 which will be a circle if Csin?@=B; the equation then is 
 
 he cosO\* _ cos’ Qa Lyi ol pqh@e 
 y +(« 7) = B tB BBC’ B’ 
 
 Make CORE Jeet 
 B BY BC” 
 
 then O is the center of the circle, and its locus is a straight 
 
 line DO parallel to AC. 
 
 Hence B and C must have the same sign, as in the elliptic 
 paraboloid; also B is < C, therefore the cutting plane is perpen- 
 dicular to the principal section whose latus rectum 1s the least. 
 In the case of the hyperbolic paraboloid, since B and C are of 
 different signs, no plane can be drawn which shall intersect it in 
 a circle. 
 
 47. To find the relation among the coefficients of the 
 general equation of the second order, when it represents 
 surfaces of revolution. 
 
 Let the surface of the second order be one of revolution 
 about Ox, then its equation referred to the axes Ox, Oy, Oz, 
 is 2°+y? =ax’ +6, a and b being constant quantities, (Art. 20.) 
 
83 
 
 Let the co-ordinates be transformed into others parallel to 
 the axes O2', Oy’, Oz’, the positions of which are determined by 
 the angles @ and @,, (Art. 40.) 
 
 .. (v’ sin@ + 2’ cos0)’ + $(« cos —z' sin8) sind + y/' cos pt? 
 = a}(x' cos0 — 2‘ sin@) cos — sind}? +6, 
 or ¢?(sin?@ + cos? sin’ @ —acos Ocos’p) +y"(cos p— asin’ P) 
 ++ 2’ (cos’6 + sin’ @ sin’ — a sin’ 0 cos’ p) 
 +2y'2'(—cos sin d sin 8 — acos sing sin®@) 
 + 2x'2’(sinOcosO — sin@cos Osin® + acos@sinO cos’ q)) 
 
 + 22'y' (sing cos cos 0 + asin d cos @cos@) =6; or if 1+a=a, 
 x” 5 1 — a(cos@ cos p) +y* (1 —asin’) + 
 z”41—a(sinOcosd)’} —y'z'acos2q@sinO + 22’ asin20cos p 
 + xy’ acos2@cos0 = b, 
 which, compared with the general equation of the second order, 
 
 Ax? + By? +Cz" 429A’ ye 4+2B a'r +2Cr'y' =D, (1) 
 
 e . - ° . v 
 gives six Independent equations; these are, making Toe 
 
 a 
 nA=1—acos’@cos*d, nn A'= — 5 coszpsind, 
 
 : rye 
 n B=1—asin* ~, n B= 5am 20cos dp, 
 
 'C=1—asin?0cos® 'C=* 0; 
 nC=1—asin'Gcos ¢, n = 5cos2pcos : 
 
 and since there are only four unknown quantities, we must 
 have two equations of condition, which are found thus; 
 
 2 
 10 , a . , (a) J 
 n°AA’ =n A’ + 3008 9 cos’ @ cos2psinO=n A +n’*BC, 
 
 therefore sme = big 
 A n 
 ee ed AO CC—A'B 1 
 similarly Sap eevee Seart 
 
 therefore eliminating 7’, we find the equations of condition 
 BC AC AB 
 A-—-——-=B-—- —-=C--—- 
 A’ B C 
 
84 
 
 Cor. 1. Since cosra” = cosOcos, cosay’ = — sing, 
 cosaz = — sin@cos@, (Art. 40.), if 2° =mz’, y' =nz', be the 
 equations to the axis of revolution (that is, the axis of x) re- 
 ferred to the axes of 2’, y’, z’, and therefore, 
 
 , , 
 cosxx =mecosrz, cosry =ncosxz (Art. 7.); 
 
 then cosOcosp=mcosrz = —msin@ cos, 
 — sing = ncosrz = — nsinOcosd, 
 
 by tan wi ex 
 
 A sin@ eT chs 
 
 as appears from the equations of condition. Hence the equa- 
 tions to the axis of revolution are A’r’ =C2z', By’ =C2’. 
 
 Cor. 2. A shorter, though less direct mode of solving this 
 question, is to substitute values of p and q, derived from equa- 
 tion (1), in the general equation to surfaces of revolution (Art. 85). 
 Then since the resulting equation must be identically true, by 
 equating to nothing the coefficients of 2°, y*, 2°, we find the 
 equations to the axis of revolution; and by equating those of 
 LY, 2, yz, to nothing, we find the equations of condition. 
 
 48. Problems on surfaces of the second order, illustrating 
 the preceding propositions. 
 
 Pros. 1. Three straight lines, mutually at right angles, 
 and meeting in a point, constantly pass through a plane curve 
 of the second order; to find the locus of their point of 
 intersection. 
 
 Let Aw’+ By’=C, z=0, be the equations to the plane curve, 
 and a, (8, y the co-ordinates of the point of intersection of the 
 three lines, and let the curve be referred to that point as origin, 
 and to the three straight lines as axes; let the cosines of the 
 angles which the axis of x makes with the new axes of 2’, y’, z’, 
 be denoted by m, n, 73; and similarly, let the corresponding 
 quantities for the axes of y and z be respectively m’, n’, 1’, 
 m, 2, 7,3 therefore by formule of Art. 39, the transformed 
 equations to the curve are mx +ny' +72 +y=0, 
 
 Ams +ny +r2' +a) + Bi'e' + n'y’ +r27+BY=C. 
 
 In order that the axis of 2 may meet the curve, these two 
 equations must agree in giving the same value for 2’, when 
 y and 2 =O; 
 
85 
 
 J“. A(mx’ +a) + Bim'e' + BP =C, ma’ t+y =0; 
 
 and substituting this value in the former equation, 
 A(m,a—my)’ + B(m, B—m'y)? = Cm? ; 
 similarly A (n,a—ny) + B(n,B— n'y)’ = Cn/ for axis of 7, 
 A(ra—ry) + Bir, B—r' yy =Cr? for axis of 2’. 
 
 Hence adding these equations together, and taking account 
 of the equations of condition (Cor. Art. 39), 
 
 A(a’ +’) + B(B’ + y*) =C, 
 
 the equation to a surface of the second order. 
 
 Cor. If the given curve be an ellipse whose equation is 
 ar i y 
 aucrdl 
 
 = 1, 
 
 the equation to the surface becomes, replacing a, 6, ry, by x, y, 2, 
 b? x? + a°y? + (a’ +b") z* =a70*, that to an ellipsoid ; if an hyper- 
 bola, the equation to the surface is b?x?—a’y’—(a° —b") 22 =a' Ll’, 
 that to an hyperboloid of one or two sheets, according as a< or >6. 
 In the case of the ellipse, if we remove the origin to the ex- 
 tremity of the major axis, by writing «—a@ for a, the equations 
 become 
 
 ox y? 
 aig Oem 
 
 2 
 
 a8 
 C) 
 
 ba® + a®y® + (a’ + b’) 2? =2ab'x 3 now let b° = 2ap—p”, 
 
 (p being the distance of the focus of the ellipse from its vertex), 
 and make a infinite, .. y°=4pa, y°+2*°=4pz; that is, when - 
 the curve is a parabola, the locus required is the paraboloid 
 which it would generate by revolving about its axis. 
 
 Pros. 2. Three straight lines mutually at right angles, and 
 meeting in a point, are applied to a surface of the second order ; 
 to find the locus of their point of intersection. 
 
86 
 
 Let the equation to the surface be 42? + By? + Cz°=D; 
 then referrmg it to the three straight lines as axes and to their 
 point of intersection as origin, and using the notation of last 
 Problem, the equation becomes 
 
 A (mx + ny +12! +ak+ Bima! +1'y' +r 2+ By 
 + C(m,x +0, +r,2’ +y) =D, 
 or xv? (mM A+m? B+m7C)+y? A t+n? B+270) 
 +22? A +r? B+r2C) + 22'y' (mnA+m'n' B+mn,C) 
 + 2a'2'(mrA+m'r B+mr,C)+2y'2 (nrA+n'r B+n,7,C) 
 +22 (mAa+m BB+m,Cy) 
 + 2y (nda +n BB+nCy) +227 (rdat+r7 BB +7,Cy) 
 +Aa’? + BR? + Cy’ —D=0. 
 Now making successively 7,2’, =0, 2’,2’,=0, 2,y',=0, in 
 
 order to determine the points where the axes of 2’, y’, z’ meet the 
 
 surface; and calling Aa’ + BB’+ Cy? —-D=G, we have 
 vA +m B+m2C) +22 (mAatm BB+m,Cy)+ G=0, 
 ynA+ n? B+ n?C)+2y' (nAa+n BB+n2,Cy)+G=0, 
 2°? Atr? B+r?C) +22 (rdAat+r BB +r, Cy) +G=0. 
 But since the axes of xv’, y’, 2’ touch the surface, the two 
 
 points in which any one meets it coincide m one, and therefore 
 the two roots of each of these equations are equal ; 
 
 o. (mA+m?B+m2C)G = (nmAat+m BB+m,Cyy, 
 (n° A+n® B+22C)G = MmAatn BB+n,Cy)’, 
 (PAL? B+r? OCG = (rAatr BB+r, Cy). 
 
 Hence, adding together these three equations, and taking ac- 
 count of the equations of condition (Cor. Art. 39.), 
 
 (A+ B+C) G=(Aa)’+(BBY+(Cyy, 
 or (4+B+C)(Aa’?+ BB’+ Cy’ — D)=(Aay+ (BB)? +(Cy)’, 
 or, finally, 
 A(B+ C0) a°+B(A+ 0) B?+C(44+ B)y’=D(A+B+O), 
 
 the equation to the locus, which is therefore a surface of the 
 second order concentric with the given one. 
 
87 
 
 Cor. In the case of an ellipsoid this equation becomes, 
 replacing a, B, y, by x, y, 2, 
 xv? (6? +c?) + y® (a® +0?) + 2? (a? + BD) =(aby + (ac)? + (b0)’, 
 2 2 2 
 
 a rs 
 ig th a ee 
 2° me We a 
 
 if g, h, k, represent the three altitudes of the triangle formed by 
 joining the vertices of the ellipsoid (Art. 16. Cor. 2.). If we 
 remove the origin to the extremity of the 4 axis a and then make 
 a infinite, as in Cor. Art. 24, the equations become 
 
 a dean? yt eaHA(ptp)xr+4app’; 
 
 if therefore the given surface be a paraboloid, the locus required 
 is a paraboloid of revolution. 
 
 If the given surface be an hyperboloid of one sheet, making 
 c’ negative, the equation to the locus is 
 
 (b°—c*) a? +(a— 0?) w+ +0") 2 =0°b? — Cc (a? +0) ;x 
 which will be an ellipsoid if a*b” > (a? + 6*) c*, and a point (viz. 
 the centre of the surface) if a®b” =(a* + b’)c*. 
 
 If c=aand be greater than 0, the equation becomes 
 (a? — b?) 2? — (a? + B*) 2? =a", 
 which is that to an hyperbolic cylinder. 
 
 If the given surface be an hyperboloid of two sheets, then 
 making 6° and c” both negative, the equation to the locus is 
 
 y? (a? —c) + 2° (a? = b’) — 2° (a? + Bb?) = (bc)? — a? (8? + c*). 
 If in the original equation we suppose 
 D=0, A=m’, B=n’?, C= —m’n’, 
 
 it becomes the equation to a conical surface of the second order, 
 and the equation to the locus becomes 
 
 (l—m’) 2? +(1 —n°) y°— (m’ -+-n’) CAGE 
 
 which is also the equation to a conical surface of the second order. 
 
88 
 
 Pros. 3. If a surface of the second order be. cut by 
 a plane which always passes through a fixed point, to find the 
 locus of the centres of all the plane sections. 
 Let Az?+ By’ +Cz*=D be the equation to the surface, 
 z—c=A'(x-—a)+ B (y — b) the equation to the plane, 
 
 a, b, and c, being the co-ordinates of the point through which it 
 passes ; eliminating y, 
 
 B etn 
 
 Re le—e- A @—a)+ Boy=D, Aaa (1) 
 
 is the equation to the projection of the section, on the plane of xz. 
 Let a, 8, ty, be the co-ordinates of the centre of the 
 
 section, and transport the origin of the co-ordinates to that. point 
 
 by writing 2 +a for x, and z ++ for z; then the coefficients 
 
 of x and z must each vanish in equation (1), 
 
 B 
 “. Aat Be aa) A*—(Bb+y—o A} = 0; 
 
 B 
 Cy + Be {(y—¢)+ B'b—A'(a—a)}=0. 
 
 Multiply the latter of these equations by A’, and add it to the 
 former, 
 
 ; : Aa 
 “. Aat+CAy=0, or A=— Cy? 
 
 B , 
 similarly B’ = — ie . Also y-c=A'(a—a)+ B(B-d); 
 
 Cy 
 " Aa BB mt 
 ~yoct Be Oe Caer 
 
 or Cy(y—c) + Aa(a—a) + BB(B—b)=0; 
 which is the equation to a surface of the second order similar to 
 
 the proposed one, the co-ordinates of its centre being ~, 
 
 2 
 It is manifest that the two surfaces intersect in a plane 
 whose equation is Ax + By + Cz= D. 
 
39 
 
 Pros. 4. To find the locus of the middle points of 
 chords of a surface of the second order, all passing through 
 a given fixed point. 
 
 ‘Take the given point for the origin, and any plane drawn 
 through it, and the centre of the surface, for the plane of ry, 
 and let the axes of the co-ordinates be parallel to three con- 
 jugate diameters ; therefore the equation to the surface is 
 
 av? + by’ +c2°+ 2ar4+ 2b'y+d=0. 
 Let r=mz, y=nz be the equations to any chord; therefore, 
 substituting for x and y, 
 
 (am? + bn? +c)z2° + 2(am + Un)z+d=0; 
 the values of z in this equation are the co-ordinates of the ends 
 of the chord, therefore the co-ordinate of its middle point 
 am + bn 
 
 ie SE 
 am + bn’ +e 
 therefore putting for m and mn their values, the required 
 equation to the locus is 
 
 ax? + by? +c2 +ax + Uy =0; 
 hence the locus is a surface of the second order, similar and 
 similarly situated with the given one, and passing through jts 
 centre ; and also through the origin of the co-ordinates. 
 
 L= 
 
 Pros. 5. ‘To prove that the curve of contact of a surface 
 of the second order, and of a cone the vertex of which is given, 
 is a plane curve; and to shew that if the plane of contact 
 pass through a fixed point, or a fixed line, or touch a surface 
 of the second order, then the locus of the vertex of the cone 
 will be respectively a plane, a straight line, or a surface of the 
 second order, and conversely. 
 
 bet ax’ + by’? +cz*+2exr=0 be the equation to the sur- 
 face, which may represent all surfaces of the second order, 
 (Art. 44. Cor, 3.) Then the equation to the tangent plane at the 
 point ry Zz, Is 
 are", 
 
 by / 
 (Qi) ae Cyi=9), 
 
 a eS an 
 
 C2 
 or (ax +e) x’ -+- byy' +ez2 +er=0. 
 M 
 
90 
 
 Now because the cone and surface touch one another, at 
 the points of contact their tangent planes coincide; and since 
 the tangent plane of a conical surface always passes through its 
 vertex, if a, 8, y be the co-ordinates of the vertex, they must 
 satisfy the above equation, 
 
 “. (ar+elatbyB+ czy +er=0......(1), 
 
 which gives the relation among the co-ordinates x, y, z, of the 
 points of contact, and shews that they all lie in the same plane. 
 The curve of contact is therefore a plane curve, resulting from 
 the intersection of the plane whose equation we have just 
 found, with the given surface. 
 
 Next let 2’, y', 2’, be the co-ordinates of the fixed point, 
 through which the plane of contact always passes; then they 
 must satisfy its equation (1), 
 
 “. (ax +elat by'B ob cz ¥ +er =0, 
 
 which gives the relation among the co-ordinates a, 9, ry of the 
 vertex of the enveloping cone, and shews that it moves in one 
 plane; and that that plane is the plane of the curve of contact 
 with the given surface, of a cone whose vertex is in the given 
 fixed point; for its equation may be put under the form 
 
 (aate)x +bBy +cys +ea= 0. 
 Again letr=mz+p, y=nz+yv, be the equations to the 
 line through which the plane of the curve of contact always 
 
 passes. Then in order that the plane may contain this line, we 
 must have 
 
 (aat+e)(mz+p) teatb(nz +r) PB +czy=0, 
 independent of the value of z5 3 
 “. (aate) m+bBn+cy=0, (aate)utbvB +ea=0, 
 
 which are the equations to the straight line in which the vertex 
 moves, its co-ordinates being a, (3, +. 
 
 Lastly, let the plane of contact touch a surface of the second 
 order whose equation 1s 
 
. OF 
 
 mx +ny?+r2°+2sx=0......(2)3 
 
 “5 , , : ° . 
 then if x, y, 2%, be co-ordinates of any point in a plane 
 touching this surface, its equation is 
 
 (ma+s)a+nyy +rz2z +sx=0 
 But the equation to the plane of contact is 
 (aate)a +bBy' +cy2’ +ea=0; 
 
 therefore, in order that these planes may coincide, we must have 
 
 mrts aate ny bB sx ea 
 
 — ——— a ie 
 
 > 
 rz ORD WAT FS Rony UM Ae aL 
 
 and it remains to eliminate 2, y, 2, between these three 
 equations, and equation (2). Now equation (2) may be put 
 under the form 
 
 = A oe. Ca Sz 
 2 Ee —=0; 
 sr si Li: Gi C207 2 
 
 which by substitution becomes 
 
 2 ea 
 Perks oe 
 n rz cy 
 
 2 2 9 
 or m(ea) += (0B + —(cry)® ao gee S=40p 
 
 Q 
 mea s aat+e scy 
 But ~ —+-—= ob D032h =a(sa— me) +se; 
 Sy CV wre cy rz 
 
 2 Q P 
 . m(ea)? + 5 (6B) +— (cr) + 2ea’ (sa— me) + 2se°a=0, 
 
 3 g 2g 
 ° 9 § 2 $ 2 oO 
 or a (Qesa—me)+ —(bB) + (oy) +2sea =O, 
 n 
 
 the equation to a surface of the second order, in which the vertex 
 moves. 
 
92 
 
 Cor. If the surfaces are. concentric, we may arrive at 
 a more simple result by taking their center for the origin. In 
 that case, if the equation to the given surface be 
 
 Av’ + By’ +C2=1, 
 the equation to the plane of contact is Aa2’ + BBy'+Cyz=1. 
 
 Let this plane touch a surface whose equation is 
 A's’ + By? +C' 2? =1...... (3), 
 
 and consequently coincide with its tangent plane, the equation to 
 which 1s 
 A’ca’' + Byy + Cex’ =1; 
 therefore Aa=A’r, BB= By, Cy=Cx. 
 
 Hence, eliminating x, y, %, between these equations and equa- 
 tion (3), we have 
 
 AS @ HF 2k 1, 
 
 TEP belli SER MAY 
 
 for the equation to the locus of the vertex of the cone; which, 
 in the case of two ellipsoids whose $ axes are a, b, c, a’, b,c’, 
 
 becomes is . 
 CCE: 
 
 Pros. 6. ‘If the surface have a center, the line joining 
 that point and the vertex of the enveloping cone, passes through 
 the center of the curve of contact. 
 
 Let C (fig. 41), be the centre of the surface, A the vertex 
 of the enveloping cone, QRS the plane of the curve of contact ; 
 and let any plane drawn through A and C, cut the surface in the 
 line of the second order PQG, and the cone in the straight lines 
 AQ, AS; then C is the center of the section PQG, and AQ, AS 
 are tangents to it. Also by a property of lines of the second 
 order, QS, a line joining the points of contact, is an ordinate to 
 the diameter PG which passes through the intersection of the 
 tangents; therefore QS is bisected in V, and CV. CA=CP’. 
 
93 
 
 Since therefore all chords of the curve of contact, drawn through 
 the point where a line joining the vertex of the cone and the 
 center of the surface meets its plane, are bisected in it, that 
 point is the center of the curve of contact; and the portion of 
 the line intercepted between the surface and its centre, 1s a mean 
 proportional between the parts intercepted between the center 
 of the surface, and the plane of the curve of contact and vertex 
 of the cone. 
 
SECTION V. 
 
 ON THE CURVATURE OF SURFACES, AND CURVES 
 OF DOUBLE CURVATURE. 
 
 ———— 
 
 Arr. 49. To find the requisite conditions for a contact 
 of the first, second, &c. order. 
 
 If two surfaces, referred to the same origin and axes, pass 
 through the same point, the co-ordinates of which are x, ee 
 and if we change x into y+/A, and y into y +h, the equation to 
 the first surface will give for the value of the new ordinate, 
 
 z+ pht+ qgkt+hcrh’ + 2shk + tk?) + &c. 
 and the equation to the second surface 
 z+ Ph+ Qk+4(RW + 2SAk + TR) + &e.; 
 
 the distance of the surfaces, measured in direction of their or- 
 dinates, will therefore be expressed by 
 
 (P—p)h + (Q—qg)kt+ 4 §(R—-1)? +2(S—s) hk 
 +(T-—t)k’}+&c. 
 
 If we suppose the equation to the second surface to contain 
 a certain number of arbitrary constants, we may determine them 
 so as to make the first terms of this distance vanish; and it will 
 follow that any other surface, for which these terms do not dis- 
 appear, cannot be situated between the two former, with refer- 
 ence to the points which are contiguous to their common point; 
 at least so long as we take 4 and k so small, that the sum of 
 the terms of the first order may be more considerable than that 
 of all the terms of succeeding orders. When we have P — p=0, 
 Q—q=0, the surfaces will have a contact of the first order; 
 if besides these, we have R—r=0, S~s=0, T—t=0, the 
 contact will be of the second order, and so on. 
 
95 
 
 Let therefore V’=0 be an equation between three variables 
 av’, y, 2 and a certain number of arbitrary constants; by the 
 determination of the constants, we may make this surface have 
 with one completely given (the co-ordinates of which we will 
 represent by x, y, 2) a contact of an order which will depend 
 on their number. The first condition to be satisfied 1s, that the 
 surfaces may have a common point, that is, that by changing 
 x’ into v, and y’ into y, in V’ =O, we may find z’=z. Besides 
 this, for a contact of the first order, we must have 
 
 oe Oder dz Oe : 
 dr (dri dy.) dys 
 and for one of the second, 
 
 da! hi dha ppzio-o> diz d'z dz 
 dx? da*’ dx dy’ iff dxdy’ dy” ~ dy” 
 
 Hence it appears that a contact of the first order requires 
 three arbitrary constants, and one of the second, six; and in 
 order to have a contact of the n order, there must be 
 
 (n +1) (n + 2) 
 
 . disposable constants. 
 
 Ex. 1. Let the surface represented by V’=0 be a plane, 
 which, since its equation contains only three arbitrary constants, 
 can in general have only a contact of the first order with a sur- 
 face completely given. We have therefore 
 
 z= Ad’ + By +C, a Dy Cah inl yi 
 dx dy 
 
 from which there results between the constants A, B, C, and the 
 co-ordinates 2, y, z, the following relations, 
 
 subtracting therefore the value of z from that of 2’, and putting 
 for A and B their values, the equation to the tangent plane at 
 a point ryZz, Is 
 
96 
 
 , idiz ifn; dz=", . 
 2-2 = —(r — r) + —ly — y). 
 » nde ( dy Aaa 4 
 Ex. 2. The equation to a sphere is 
 (a ne: a)? + cy’ ~% By ue (2 “y yy 2g é, 
 and since it contains only four arbitrary constants, a sphere 
 cannot generally have a complete contact of the second order. 
 Suppose it required only to have a contact of the first order ; 
 dz’ v—a dz y —p 
 then At ee =, 
 
 x zim ry’ dy ae z—y 
 
 Hence there results between the constants a, 8, y, 6, and the 
 co-ordinates 2, y, z, the following relations, 
 
 ra y—B 
 =p, — > = 
 — ard a oy, 
 or t—at+pe-y)=0, y—-B+¢q(@-y =9, 
 which equations shew that the centers of all the spheres lie in 
 the normal to the point of contact. 
 
 é b 
 Also (z— yy {1+ p+¢} =o, orz—-y = SSSS555 5 
 : Ne a ie 
 
 (na) + (y—BY+@—yF=H8, — 
 
 5 os 
 cr cln Tra} cote ye vemcarme f 
 D+ipe-+ Ie pahig 
 y 
 
 Hence the number of spheres, which may have a simple 
 contact with a surface atthe same point, 1s unlimited; the radius 
 of the sphere being assigned, the co-ordinates of its center are 
 given by the above equations. 
 
 50. At any point of a curve surface, to find the radius of 
 curvature of any section made by a plane passing through the 
 normal. 
 
 ~ Let AO, the normal at any point A of a curve surface, be 
 the axis of z; and x Ay, the tangent plane at the same point, the — 
 plane of ry, (fig. 42). 
 
97 
 
 Let AB be a section of the surface made by any plane 
 passing through AO; AC the corresponding section of the tan- 
 gent plane, which therefore touches the plane curve AB at A. 
 Then if AO=R be the radius of curvature of the plane curve 
 AB at A, and BC be parallel to AO, 
 
 2 
 BC’ 
 Let AN=h, NC=k be the co-ordinates of C; 
 
 . BC =ph+qk+ LP a 
 oF q Coane als 
 
 AO = + limit of 
 
 + &c. 
 
 But p=0, g=03; *.° the plane of xy is the tangent plane ; 
 
 c.. h? + ke 
 che AO i= 5) limit.08 eee aa at a 
 
 a 
 
 k 
 “one + shk a ae +- &e. 
 
 1+#G) 
 
 Patek cH ee 
 
 = limit of 
 
 1 + tan’ 0 TEMA a Ie 
 or h pani! z9? if limit of 5 = tan NAC = tan 0, 
 1 
 
 7% r+2stan0+¢ tan 
 where 7, s, ¢ are the values of 
 
 dz td ah dz 
 
 dx*’ dady’ dy’ 
 at the point A, derived from the equation to the surface. 
 
 1 
 COR. = Hence ese EE ey ee 
 r cos?O0+2s cos 0 sinO+¢ sin? @’ 
 
 also if R’ be the radius of curvature of a section inclined at 
 90 + @ to AN, 
 ] 
 
 rsin @—2s cos@ sin@+¢ cos’ @’ 
 
 N 
 
 then R’ = 
 
98 
 
 ] 1 
 Sy TEE eae 
 
 that is, the sum of the curvatures of any two normal sections at 
 
 right angles to one another, is a constant quantity at the same 
 point of the surface. 
 
 51. Of all sections of a curve surface, made by planes 
 drawn through the normal at any point, to determine those of 
 greatest and least curvature at that point; and to shew that they 
 are at right angles to one another. 
 
 1 : : 
 We have Pia r cos 9 + ssin2@ -+ ¢ sin’ @, 
 which is to be a maximum or minimum by the variation of @ 5. 
 — 2r cos @ sm 0+ 2s cos 20 + Bt sin @ cos O=0, 
 
 or (t—r) tan 8 +s (1 — tan’ 0)=0.........(1)3 
 
 r—ft 
 
 5 
 
 “. cot 8 — cot? —1=0, 
 
 r—t 
 9s 
 
 ] wom AL SE bith o se ee 
 or cot @ = +— J (rt? + 45°; 
 s 
 
 ] 9 
 and 5 gig cos 0 (r + 2s tan@ +¢ tan? 0) 
 
 = cos’ 0 (: ne Mace a UIA Soy ef 0), from equation (1), 
 cos 8 sin 8 
 
 =it+scot? = 4(r+t) Tm 4 a/ (r—t)’ + 45. 
 
 If 6, @ be the two values of 0 derived from equation (1), 
 since tan@.tan@’=—1; .*.6°'=90-+963; and therefore the 
 sections of greatest and least curvature are at right angles to one 
 another. The positions of these sections are given by the equa- 
 tion 
 
 is Geile ar Se bat f(r i) + 4s" 
 2s 
 the upper sign corresponding to the section of greatest curvature; 
 and the values of the least and greatest radii of curvature are 
 respectively 
 
99 
 
 2 ; Q 
 t eomte/ Gms” Bee 1 Le iqedy yy yey Fi 
 
 52. If the axes of x and y are drawn in the planes of 
 greatest and least curvature, then 9=0, and .*. s=0, as appears 
 from equation (1) in the last Art. 
 
 1 ] 
 Hence p fale p= 73 
 and for any section inclined at an angle a@ to the plane of rz, 
 that is, to the plane of greatest curvature, 
 
 R= 1 pp 
 
 i eh as PPO PORES IS. 
 rceos-a+tsin? a p sin’ a+ 9’ cos’ a 
 
 f 
 
 from which it follows, that the radius of curvature of any section 
 is known, provided the radii of those of greatest and least cur- 
 vature are known. If p=p’, then R=p, that is, all the normal 
 sections have the same curvature. 
 
 53. Hence we may determine a paraboloid, the vertex of 
 which shall have a complete contact of the second order with a 
 given surface, at a proposed point. 
 
 Let A (fig. 43.) be the given poimt of the surface, Az the 
 normal; and let the sections of greatest and least curvature meet 
 the tangent plane in Aa, Ay, which lines take for the axes. 
 
 ° 
 
 nn 2 
 
 x 
 Also let z= 3p - a be the equation to a paraboloid, p, p; 
 
 being the least and greatest radi of curvature of the surface at A. 
 Then if AP be a section inclined at an Z (a) to xz, and AN=r, 
 the equation to AP is 
 2 (cos? a sin’ a) 
 = 7 ( aaa - ) 
 2p 2 
 which belongs to a parabola, whose 4 latus rectum, and therefore 
 
 ; 
 eee ay: . Hence the cur- 
 
 vature of every section is the same as that of the corresponding 
 section of the surface, and therefore the paraboloid has a com- 
 plete contact of the second order with it. 
 
 radius of curvature at vertex = 
 
100 
 
 54, Inthe equation to the paraboloid, suppose z constant, 
 
 and =c, therefore 
 A 2 
 
 od ¥y 
 Qoc F Qp'c 
 
 ey 
 
 this is the equation to a section of the paraboloid, or to a section 
 of the surface if ¢ be indefinitely small. Hence it appears that 
 if the surface be cut by a plane parallel to the tangent plane at 
 any point, and indefinitely near to it, the section is ultimately a 
 curve of the second order whose centre is in the normal, and 
 axes in the planes of greatest and least curvature ; also the square 
 of the diameter in which the curve is intersected by any plane 
 drawn through the normal, is proportional to the radius of curva- 
 ture of the corresponding section of the surface. 
 
 This curve is called the Indicatrix of the surface, because 
 it indicates the directions of the curvatures; if at any point it 
 be an ellipse, p and p must have the same sign, that is, the 
 curvatures are in the same direction; if an hyperbola, the cur- 
 vatures are in opposite directions ; and if a circle, p=p, and the 
 curvature of every normal section is the same. For instance, 
 it has been shewn that an ellipsoid may be generated in two 
 ways, by a circle of variable radius moving parallel to itself; 
 consequently there are four points on its surface at which a plane, 
 parallel and indefinitely near to the tangent plane, cuts it in a 
 circle; that is, the indicatrix at those points is a circle, and 
 therefore the two radi of curvature equal. ‘These points are 
 called wmbilict, and are symmetrically placed in the four angles 
 of the principal section containing the greatest and least axes ; 
 their co-ordinates have already been determined (Art. 46. Cor. 2.) 
 It is manifest that at these points, a sphere can have a complete 
 contact of the second order with the surface. In the case of a 
 spheroid the poles are umbilici. 
 
 55. ‘To find the radius of curvature at any point of an 
 oblique section of a curve surface. 
 
 Let Ow (fig. 44.) be the common tangent of the normal 
 section OP, and the oblique section OP’, of a curve surface; 
 Oz, Oz perpendiculars to Ox in the planes of the sections ; 
 R and R’ the radii of curvature at O, of OP, OP’; NP, NP’, 
 
 perpendiculars to Ox; 
 
101 
 
 Let ON=h, NP =z, be co-ordinates of P, also let h, k, 2’ 
 be co-ordinates of P’, and 2zOz'=@; 
 
 , 
 ‘in a) z sec@ 
 — = limit of —— = limit of 
 
 WR NP : 
 
 k 
 nik slelich to ie 
 = sec@. limit of 
 
 rte (yi 
 
 =sec@. limit of ————____________- = sec 9 
 
 *.* limit of ~ =O, since Ox is a tangent to the projection of OP’ 
 
 h 
 on the plane of ry; °. R’=Rcos 0. 
 
 Hence it follows that the osculating circles of all the sections 
 of a curve surface which have a common tangent, belong to the 
 surface of a sphere, the radius of which is the radius of curvature 
 of the normal section passing through the same tangent. 
 
 56. Having given a point on a curve surface, to find the 
 directions in which we must pass to consecutive points, in order 
 that-the corresponding normals may intersect. 
 
 Take the normal at the given point for the axis of z, and 
 the tangent plane for that of xy, and let h, k, / be co-ordinates of 
 a point m the surface; then the equations to the normal at 
 that point are 
 
 a —h+p/(2—l=0, y-k+ 9,2 - {)=0, 
 
 where p,, q,, denote certain functions of h and k, (p and q de- 
 
 dz dz NG? 
 noting the values of —, — at the origin, and therefore each 
 5 dx’ dy ft 
 
 =0). Then in order that this line may meet the axis of z, its 
 two equations must: agree in giving the same value for z’, when 
 x’ and y’ =0; that is, 
 
102 
 
 h k 
 z2=- +d, and 2 =-+4, 
 4 4 
 must be the same; but if the point be consecutive, then the 
 ultimate values of z’ must be the same when / and k vanish, 
 
 h 1 
 
 that is 2 = limit of ———__—_———— = ————— 
 f d ‘tan @’ 
 yaya 
 dix dy 
 d 4 hi it f k tan 0 
 an ESS OE cy 7 am ek Re a a mn. Se 
 d ttan@’ 
 ABALONE Te 
 dx dy 
 
 must be identical; .°. s-+¢tanO0=7 tan@+s tan’@; 
 
 8 is the angle which the line, joining the projection of the consecu- 
 tive point and the origin, makes with the axis of a; and the above 
 equation for determining it being the same as that for finding the sec- 
 tions of greatest and least curvature, it follows that there are two 
 directions at right angles to one another in which we may pass to a 
 consecutive point, so that the normals shall intersect, and that 
 these directions coincide with the sections of greatest and least 
 
 : l 
 curvature. Also the value of baa: 9? shews that the 
 sco 
 
 points, where the normal is intersected by consecutive normals, 
 are the centres of greatest and least curvature, (Art. 54.). The 
 part of the normal intercepted between these points 
 
 =p'—p=pp' (---) 
 
 Pp 
 4 Se aD Wik ru Fea 
 — cart Cow et Gh ene 2 7 SN A, See Py 
 Art — 4s° Jr y+ 4s pea 
 
 57. This property of consecutive normals enables us in 
 some cases to determine at once the sections of greatest and least 
 curvature ; for instance, im surfaces of revolution, the plane of the 
 generating curve is necessarily one of these sections, because in 
 it consecutive normals intersect; and therefore a plane through 
 the normal perpendicular to it, is the other. The first series of 
 consecutive normals intersect in the evolute of the generating 
 curve, the second in the axis; hence in a surface of revolution, 
 
103 
 
 the loci of the centres of the two curvatures, are ithe axis of the 
 figure, and the surface formed by the revolution of the evolute 
 of the generating curve about that axis. Hence we can find 
 immediately the radius of curvature of any section of a surface of 
 revolution, as will be seen in the following example. 
 
 Ex. To find the radius of curvature of any normal section, 
 at a given point of an oblate spheroid. 
 
 Let /= latitude of given point P= 2 PGA, PG being a 
 normal (fig. 45.) ; 
 ". radius of curvature of meridian =p = Lda 
 - (l—eé sin’ 13’ 
 
 a 
 
 Bars, TEAC CUE 4 
 nha ie ee sin? l 
 
 radius of curv. perpendicular to meridian =p = 
 
 Let the section PQ make an Z(a) with the meridian, and 
 
 let its radius of curvature = R, 
 
 a Spy MRR nA es; a" 
 P sin” at+p cos a 
 oe |e" 
 p (1 —e* sin’/) cos* a-+-(1 — e’) sin® a 
 
 a l1—e 
 om . iS 2 4 2 al 
 /1—e’ sin’ / 1+-e cos /cos a—e 
 58. To determine the lines of curvature of a olven 
 surface. 
 
 A line of curvature is a series of points on a surface, deter- 
 mined by the condition that the normals at any two consecutive 
 ones intersect. Every point of a surface is situated on two 
 curves of this kind, whose directions coincide with those of 
 greatest and least curvature, and cut one another at right angles. 
 Thus in a surface of revolution, every section through the axis, and 
 every section perpendicular to the axis, is a line of curvature. 
 
 The equations to the normal, at a point xyz of a curve 
 surface, are 
 
 v—at+ pe —2)=0, y—ytq@ —2)=0; 
 
104 
 
 and at a point x +h, y +k, z+, the equations to the normal are 
 
 vw —r—h+p(2-2z-)D=0, yy —y—k+9(¢-2z—-D)D=0; 
 P,» 1, denoting the same functions of r+, y +k, that p and q 
 are of x and y. 
 
 If these two normals intersect, their equations must be satis- 
 fied by the same values of 2’, y’, 2’, which will be the co-ordi- 
 nates of their point of intersection. Subtract therefore the first 
 equations from the second, then 
 —h+(p,—p)(2'—2)—pl=0, —k+(q¢,-(2% —2z)—9¢,/=0, (). 
 
 6 a oii , 
 and these equations must agree in giving the same value of z, 
 
 it a ia 
 
 k+ql 7 sae, 
 Now let the points be consecutive, therefore 
 h+(ptrh+sk)(ph+qhk) wg rh+sk 
 —______________=— == limit of ———_ , 
 k+(q+sh+tk)(pht+qh sh+tk 
 
 retaining in the developments of p,, q,, 4, only those terms which 
 
 limit of 
 
 involve the simple powers of A and k; or if limit of = = tan@, 
 
 h 
 l+p +pqtan0 _ r+stan0 
 tan? + pq+q° tan@  s+¢tand’ 
 9 is the angle which the line joining two consecutive points of 
 the projection of the line of curvature, makes with the axis of x; 
 wy) 
 
 2 : 
 and therefore tan @ = —* (wx and y being co-ordinates of that 
 
 dx 
 
 projection). Hence, substituting and reducing, 
 2 dy\* Q oy 47 2Y 
 UI g")s— pats (-) Tit g r= (1 + pty 
 —{(i+p*)s—paqrt =o. 
 
 It remains to substitute for p, q, 7, s, ¢, their values 
 in x and y derived from the equation to the surface, and 
 to integrate; the result will be the equation to the projec- 
 tion of the lines of curvature. Since it is of two dimensions 
 
 SMT, We 
 in — , Its integral, when completed by the arbitrary constant C, 
 
 dx 
 
105 
 
 will be of the form 
 CFC, Y+o@, y) =0. 
 
 Suppose the line of curvature is to pass through a point for 
 
 which r=a, y=5, therefore 
 C? + Cf(a, b) + f(a, b) =0, 
 
 which will give two values of C; and by substituting them suc- 
 cessively in the complete integral, we shall have the equations 
 to the two lines of curvature passing through the given point. 
 
 59. ‘To determine the lines of curvature of an ellipsoid. 
 
 Let the equation be 
 
 Tae nr aoe eacy ; 
 ste C2 AYE 1G Oe a GY 
 Re Cae Cine Biz 
 dp >? 2 Pp oa 
 Ss reais Aarne ALA aap eras 
 dq Y 
 fer ore BAC aCe 
 
 therefore, by substitution, 
 dy\* 
 a(B—y)ry (*) 
 ps da: 
 + {B(a—y)2*—a(B—y)y*—aB(a— Bt" — B(a—y)ay =0, 
 
 dyna, Peet 
 or <Acy Cs) +(°— Ay? —B) —2y=03 
 
 aC PamRANNS: ¢. 2a Gar HD) eri 
 Bla—vy) a= 
 O 
 
 calling 
 
106 
 
 ; : dy £U. ; ; 
 To integrate this equation, let i pati , u being a function 
 | dx 
 
 of x. 
 rue ru Ay! 
 Coke AS 1) + "Gr “— B)=0, 
 y( y y A) 
 or Av?w—y?+ua’?— Ay? — By = 0, 
 
 or y+ dAu=22u(l + Au) — Bu, 
 
 Bu 
 1+Au’ 
 
 Frat, eee 
 “YY =ru— 
 differentiate, then 
 
 9 CY IN eal ob o du gy game du 
 tg CROCCO ERS: de (1+ Au) dx’ 
 
 or “(2° ida i oT) = (> 
 
 B 
 Ai ee fe ere 
 ae Otay c(: 1+ 7) 
 the complete integral, C being the arbitrary constant ; 
 1 ots 
 and y?=u(2°—2r,/ B)= qe EB) B—x), 
 
 or (ya/ AY + @ — ./ BY = 0, 
 
 the singular solution, which resolves itself into the two 
 
 ees ait h" 
 y=0, r= +./ B= +a —s) 
 ray ae 
 
 C 
 
 and indicates the umbilici of the surface (Art. 54.), the normals 
 at which, as we know, are intersected by the normals drawn 
 from consecutive points in every direction on the surface. 
 
107 
 
 The complete integral represents an ellipse or hyperbola, 
 according as C is — or +3 hence the projections of the lines of 
 curvature passing through any point on the surface of 3 an ellipsoid 
 are an ellipse and hyperbola. 
 
 In the first case, if m and 2 be the semi-axes of the ellipse, 
 
 n ; B 
 —= i es? 
 m* : 1— AC 
 therefore m aud n are connected by the equation 
 m An? , 
 SB .mewbin’ 
 in the second case, 
 n° B m  An® 
 — oC, m° wince. Ns So ve a oto =~ i 
 m- 1+AC B B 
 
 Hence the semi-axes for each ellipse in the projection are the 
 co-ordinates of a pomt in the same given hyperbola, and for 
 each hyperbola, the co-ordinates of a point in the same given 
 ellipse; these are called the auxiliary ellipse and hyperbola; 
 they are concentric with the ellipsoid, and their semi-axes, 
 (which are the same for both curves, and coincide in direction 
 with those of the ellipsoid) have the following values 
 
 shin Ws 
 ogee : B Je — b° 
 % axis major = Ri = seas 
 
 c 
 
 “R 2 
 5 Ve 
 oO axis minor = Omm 2° 
 
 C 
 
 60. Hence we have the following construction for the 
 lines of curvature of an ellipsoid. Let ACB (fig. 46.) be a 
 principal plane (that of vy), and with the axes found above 
 construct the auxiliary ellipse and hyperbola GO, HO. From 
 any point I in the hyperbola, and z in the ellipse, let fall perpen- 
 diculars on the axes CA, CB; and with semi-axes CN, CM, 
 construct the ellipse MCN, this is the projection of a line of the 
 first curvature; and in the same manner the projections of any 
 number of lines of the first curvature may be constructed. 
 
108 
 
 Again with semi-axes mz, nz, construct the hyperbola 2Q, this 
 is the projection of a line of the second curvature; and similarly 
 any number of projections of lines of the second curvature may 
 be constructed. As the points I and 7 approach O and G, the 
 ellipse and hyperbola continually approach the lines CO, CB, 
 and are at last confounded with them; therefore CO, CB are 
 the projections of lines of curvature, and consequently the curves 
 in which the surface is cut by the principal planes of xz, yz, 
 are lines of curvature; also if in the equation to the auxiliary 
 hyperbola we make m=a, we find n=), therefore the ellipse 
 AB is included in the construction, and is itself a line of curva- 
 ture. Hence the intersections of the surface with its three prin- 
 
 zg 2 
 
 2 cw 
 cipal planes are all lines of curvature. Since CO=a VV =—; 
 C 
 
 and is therefore < a, the point O falls within the ellipse A.B; it 
 is the projection of an umbilicus of the surface (Art. 54.) round 
 which the lines of both curvatures are symmetrically disposed, 
 and towards which they all turn their concavities. 
 
 61. If we take the plane containing the greatest and least 
 axes of the ellipsoid, that is, the plane of xz, for the plane of 
 projection, we shall find the construction easier; to find the 
 equation, we must eliminate y between the equations 
 
 4 
 
 y 2 ’ g B 
 
 _— cos a dy = — ——— }: 
 aya , andy? = C (x*-——-); 
 
 rete $m pte rey, p eit kanes 2) 
 this gives 2° = 7 (& + C) Ca rT), 
 
 observing thata = AB + B; 
 
 ts 
 , Wh 
 a 
 
 <1 Mo 
 
 which, by reason of the determination already made of C, is 
 always the equation to an ellipse; that is, the lines of both cur- 
 vatures are projected into ellipses; and if m’, n’ be its semi-axes, 
 
 1+ AC’ 
 . . . B 72 : 1o 
 ries § elimmating C; — m~ + Ae n= 1, 
 an ay 
 
109 
 
 which is the equation to an ellipse, whose semi-axes are 
 
 2 Z 2 Zz 
 Rael afa—e 
 a a be? C Be ah 
 
 a C 
 
 In this case, therefore, the semi-axes of each of the ellipses 
 of the projection are the co-ordinates of a point taken on another 
 given ellipse, which is the same for all. 
 
 Hence let CA, (fig. 47), be the principal section of the 
 ellipsoid containing the greatest and least axes; and concentric 
 with ‘it, describe the auxiliary ellipse 2X with the semi-axes just 
 found, and which we observe are greater than the corresponding 
 semi-axes of the ellipsoid a and c._ From auy points J, 2, let fall 
 perpendiculars on OX, OZ, and with these as semi-axes construct 
 the ellipses MN, mn; then MN, mn, are the projections of lines 
 of curvature. If in the equation to the auxiliary ellipse we make 
 m =a, we find n =c3 therefore the principal section is itself 
 included in the construction, and if from the points A and C we 
 draw tangents meeting the auxiliary ellipse in D, the perpendiculars 
 dropped from every point in DZ will serve to construct the pro- 
 jections of lines of the first curvature, and perpendiculars dropped 
 from the points in DX will serve for the construction of the 
 projections of lines of the second curvature. It is not dif- 
 ficult to shew that if XZ be joined, all the ellipses in the pro- 
 jection are touched by it, and consequently they are all inscribed 
 in the same equilateral parallelogram. 
 
 62. To determine the radii of curvature at any point of 
 a surface, in terms of the co-ordinates of that point. 
 
 We have seen that consecutive normals in the planes of 
 greatest and least curvature intersect, and that their intersec- 
 tions coincide with the centers of greatest and least curvature. 
 Hence if 2’ be the co-ordinate of the intersection of consecu- 
 tive normals at a point xyz, and @ the angle which a plane of 
 curvature makes with that of xz, then taking the limits of equa- 
 tions (1) in Art. 58, 
 
 (r+s tan 0) (2) —2z) = 1 +p? +pq tan8, 
 (s+ é tan 8) (z —z) = tanO + pq + gq? tan 0; 
 
110 
 
 therefore, eliminating tan 0, 
 
 l+p?—r (2°—2z) (2) ~z)s—pq 
 ()-z)s—pq  1+q°-—(2 —2)t’ 
 or (rt —s°) (2° = 2 —{O+p) t-2pqst+ tq) r} (i —2) 
 +1+p'+q=0; 
 z'—z is the projection of the radius of curvature on the axis 
 of z; 
 R=- (@'-2)/itpre. 
 
 Hence if we find the two values of 2’ —z from the above 
 equation, and substitute them in the value of &, we shall have 
 the two radii of curvature of the surface; also knowing z — z, 
 the other co-ordinates a and yf of the centers of curvature are 
 found from the equations to the normal 
 
 xv—-axrt+ p (2 —z)=0, yy + q(z—z)=0 
 
 Ex. 1. To determine the radi of curvature of an ellip- 
 
 solid. 
 x y 2" 
 Let the equation be — ad) +—=1; 
 
 Bo+¥ 
 
 $ = mre A “ = ‘ae —_ — ae be. 2 
 
 +P bic a ah Ge az (By), 
 
 Me ses nog Wing 
 ci aA yeahh Be pape y) 
 
  (+p)t ae eee “+ (%) ( yes xt, 
 
 — I2ngs:= — ae ={- (4) =e Val 
 
 (+q)r=— i (B-y'+ 010 ~y). 
 
 But in taking the sum of these three equations, the co- 
 
 efficient of (7) within the brackets is 
 z 
 
111 
 eee) Cs) (ogre) eG) 
 + pt — 2pastte)r= 
 
 2 
 7 applet By “af micogt piety % DS 
 
 Also rt-s* = 
 
 a Years 
 — afd (3 + appz? 
 and 1 + p> + q = aby +4 a +5). 
 
 Let AMa=at+PpPty=a t+ 6? + c’, 
 D = the distance of the point in question from the center, 
 =) dD feos x |. y? + a” 
 
 P = the perpendicular from the center on the tangent plane, 
 Aha: 5 2 
 i SE = +o, (Prob. 2; Art. 20.); 
 *y: 
 therefore, by substitution and reduction, 
 
 CxS = 2) + Ta?— DG! - 
 
 But R= —-(2- 2). /1+p°+¢= aaa 
 
 or SPY. _ PR; 
 
 I 
 = 
 
 pe 
 
 A’ — D’ a [3 
 ib ae 
 
 Let R’, R” denote the values of R in this equation, then 
 
112 
 
 A -— Dp 
 P ) 
 
 abe 
 P? 
 which are the simplest terms to which the question can be re- 
 duced; the first shews that at points where the tangent planes 
 are equidistant from the center, the product of the radii is con- 
 stant; and the second that at points which are themselves equi- 
 distant from the center, the sum of the radii varies inversely as 
 the perpendicular on the tangent plane. Also the volume of the 
 ellipsoid may be expressed by the two radii of curvature at any 
 point, and the perpendicular on the tangent plane, for it 
 
 RR" = ( yy Bet ea 
 
 Anrabe a na a7 
 i gee w/ RURE 
 
 Hence we may determine the locus of points on the surface 
 of an ellipsoid, at which the two curvatures have any given re- 
 lation. For instance, if they are equal, or 
 
 abc 
 ‘ = Re th 9) ee ee 
 R en Pp A 
 
 © 
 | 
 Q 
 
 2 
 
 BGP UNG Oe NY Of i an 
 aBy (Gt etc )setBty ty 2, 
 which becomes by reduction, 
 
 2g x 2 x 2 
 (a-B+ 4 (B-y)-= (a9) Ts me: (a—y)(B—y)=0; 
 
 this manifestly resolves itself into the two 
 74 
 
 a-B + Bry — 7 -Y=0 ry =0; 
 
 but we cannot suppose 2 =0, for that gives y equal to a negative 
 quantity, since a and ‘y are supposed to denote respectively the 
 squares of the greatest and least semi-axes; therefore we must 
 have . 
 
 $ a— a’ — b° 
 y=0, and x =a tthe +a —z> 
 
 a= Leal 6 
 
 which are the co-ordinates of the umbilici, and agree with Art. 46. 
 Cor. Qo. } 
 
113 
 
 Also the radius of curvature in this case | 
 
 abe i é r a b° \. | 
 = = a@d0ci-— — = 5h 
 q* a‘ i) \ 
 
 ? ac 
 
 at which we may arrive immediately, by observing thatut_is the 
 radius of curvature of the principal section in the plane of 42, 
 
 3 b 
 Bor 
 at a point whose abscissa = a : 
 
 ae: 
 a=—C 
 
 Ex. 2. To find the radu of curvature at any point of a 
 paraboloid. 
 ry : BATT, 
 If — + r -+- 2z =O be its equation, it may be shewn, 
 a 
 
 as in the preceding example, that the maximum and minimum 
 radii of curvature at a point vyZz are 
 
 = 2 ‘ — Oe 
 Ir e+ 1 fee 92 
 a” b 
 
 g 
 (F) 0 Gree 
 
 Hence if the radii be projected on the axis of z, the sum of 
 ithe projections = a+b—2z. 
 
 I+ 
 
 Cor. 1. The roots of equation (1) have the same or dif- 
 ferent signs, according as rf —s° is + or —; therefore the cur- 
 vatures are in the same or different directions, according as 
 rt — sis + or —. Hence the curvatures are always. in the 
 
 YEN Mie shy tr Sha\3 
 same direction in the ellipsoid, for rt—s* = ( :) , 
 abez 
 
 Cor. 2. In order that the two radi of curvature may be 
 equal, but of different signs, we must have 
 
 (l+p)t—2pqs+(lt+q)r=0. 
 That they may be equal and of the same sign, we must have 
 
 {l+p)t—@pqst+(i+q)r}? 
 P 
 
tl4 
 
 = 4(1 +7)” -} q°) (rt —s*}=4 (1 +p 1p” —p*q') (rt —s*), 
 which is manifestly equivalent to the double equation, 
 
 itp py _1+¢° 
 
 r sabowrgt vit 
 
 By substituting for p, g, 7, &c. their values in these equations, 
 we may determine the locus of the points ou a given surface at 
 which the curvatures are equal, and in the same or opposite 
 directions. If they could be integrated, we should obtain the 
 equations to surfaces having the above-mentioned properties at 
 every point ; the first also represents that surface which has its 
 area a minimum within given limits. 
 
 63. Preparatory to finding the radius of curvature, and 
 evolutes of a curve of double curvature, consider figure (48), 
 where for the curve is substituted an equilateral polygon mm'm’’..., 
 and through the middle points of its sides are drawn planes re- 
 spectively perpendicular to them, which intersect, two and two, 
 in the lines kA, kh’, Rh", &c. Then the plane which contains 
 the two consecutive sides mm’, mm’, is perpendicular to each 
 of the planes gh, g’f’, and therefore to their common intersec- 
 tion kh; let kh meet this plane in the pomt g, then q is equi- 
 distant from the three angles m, m’, m”', which is also true of 
 every point in the line kA. The lines kA, kh’, &c. will be 
 parallel only when the sides of the polygon mmm’... are in the 
 same plane; in other cases, if they be produced till each meets 
 its consecutive, they will form a polygon hop..., the angular 
 points of which are pee e from four consecutive angles of 
 the first polygon. mim m’.... The point o for instance, since it 
 is situated in kh, is equidistant from m, m’, me; and again, being 
 situated in kh’, it is equidistant from m’, m’’, m’”’; that is, it is 
 the center of a sphere passing through four Festiva angles 
 am, my, m', mm". | 
 
 This being true when the number of sides of the polygon is 
 indefinitely increased; it follows that the normal planes of a 
 curve of double curvature generate, by their perpetual intersec- 
 tion, a curve surface; also since the plane kh’ may be turned 
 about A’h’ till it is in the same plane with kh”, and again the 
 
115 
 
 system fornied by these two may be turned about kA” till they 
 are in the same plane with kh’, and so on, this surface is 
 capable of being spread out upon a plane without rupture or 
 duplication, and is therefore called a developable surface. 
 
 On the same supposition, the polygon hop...becomes a curve 
 of double curvature, to which all the lines ko, k’p, &c. are 
 tangents, that is, hop... is constantly touched by the straight line 
 which generates the developable surface, and is therefore its 
 limit, for no part of the surface can fall within the space towards 
 which the curve is concave; moreover its tangent, being pro- 
 duced both ways from the point of contact, generates two sheets 
 of the surface which are united and terminated in the curve ; 
 hence it is called the edge of regression of the developable 
 surface, from the analogy it bears to a point of regression in 
 a plane curve. ‘This curve is also the locus of the points of 
 intersection of three consecutive normal planes, or of the centers 
 of spheres passing through four consecutive points of the first 
 curve mm m’..., that is, it is the locus of the center of spheri- 
 eal curvature. 
 
 The plane ggg’ passing through two consecutive tangents 
 to the curve, and which is perpendicular to the intersection of 
 two consecutive normal planes, is the osculating plane, whose 
 equation was found in Art. 33; and the point g, in which 
 that intersection meets it, 1s called the absolute center of curva- 
 ture. 
 
 64. To find the equation to the surface, generated by the 
 perpetual intersection of the normal planes of a curve of double 
 curvature. If y=@r, z=fx, be the equations to a curve of 
 double curvature, and if we denote 
 
 dz d’z I ” 
 RAY Pr ss Ya ahs 1 
 / pie Bs 
 
 we have seen (Art. 33.) that the equation to the normal plane 
 t,—2+(y,—y) y +(zZ,—2) 2 =H0,eee eee eee ee (1), 
 and its differential with respect to x, namely, 
 
 (y,—y) y’ +(2,—2) 2” -—U+y? +2") =0,...(2), 
 
116 
 
 are the equations to the line of intersection of two consecutive 
 normal planes; 2, ¥,, Z, denoting the co-ordinates of any point 
 in that intersection. If therefore we elimmate x between them, 
 we shall have the equation to the surface generated by the line 
 of intersection in all its successive positions. On this surface lie 
 all the evolutes of the curve, as will be shewn; it will be a 
 cylinder if the given curve lie all in one plane, and a cone, with 
 its vertex in the center, if the given curve be traced on the sur- 
 face of a sphere. 
 
 65. To find the radius, and co-ordinates of the center, of 
 spherical curvature of a curve of double curvature. 
 
 The center of spherical curvature, or of the sphere which 
 passes through four consecutive points of a curve, is the pomt 
 of intersection of two consecutive generating lines of the surface 
 considered in the last article; hence we must join to equations 
 (1) and (2) of last article, their differentials with respect to x. 
 
 If therefore x, y,, z, be the co-ordinates of the center of 
 spherical curvature, we have in order to determine them, the 
 equations, 
 
 x—xt(y,—y) y +(2,—2) 2 =0, 
 —y) yy +@,—2) 2 —U+y? +2=0, 
 
 Meld 4 
 
 (y,—-y) y+ (z ple z) Ad aa 3 3(y/y” 42"2 ys 0, 
 and the radius of the sphere 
 
 = /(t,—2) + (y,-¥) +(z,- 2). 
 
 Moreover if from these three equations we eliminate x, there 
 will result two equations between w,, y,, z,, which will be those 
 to the locus of the center of spherical curvature, or to the edge 
 of regression of the developable surface generated bythe in- 
 tersections of the normal planes. 
 
 Cor. If for the sake of simplifying the results, we take 
 the origin on the curve, and the tangent at that point for the axis 
 of x, then | 
 
rR iy: 
 x=y=z=0, and z= 0; 
 hence the equations become 
 x, = 0, yy" at 2,2” = ve yy" 44 2, yt a Oh 
 
 Hence the center of spherical curvature lies in the normal plane, 
 and its co-ordinates in that plane are 
 7” yf! 
 
 Se 2.5 
 Y, TTT) TTT) / wlth 0 LLG LES 
 Yr 2 7etfaee ae Y Veti2 
 
 7 MING y 7 WAS 
 also R = NAC Weg) 
 
 View (isis ipa oe ® 
 a ek’ 
 
 z 
 
 66. To find the radius, and co-ordinates of the center of 
 the absolute circle of curvature. ‘The absolute center of curva- 
 ture is the point where the intersection of two consecutive 
 normal planes meets the osculating plane. 
 
 Now spe 2H isa a b; 
 J y 
 
 and ..a+2 =by'; also 142” +y°= 
 the equation to the osculating plane is 
 z,—2= —a(@,—2) +b (y,—y)3 
 
 and the equations to the line of intersection of two consecutive 
 normal planes are 
 
 L,—-xr=a (2) ¥,— y= — b(z —2) +o CArt. 33: ); 
 
 therefore at the point where they intersect, 
 
 oe © 
 tf 2 
 
 z,-z=—a (z,—-2z) + oy — (z,-2z) + 
 
 teh Osa | 
 “. 2—z2= ai eal aupernspresrese*\lls 
 
 MiG a OA Rta) 
 
 sinilarly 2,.—27 = — — . ——_;+—— 
 yeve y Lea + 6° 
 
118 
 
 as va 
 
 a sore Vf ee yf oy 
 
 ar* (ay +b +(b2' +y'¥+ (az —1)° 
 peal (*) “Mey hit celia 
 =(2) Cane ea oy ates peeby byt oat), 
 TE (t+a?+0°y 
 But a+: — by’ =0; 
 . Baby! 4° by!2— ax) =a? + 2? + by"; 
 PG i citer: Di (+4? +27) 
 ii 
 
 172 / rye , 
 
 T4@4e 427% 4+ Q'y—zy 
 
 .(3)3 
 
 (I+ y? + 2°)8 
 AE wk 2? i (2""y/ — a vy. 
 
 this is the value of the absolute radius of curvature, and the co- 
 ordinates of its center are given by equations (1), (2), and (3). 
 
 or R= 
 
 Cor. 1. If we make s, the length of the arc, the inde- 
 pees variable instead of x, we find by the common pioces: 
 of changing the independent variable, 
 
 ] 
 R me M2 12 my? 
 SU | en me 
 
 9] 
 = 
 
 ‘ ha ; 
 where 2 denotes 72? &c.; also the expressions for the co- 
 S 
 
 ordinates of the center assume the following simple forms, 
 
 x! 
 L-L6 => Se ee se 
 y We Meee, pe4+ yea”? 
 ey +2 +y +2 
 ‘, 
 “ 
 gies 
 
 bad (10 (dev Mbt 
 
 oy"? +2 
 
 Cor. 2. If, as in the last article, we take the origin on the 
 curve, and its tangent line for the axis of x, then 
 
 rey=2=0, andy = 25 03" 
 
119 
 
 ° . id, 47 
 
 the equation to the osculating plane becomes z,y =y,z , and 
 the equations to the line of intersection of consecutive normal 
 planes, 
 
 J ” 
 Ly = 0, YY + ® 2 =) ) 
 y" Mt 
 ve 
 An ay = rear are ee R Ce pan eT 
 e Q 729 12 w/f2 9 3 
 Y, y 2 i i / y ae z Rie or 22 
 
 67. To find the points of inflexion of a curve of double 
 curvature. A curve of double curvature may either have three 
 consecutive points in the same plane, or two consecutive points 
 in the same straight line; the first 1s called a semple inflexion, 
 that is, when the curve becomes plane, and therefore the radius 
 of spherical curvature infinite; and the second a point of double 
 inflexion, that is, when the curve becomes a straight lme, and 
 therefore the radius of absolute curvature infinite or evanescent. 
 
 Let ¥, 2, Yo Zo, denote the co-ordinates of a curve of double 
 curvature corresponding to the abscisse r-+h, x+2h; and 
 
 — Ar + by + C, 
 
 the equation to the plane in which three consecutive points lie ; 
 then at a point of simple inflexion, the three equations, 
 
 z=Axv+ By + C, 
 = A(x+h) + By, + C, 
 Za A(r+2h) + By, + C, 
 must be simultaneously true when / vanishes ; hence subtracting 
 
 the first from the second, and the second from the third, and 
 taking the limit, 
 
 dz Hh dz dy 
 
 5 el) f Sag pe ait Ite 
 
 dx eas ae 1s seb?) by 8 ne dx 
 Oe pty os pty 
 
 dx” dx? dx’ dxr*’ 
 
 but the latter of these equations is 
 
120 
 d*z 
 we +h he ME: ° = B(SE ec + eee 
 
 which, by virtue of the first becomes ultimately, 
 ag = BZ dy 
 dx” dx’ 
 
 hence substituting for B its value, 
 
 dz ay | dy d*z ; 
 dx dx? dx dx* 4 
 
 is the equation for determining the abscisse of the points of 
 simple inflexion. 
 
 If in Art. 65, we actually determine the values of 7,—2, y,—y, 
 z,—2, we find for their common denommator the expression 
 gly" — ee : 
 hence the condition we have just found expresses that the center 
 of spherical curvature is at an infinite distance, as it ought to 
 
 do. 
 
 At points of double inflexion, the absolute radius of curva- 
 ture changes its sign, and the curve, after being concave in one 
 direction, crosses its tangent line and becomes concave in the 
 opposite direction; hence also its projections on the planes of 
 xz and xy cross their tangents, and have corresponding points 
 of contrary flexure ; 
 
 dz d’y 
 
 O, or © 
 
 ON Ste : OT ies 
 
 = O; or ©3 
 
 which are the equations for determining the abscisse ; and which 
 we observe make the expression for the absolute radius of cur- 
 vature infinite or evanescent, as ought to be the case. 
 
 68. The tangent lines of the given curve mm’... (fig. 48.) 
 will generate a developable surface, whose edge of regression 
 is the curve mm...; and since the tangent line is the intersec- 
 tion of two consecutive osculating planes, this surface would 
 also be generated by the perpetual intersection of the osculating 
 
121 
 
 planes. But the lines of intersection of the normal planes are 
 perpendicular to the osculating planes; therefore the angle 
 between two consecutive lines of intersection, that is, two con- 
 secutive tangents of op..., 1s equal to the angle between two 
 consecutive osculating planes of mm’... Again, the angle between 
 two consecutive tangents of mm’... is equal to the angle between 
 its corresponding normal planes, that is, to the angle between 
 two consecutive osculating planes of op... If therefore we call 
 the angle between two consecutive tangents of a curve of double 
 curvature its first flevion, and that between two consecutive 
 osculating planes its second flexton, we may enunciate the 
 above properties as follows. The first flexion of a curve of 
 double curvature is equal to the second flexion of the edge of 
 regression of the developable surface of its normal planes; and 
 the first flexion of the latter, is equal to the second flexion of the 
 former. 
 
 69. Every curve, whether plane or of double curvature, 
 has an infinite number of evolutes, all of which he on the de- 
 velopable surface generated by the perpetual intersection of the 
 normal planes. 
 
 Recurring to fig. 48, since kA is perpendicular to the plane 
 through two consecutive sides mm’, m'm’, of the polygon, any 
 point f in it is equidistant from the middle points of the sides 
 g and g’. If therefore we draw gf in the first normal plane, 
 and 2’ ff’ in the second, the lines gf, g’f are inclined at the same 
 angle to kh; and if with centre f and radius fg a circle be 
 described, it will touch the two consecutive sides mm’, mm’ in 
 their middle points g and g’, and when the describing radius 
 comes to g’ it will be confounded with g’f’. In like manner if 
 we draw g” f'f” in the third normal plane, the lines gf’, gf” 
 are equally inclined to kh’; and a circle described from f” with 
 radius f’g’ will touch the consecutive sides m’m’, m’m’” in their 
 middle points, and the describing radius will be confounded 
 with of” at the pomt 9”. By continuing this construction the 
 polygon fff"... will be formed, by unwinding a thread gf from 
 which, circular arcs, touching every two consecutive sides in their 
 middle points, will be described. Therefore when the number 
 of sides is indefinitely increased, the polygon fff’... will become 
 
122 
 
 a curve traced on the developable surface formed by the in- 
 tersections of the normal planes; and by unwinding a string 
 from it, the proposed curve of double curvature will be traced 
 out. Also since the direction of the initial radius gf is arbitrary, 
 any other direction would have produced a curve endowed with 
 the same properties as Shit aoe it follows therefore that every 
 curve has an infinite number of evolutes all contained on the 
 surface generated by the perpetual intersections of its normal 
 planes; in the case of a plane curve all the evolutes, except that 
 in its own plane, are curves of double curvature, traced on the 
 right cylinder whose base is the evolute in its own plane. 
 
 Since 2 of kh = g"f'k' = Nf f", if the plane hk’ were 
 brought into the same plane with /’k” by being turned about 4'A’, 
 ff would be brought into the same straight line with f’f"; and 
 since the same may be proved successively of all the other sides 
 of the polygon ff/f”..., it follows that if the surface formed by 
 the intersections of the normal planes were developed, the evo- 
 lute ff’ f’... would become a straight line, and therefore would 
 be formed by stretching a thread in "ite direction gf and applying 
 it freely to that surface. Hence if from any point in a curve of 
 double curvature, any line be drawn touching the surface formed 
 by the normal planes, and be plied freely on that surface, it 
 will form an evolute of the proposed curve; and will be the 
 shortest line that can be drawn between any two points of the 
 surface, through which it passes. 
 
 70. It is to be observed that the absolute centers of cur- 
 yature are not situated on an evolute of the curve. For if g’q’ 
 be perpendicular to 4’A’, then it cannot pass through the point g; 
 
 because it would then coincide with g’q, and therefore cut kh 
 
 at right angles, which would require that kA and k’h’ should be 
 parallel; and this can never happen as long as the curve is of 
 double curvature. Hence the consecutive absolute radii of cur- 
 vature gg, gg’, since they are in different planes, and do not 
 meet one another in the common intersection of those planes, 
 therefore they do not meet one another at all; and therefore 
 cannot be consecutive tangents of the same curve; that is, 
 ‘the locus of the absolute center of curvature is not an evo- 
 lute. 
 
123 
 
 71. To find the equations to any proposed evolute of a 
 curve of double curvature. 
 
 As we know the equation to the surface on which all the 
 evolutes lie (Art. 64.), it only remains to find the equation which 
 particularizes a given evolute. Let x, y,, be the co-ordinates of 
 a point in the projection of an evolute, 2, y, the co-ordinates of 
 the corresponding point in the projection of the curve; then a 
 tangent to the projection of the evolute, must pass through the 
 corresponding point in the projection of the curve ; 
 
 ay) 4 Ry 
 ke w,—- x 
 
 This joined to the two equations 
 x,—a+(y,—yy' +(z,- 22 =0, 
 (yyy +@,- 2) -A+y?+2'")=0, 
 
 which belong to the generating line of the developable surface 
 containing the evolutes, will ‘give by eliminating z the two equa- 
 tions to the evolute, one of them being a differential of the first 
 order; and its integral will introduce an arbitrary constant, by 
 means of which the evolute may be made to satisfy that condition 
 which particularizes it; as, for instance, to pass through a given 
 point of the developable surface on which it is traced. 
 
 72. Asan example of this theory, take the curve of double 
 curvature, resulting from the intersection of a sphere and cylin- 
 der, considered in Prob. 1. Art. 34. 
 
 Its equations are 
 ‘ 2 2 
 
 y =2axr—2xr, 2 = 4a° — Zaz, 
 
 r) — Ay ” 
 
 a a a 
 ae = =- =, 2.> 7 Z2= - =, 
 ¥y > £Y y° oh 23 
 
 y 
 
 Hence the equation to the normal plane, as we have seen, is 
 
 obey, = Zante al Wy 
 y z 
 
124 
 
 To obtain the equations to the line in which this is inter- 
 sected by the consecutive normal plane, we must join to it, its 
 differential with respect to 2, namely, 
 
 i Oxted e(2).5 
 
 and to find the equation to the surface generated by the per- 
 petual intersection of the normal planes, we must eliminate x, ¥ 
 and z, between these two equations, and the equations to the 
 curve. | 
 
 Now from equation (2), 
 
 (- ar ove or 7=2(%)’, 
 
 and equation (1) may be written 
 
 —_—-— — — 
 
 ANT Maha ABO S77) z 
 
 EN bkiee 5! AERO 
 
 Hence substituting for — its value, and reducing, we find for the 
 a 
 
 BUY GAN Vint as Ht, 
 
 equation to the surface 
 
 (x, -y, \/ 2, rainy )=z, a Rae)! bier 
 
 Again, to find the expression for the radius of curvature, 
 
 RL 
 i+y?4+2°= apa) 
 
 Cy" Y +2" + (y" 2 = 2" yf = 'G ) (10a +32); 
 
125 
 
 Qatex)t 
 
 .. radius of curvature = —————— , 
 x] 10a + 32x 
 
 73. The following Problems are examples of finding the 
 involutes of curves of double curvature. 
 
 I. On a given surface, to trace a curve of such a nature, 
 that its involute shall be a plane curve. 
 
 The curve of equal inclination, that is, one whose tangent line 
 is inclined at a constant angle to the plane of ry, will have this 
 property. For if s denote the length of the arc, intercepted 
 between the plane of xy and a point whose ordinate is z, and 
 a =the constant angle which the tangent line makes with the 
 
 dz 
 
 plane of ry, then hoe sina; .. Z=Ssina, ands=zcoseca 
 = the length of the tangent line between the point of contact 
 and plane of ry; hence if a string be applied to the curve of 
 equal inclination, and then be unwound from it, beginning from 
 the plane of xy, and be kept stretched in the direction of a 
 tangent, its extremity will be always in the plane of xy, and 
 will describe a curve, which is manifestly the involute of the 
 projection of the curve of equal inclination on the same 
 plane; for the projection of the tangent line will touch the pro- 
 jection of the curve, and will be of the same length. 
 
 To find the equations to the curve of equal inclination. 
 
 Let dz=pdx+qdy, be the equation to the surface on 
 which it is traced, o=length of its projection on the plane 
 of ry; 
 
 and it remains to substitute for p and q their values in r and y, 
 and to integrate this equation. 
 
126 
 
 II. When the given surface is one of revolution, the equation 
 to the projection of the curve of equal inclination may be more 
 readily obtained by polar co-ordinates. 
 
 Let BP be the curve, (fig. 49.) BQ its projection on a 
 
 plane perpendicular to the axis of revolution ; 
 
 arc bi =< Gwen 27, 2. bo Ge 
 
 os — = tana, or (£) = tan” a (52) =tan?a +r a) 
 
 which may also be put under the form 
 
 =) An r 
 —$e= tan a, 
 dr 
 
 > 
 r—p 
 
 2 
 
 where p is the perpendicular from C, on the tangent to BQ at 
 the point Q. 
 
 Now from the equation to the generating curve, we know 
 
 ft a 
 — in terms of r, and hence obtain the differential equation to 
 r 
 
 the required projection. 
 
 Ex. 1. Let the surface be a paraboloid, and the origin in 
 the centre of its base; c = its altitude, a = 3 latus rectum; 
 
 ae . " e r? 
 “.(c—2)2a=r, —=—-, .. y=tana.>—,, 
 dr a a r—p 
 2 
 or 9 — p? = a’ tana, 
 
 the equation to the involute of a circle whose radius = a tan a. 
 Hence the generating point cannot approach the axis nearer than 
 atan a; in fact, it must then move in a plane passing through the 
 axis in order to make an angle a with the plane of ry; it will 
 thence return towards the base, and there will be a cusp in its 
 path corresponding to the cusp in the projection; also the 
 portions on each side of the cusp will be similar and equal; and 
 the whole length of the path, between leaving the base and 
 returning to it 
 2c tan a 
 
 1 
 Z 
 = 2z coseca=(2c —a tana) —- = — —a : 
 sina sina cos a 
 
127 
 
 The locus of the pomt in which the tangent line meets the 
 base of the paraboloid, will be the second involute of the circle 
 whose radius = a tana; its equation will therefore be 
 
 pt+atana=,/r? =a? tan?a. 
 
 Ex. 2. Let. the surface be a right cone, and the origin 
 in its vertex; 
 
 2 
 r 
 
 ‘ bf ; Pa ee 2 f 
 ee r = ztan PB, .. cot 6 = tana. ~=—3; 
 
 Ler 
 
 or p=ra/1— tan’a. tan?p, 
 
 the equation to an equiangular spiral. Hence it follows that 
 this curve cuts the generating line of the cone under a 
 constant angle, that is, it is the conical helix. Hence if a 
 thread be applied to the surface of a cone, according to a helix 
 traced on it, and then be unwound beginning from the vertex, its 
 extremity will always be found in a plane perpendicular to the 
 axis through the vertex, and will trace out an equiangular spiral 
 on that plane, which is also the involute of the projection of the 
 conical helix on the same plane, and therefore similar to it. 
 
 Ill. Ifa uniform and flexible string be suspended from two 
 points in the surface of a vertical cylinder whose base is a circle, 
 it will form itself into a curve of double curvature, such, that 
 its involute lies on the surface of a sphere. 
 
 Let A (fig. 50.), be the lowest point of the catenary, AB 
 the radius of curvature at that poimt =a; take the horizontal 
 plane drawn through B for the plane of ry, and the axis of the 
 cylinder for that of z, and let 
 
 CM=yr1, MN=y, PN=~z, 
 
 be co-ordinates of any point P in the catenary, 
 
 Cheb, arc BNi=.s: 
 
 : $ oain- 12 ® sin 
 pat bc = = ~? sj Le 
 a a 6 
 
 z 
 then - =H(e+e \=4(e 2 
 a 
 
 and = y= ee — vr", 
 
128 
 
 are the equations to the curve, e being the base of Napier’s 
 system of logarithms. 
 
 Draw NT’ perpendicular to the tangent PT and join 
 CT; then PT is perpendicular to the plane CT'N, for a line 
 drawn through N parallel to PT, and therefore coinciding with 
 the tangent plane of the cylinder would be at right angles both to 
 T'N, CN, and therefore to the plane passing through them; and 
 CNT is aright angle, therefore 
 
 CT’ = CN’?+ NI°=CB’ + BA’ =CA’*, 
 
 because the perpendicular upon the tangent from the foot 
 of the ordinate, is constant by the nature of the catenary. 
 Hence the locus of 7’ is a sphere radius CA, and since C7'P 
 is a right angle, therefore PT touches the sphere. But 
 PT =arc AP; if therefore a string be unwound from AP 
 beginning from A, and be always kept stretched in the direction 
 of a tangent, its extremity will trace out a sphere and the string 
 itself will be a tangent to the sphere. 
 
 If x’, 7’, 2, be co-ordinates of T, and 2 BCN=8, since 
 Nt=,/a’—2z”, the equations to the locus of T are 
 
 x’ =b sind — ./ a? — z” cos 6, y =b cos0+,/a’ — 2” sin 8, 
 
 Qe Ug MSs 
 where @.= 5 log (ee = ig (Pa Jai: ) 
 a z 
 
 . , 2g 
 since 272 =@. 
 
SECTION VI. 
 
 ON SURFACES IN GENERAL. 
 
 74. "To find the general equation to cylindrical surfaces. 
 
 A cylindrical surface is generated by a straight line which 
 moves parallel to itself, and always passes through a given 
 curve. 
 
 Let the equations to the generating line in any position be 
 r=az+t+a, y=bz+P, a and B being variable quantities 
 depending upon that position, and a and 0 constant quantities, 
 since the line is always parallel to itself. Also let y=@z, 
 z=fx, be the equations to the directrix, or curve through which 
 the generating line always passes, and 2’, y’, z’, the co-ordinates 
 of the point in which they meet; then a’, y’, z’ must satisfy 
 both the equations to the directrix, and generating line; 
 
 Ke av’ =az' +a, y =b2' + B, y =u’, d= fa’; 
 
 now by means of these four equations, we may eliminate 2’, 7, z’, 
 and there will remain 8B = Fa; but B= y— bz, a=a— az; 
 ~. y—bz = F(x — az), a relation among the co-ordinates of 
 any point in the generating line, and therefore the equation to 
 the surface which it describes. 
 
 Cor. 1. If the directrix be a curve traced in the plane of 
 ry, its equations are 
 
 2=0, y=ou; «.y—bz=h (e—az) 
 
 is the equation to the cylindrical surface; that is, we have only 
 to. change v and y into r—az, y—Obz, mm the equation to the 
 directrix. This has already been exemplified in Art, 19. 
 
 R 
 
130 
 
 Cor. 2. If for the purpose of eliminating the arbitrary 
 function, we differentiate the equation y — bz= F(x — az), suc- 
 cessively with respect to x and y, we find 
 
 —bp=F@—a2)(\~ ap), 1—bg=F w—a2)(- a9); 
 
 dz dz 
 *.. abpq=(1- 1— —+b— =1, 
 abpg=(1—ap) (1—6q), or aT + a 
 which is the differential equation to cylindrical surfaces; we 
 may however obtain it more easily by the consideration of the 
 tangent plane, as follows. 
 
 75. To find the differential equation to cylindrical sur- 
 faces. 
 
 One of the distinguishing properties of cylindrical surfaces 
 is, that the tangent plane is always parallel to the generating 
 line. 
 
 Let x’ =az', y' =z’ be the equations to a line through the 
 origin, to which the generating line is always parallel; then this 
 line is parallel to the tangent plane, whose equation is 
 
 sx —zg=p(r—2) + q(y—-y); 
 *, (Art. 5.) ap+bq=1 Ribeca el idk Geen as before 
 oem e . P pa 9 a dy 3 ° 
 
 76. Given the direction of the generating line, to find the 
 equation to the cylindrical surface which envelopes a given 
 curve surface. 
 
 We have seen that whatever be the nature of the directrix, 
 the equation 
 
 dn dz 
 dx’ dy’ 
 
 rived from the equation to a cylindrical surface. But at the 
 
 points where the cylinder touches the surface, the values of 
 
 a2 dz 
 
 —, >, are the same for both; therefore at those poimts the 
 dx dy 
 
 must subsist between the differential coefficients e- 
 
idl 
 
 above relation subsists between the differential coefficients de- 
 rived from the equation to the surface; that is, the co-ordinates 
 of the points of contact are such that the equation 
 
 ome 1 Gini 
 a Pea , is satisfied. 
 
 Hence if we differentiate the given equation to the surface, and 
 . ce Py ay: 
 write the values thence obtained of — , —— in equation (1), the 
 dx” dy 
 
 result together with the equation to the surface, will be the 
 equations to the directrix; and we know the direction of the 
 generating line, and therefore can find the equation to the 
 required surface by Art. 74. 
 
 Ex. To find the equation to the cylindrical surface which 
 envelopes a given oblate spheroid. 
 
 Let the center of the spheroid be the origin, and the plane 
 of xz that which contains the axis of the spheroid, and a line 
 through the origin to which the generating line of the cylinder 
 is always parallel; therefore the equations to the generating line 
 in any position will be 
 
 2 = az + a; y= Pp. 
 
 Let 2° + y? + n°’z° = cc’ be the equation to the spheroid; 
 
 _ dz vo 
 Ye UO Tee 
 : eipey ets ; ax 
 therefore by substitution in equation (1), since 6=0, 1+ —— = 0, 
 nz 
 
 ¢ 2 2 2.2 2 : 
 or 2 2z-+- axr=0, and x +y +n 2° =c, are the equa ions to 
 the directrix, between which, and the equations vr = az + a, 
 y = , eliminating x, y, z, we find 
 
 3? (n? + a’) +n’? = c° (n? + a’). 
 Therefore, restoring the values of a and #3, 
 y? (nr + a®) +n? (w— az) =C (n° +0") 
 
 is the equation to the surface. 
 
132 
 
 c . ; 
 If we make z = — -, we shall obtain the equation to the 
 n 
 
 curve in which the cylinder meets the horizontal plane on which 
 the spheroid stands; this gives 
 
 y? (n® + a”) + (ax + acy = c (n® + a’), 
 aqey. 
 : (2 cares 
 e's, te 
 oa be — 
 c i LONG 
 Cc + (=) 
 n 
 
 the equation to an ellipse, the origin, that is, the point in which 
 the spheroid stands, being the focus. 
 
 3 
 
 77- To find the general equation to conical surfaces. 
 
 A conical surface is generated by a line which passes through 
 a given point, and always meets a given curve. 
 
 Let the co-ordinates of the given point or vertex be a, b,c; 
 “t-aza(z—c), y—b=B(z—o), 
 
 are the equations to the generating line in any position, a and 
 being variable quantities depending upon that position. Also 
 let y= x, z=fx, be the equations to the directrix or curve 
 through which the generating line always passes, and 2’, ai, en 
 the co-ordinates of the point in which they meet; then 2’, y’, 2’ 
 must satisfy both the equations to the generating line and direc- 
 trix 5 
 . wt -a=a(z'—c), y¥ -b=B(!—0), Y=ou, 2 =f’. 
 
 Now by means of these four equations, we may eliminate 
 , 4 ° . 
 a, y, 2, and there will remain B= Fa; 
 
 iG ee Pete ee bee (Ee 
 
 ’ ’ 
 Bharat © ye) ae en Se ml 6 
 
 a relation among the co-ordinates of any point in the generating 
 line, and therefore the equation to the surface which it de- 
 scribes. 
 
133 
 
 Cor. 1. If the directrix be a curve traced in the plane of 
 xy, it equations are z=0, y=g2; 
 
 _ O2—cy =9 (=) 
 
 daha od aa a z—c 
 is the equation to the conical surface; that 1s, we have only 
 G2 tr P02-~ Cy 
 
 to change xv and y into , in the equation to 
 
 z—c 7’ z-e 
 the directrix; this has already been exemplified in Art. 19. 
 Cor. 2. If from the equation 
 y—b F é = *) 
 Z—Cc Z—C 
 we eliminate the arbitrary function by differentiation, as in 
 Cor. 2. Art. 74, we arrive at the equation 
 
 dz dz 
 z—e = —-(r—a) + — (y— 9), 
 dx dy 
 which is the differential equation to conical surfaces ; but which 
 may be obtained more easily by the consideration of the tangent 
 plane, as will be seen in a following article. 
 
 78. One of the most useful applications of the last article, 
 is to find the projection of a curve upon any plane. 
 
 If we take a line drawn through the place of the eye, per- 
 pendicular to the plane of projection, for the axis of z, and 
 consider the place of the eye as the vertex, and the curve to 
 be projected as the directrix, the equation to the optical cone is 
 
 eer ae 
 
 which results from the general equation to conical surfaces in 
 the last article, by making a=) =0, and changing the sign of c, 
 that is, supposing the vertex to be situated below the plane of 
 xy. If in this equation we make z constant, we obtain the 
 equation to the stereographic projection of the given curve upon 
 
134 
 
 any plane perpendicular to the axis of 23; and if in the result 
 we make c¢ infinite, we find the equation to the orthographic 
 projection. 
 
 Ex. To find the stereographic, gnomonic, and ortho- 
 eraphic projections of any circle of a sphere. 
 
 Let the center of the sphere be the origin, and a plane 
 perpendicular to the plane of the circle to be projected, the 
 plane of rz; then 
 
 Boe as aateal 
 z=axrt+b 
 
 are the equations to a circle of the sphere, forming the directrix 
 of the optical cone ; 
 
 also r= a(z+c), y=Bz+4+ 0), 
 
 are the equations to the generating line; hence, eliminating 
 x, y, Z, between these four equations, we find 
 
 b 
 = (aie ac) et ao, fi pein eed. 
 l—aa 
 b+c b+e 
 Faith? aa hard is ae y = he aera 
 
 “. (b+c) (a° +B) + (acat bY = r°(1—aa); 
 or, restoring the values of a and [, 
 (b+c) (a? +y*) + (aertbet+o) = r(z+ce-az)y, 
 the equation to the optical cone, from which those to the pro- 
 
 jections may be deduced. 
 
 1. In stereographic projection, the plane of projection 
 passes through the centre, and the eye is situated in the surface 
 of the sphere, therefore, z=0, c=7; 
 
 oo (O+7rP @ ty) +9 (ax + bY = 7° (r—ax)’, 
 Qar” > r—b 
 
 yee ‘Fp 
 
 or x +y + 
 
 the equation to a circle, whose radius 
 
135 
 
 7 73 a ROL Nt 
 
 = eb Sra +a)—, 
 3 
 ar 
 
 r+b- 
 
 and distance of its center, from that of the sphere = 
 
 2. In gnomonic projection, the plane of projection touches 
 the surface, and the eye is situated in the centre of the sphere, 
 therefore, z=7, c=0; 
 
 2. Pa +y’) + Fr = 2 (v¥—az)’, 
 or 2° (0? —a’r*) + By? = 2° (rr — 6) — Qr’an; 
 
 the equation to an ellipse, hyperbola, or parabola, according 
 asb >, <, or=ar. 
 
 3. In orthographic projection c= ©, z=0; 
 Oey + (ar +by =r’, 
 the equation to an ellipse. 
 
 In the same manner it may be shewn that the stereographic 
 projection of any plane section of a paraboloid of revolution, upon 
 a plane perpendicular to its axis 1s a circle, the eye being placed 
 in the vertex; or that the stereographic projection of any dia- 
 metral section of an oblate spheroid, upon the plane of the 
 equator, is a circle, the eye being placed im the pole. 
 
 79. We may employ this method also to compare any 
 angle on the surface of a sphere with its different projections ; 
 this however may be done more simply by geometrical con- 
 siderations, as will be seen in the following problem. 
 
 Pros. The gnomonic, orthographic, and stereographic 
 projections of the rhumb-line on the surface of a sphere, are 
 three of Cotes’s spirals. 
 
 The rhumb-line is a curve of double curvature, traced on 
 the surface of a sphere so as to cut all meridians under the same 
 angle. ) 
 
 }. To find the gnomonic projection. 
 
 Let y Bx, (fig. 51.) be the plane of projection perpendicular 
 to BC, T't its intersection with a plane touching the sphere at 
 
136 
 
 a point P in the plane of xz, then T’t is parallel to By; Pt 
 a tangent to the rhumb-line at P, Q the projection of P de- 
 termined by producing CP to meet the plane «By. Then 
 Z TQt is the projection of T’P¢; 
 
 Tr a rT B 
 iy i al PTE) attire CLR 
 
 and HE ES alert = TQ: TP~ TQ" Ge 
 
 if a=the constant angle at which the rhumb-line cuts the me- 
 ridian. Let BQ=r, p=perpendicular from B on Qé which 
 touches the Dy aeasats of the rhumb-line at Q, BC=a, 
 
 adh head a tana ea 
 then 9 or > 1 =cot*a (~ + 1s 
 SPP wi ANA kes +a n 
 1 cosec a cota 
 OF ee 9 ee 
 p oe a 
 
 the equation to Cotes’s third spiral. 
 2. To find the orthographic projection. 
 
 Let y Cx be the plane of projection (fig. 52.), J’¢ its inter- 
 section with a plane touching the sphere at a point P, situated 
 in the plane of xz, then Té is parallel to Cy; Pt a tangent 
 to the rhumb-line at P, N the projection of P, then 2 T’Né 
 is the projection of J'P¢; 
 
 4833 PP* FF TP 
 and tan TNt = TN = ON’ rip or ate 
 
 or if CN = r, p = perpendicular from C on Né which touches 
 the projection at N, 
 
 j p a tana ” , i i 
 ———— = —__—:;:_ »*, —. — ] = cot®a —-—-— 
 2 2 2 Q? 2 ( =); 
 WA Typ A/C ao, P a 
 I cosec’ a cot? a 
 or as — 2 — oman 
 , at 
 
 the equation to the hyperbolic spiral. 
 
137 
 
 3. -To find the stereographic projection. 
 
 Let CO=CP be perpendicular to the plane of projection 
 x Cy (fig. 52.); jom PO meeting the plane of projection in Q, 
 which is therefore the projection of P, and Z J7'Qt of T'Pt. 
 Then Z TPQ = complement of CPO=complement of COP 
 =PQT, .. PT=QT; and since Tt is perpendicular both to 
 PT and QT, .«. 2 TQt=TPt=a; therefore since the angle 
 between the tangent and radius vector is constant, the curve 1s 
 an equiangular spiral, its equation being p=r sin a. 
 
 Cor. 1. It appears, that if an angle a on the surface of a 
 sphere have one of the arcs which contain it, a meridian, and if 
 ry denote its gnomonic, its orthographic projection, and @ the 
 arc of the containing meridian intercepted between its vertex, 
 
 and a plane through the center parallel to the plane of projection, 
 then 
 
 . ] 
 tan y = sin@ tana, .tanw = sin 0 tana; 
 
 also tana = ,/tanw.tanry, that is, the tangent of the stereo- 
 
 graphic projection is a mean proportional between the tangents 
 of the orthographic and gnomonic projections of the same angle. 
 
 Cor. 2. Hence we can determine the orthographic and gno- 
 monic projections of any angle whatever on the surface of a sphere. 
 
 Let a, a be the angles which the arcs containing any angle 
 make with the meridian drawn through their intersection; w, w, 
 their orthographic, ¢y, sy’, their gnomonic projections, 0 = height 
 of the pole ; 
 
 : in 0 (t tana’ 
 Ps Games) 228 (tana+tana ) 
 
 1—sin’@ tana tana’ 
 
 2sin@ sin (a+a’) 
 of (1+ sin® @) cos (a +a’) + (1—sin? @) cos (a—a’')’ 
 1+sin°@  1—sin’@ cos(a aa 
 
 or cot (y +f )=cot(ata’)} 
 
 “gsin0 ' Qsin@ cos(ata)§’ 
 alec cont enemas laratta Og cos tenet 
 nara y, cot(w + w )=cot (ata 2sin@ Qsnm@ cos(ata)9’ 
 
 writing ——~ instead of sin é, 
 sin 0 
 
138 
 
 1 +4sin°0 1 —sin°0 
 Let ai aarti =m, and .. ———— =,/m'-1; 
 2sin@ 2sin0 
 also let 2a represent any angle on the sphere, 2w, 2ry, its ortho- 
 graphic and gnomonic projections, and 6 the angle which an arc 
 
 bisecting it, makes with the meridian through its vertex ; 
 
 —_— cos 20 
 “. cot2y = cot2a (m+ J m? — i 7 
 
 cos 2a 
 
 cos 28 
 cot2w = cot2a (m — Jn? — 1 ): 
 
 cos 2a 
 
 These formule hold whether the containing arcs fall on the 
 same, or different sides of the meridian. 
 
 Hence cot 2w + cot 2y=2m cot2a; which shews that the 
 sum of the co-tangents of the projections is independent of the 
 position of the angle, as long as its vertex remains fixed; and 
 that the orthographic and gnomonic projections of a right angle 
 are supplements to each other. The stereographic and gnomonic 
 projections will be equal to one another, when 0= 45". 
 
 80. ‘To find the differential equation to conical surfaces. 
 
 The distinguishing property of conical surfaces 1s, that the 
 tangent plane always passes through the vertex; let a, 0, ¢ 
 denote the co-ordinates of the vertex, then whatever be the 
 dz dz 
 dx’ dy” 
 derived from the equation to the surface, must be such as to 
 satisfy the equation 
 
 nature of the directrix, the differential coefficients 
 
 dz dz 
 zen F@-a+F y-d) 
 which is the general equation to conical surfaces. If we inte- 
 
 grate this equation, we shall of course find 
 
 y—b 
 
 fo =€ 
 
 r—a 
 F(Canys agreeably to Art. 77. 
 
 81. Given the position of the vertex, to find the equation 
 to the conical surface which envelopes a given curve surface. 
 
139 
 
 Whatever be the nature of the directrix, we have seen that 
 the differential coefficients derived from the equation to a conical 
 surface must satisfy the equation 
 
 dz dz 
 ecu (i come 7 ee a + ay Ladies san alle 
 
 but at the points where the cone touches the surface, the values 
 are the same for both; therefore at those points 
 
 the above relation subsists between the differential coefficients de- 
 rived from the equation to the surface, that is, the co-ordinates of 
 the points of contact are such, that the above equation is satisfied. 
 
 ; dz dz : 
 Hence if we obtain values of —, —, from the given equation 
 
 dx dy 
 
 to the surface, and substitute them in equation (1), the result, 
 together with the equation to the surface, will be the equations 
 to the directrix; and we know the co-ordinates of the vertex, 
 and can therefore find the equation to the required conical 
 surface by Art. 77. If the vertex of the cone be considered 
 as a luminous point, the curve of contact whose equations 
 we have just found, is that which on the surface separates 
 the illumined and obscure parts; if it be considered as the place 
 of the eye, the curve of contact is the line of the apparent 
 contour of the surface. 
 
 Ex. To find the equation to the conical surface which 
 envelopes a given surface of the second order. 
 
 Let the surface be referred to a system of conjugate diameters, 
 one of which passes through the given vertex, let this be the axis 
 
 x yt a . 
 or 2; and let’ —+— Y 4 = =1 be the equation to the surface ; 
 a 
 
 Peavy. 
 dz dz 
 
 therefore, substituting the values of — , —, m the equation 
 
 dx dy 
 
 (9) 
 
 dz dz ZC 
 de” + dy since a=b=0, we ao, mats 
 
 2 2 2 
 
 F x z 
 Which’ HEeMer Neth = eee eer 
 
 CEA LSP ty 
 
140 
 
 are the equations to the curve of contact; also let 
 eins =~ Cy mien (Zt Ol, 
 
 be the equations to the generating line; then eliminating 2, y, 2, 
 between these four equations, we find 
 
 + Demne 
 
 for m, and for 7, 
 c 
 
 | ‘ . aX 
 
 the equation to the surface is 
 Q 2 
 Bree a BENG 
 ae (E 48) 
 
 This completes what was said upon this subject in Prob. 5, 
 Art. 43. 
 
 82. To find the general equation to conoidal: surfaces. 
 
 A conoidal surface is generated by a straight line which 
 moyes parallel to the plane of xy, and always meets a given 
 curve, and also the axis of z 
 
 Let z=, y=az, be the equations to the generating line 
 in any one of its positions, depending upon the variable quan- 
 tities B and as y=pr, z=fx, the equations to the directrix, 
 and a’, 7’, 2’, the co-ordinates of the point in which the gene- 
 rating line meets it; therefore wv’, y, z. must satisfy both the 
 equations to the directrix, and generating line, which gives 
 
 2 ala B, y — ax’, y’ i px’, 2 =r. 
 
 Now by means of these four Saunas we may eliminate 2’, y’,z 
 and there will remain 
 
 Beta: hit Be, a=: tak), 
 
 a relation among the co-ordinates of the generating line in any 
 position, and therefore the equation to the surface which it 
 describes. 
 
141 
 
 Ex. Suppose the directrix to be a helix, (Art. 34. Prob. 2.), 
 whose equations are- 
 J nen 
 ory =a, 2 =!\nacos —, 
 a 
 
 and let the equations to the generating line be 2=B, y =a Cid, 
 as above, then eliminating 2, y, z, x? +a°x’ =a’, 
 
 r = —————_ .*. == JL COS pea SL La his 
 a fifa” RA ie 
 
 therefore, restoring the values of a and (3, the equation to the 
 surface is 
 
 1 
 
 Z2=natan ~*~. 
 x 
 This is the surface presented. by the inferior superficies of 
 a staircase, attached to a vertical column round which it winds. 
 
 We have seen another instance of this mode of generating 
 surfaces in Prob. 5. Art. 37. 
 
 83. ‘To find the differential equation to conoidal surfaces. 
 
 Their characteristic property is that the tangent plane always 
 touches the surface according to a generating line, that is, 
 according to a line parallel to the plane of xy, and meeting 
 the axis of z. If therefore in the equation to the tangent plane, 
 we make z’ =z, we find 
 
 dz , dz , 
 Gee Tae th Gy y= 
 for the equation to the projection of the line of contact; and 
 
 in order that this line may pass through the origin, when x’ =0, 
 we must have y’=0, and 
 
 which is the required equation; and which might also have been 
 
 ; a : ; y,4), 
 obtained by eliminating the arbitrary function from z= (=): 
 
142 
 
 84. ‘To find the general equation to surfaces of revolution. 
 
 A surface of revolution is generated by a circle whose center 
 moves along a straight line to which its plane is perpendicular, 
 and whose perimeter always passes through a given curve. 
 
 Let r= az oe 
 y=bz+ 8 
 
 be the equations to the axis, or line along which the center 
 of the generating circle moves; and consequently 
 
 z+-ar+t by=c ! 
 eas realism |. 
 the equations to the generating circle. 
 
 Also let y=@xr, z=fa, be the equations to the directrix, 
 or curve which the perimeter of the generating circle always 
 meets, and 2’, y’, z’ the co-ordinates of their point of inter- 
 section; then 2’, y’, 2’, must satisfy both the equations to the 
 generating circle, and directrix; 
 
 2 ar +by'=c, @’—ayt(y — BY +22 =P, Y= oe, J=fr; 
 hence, eliminating 2’, y’, 2’, from these four equations, there 
 results c= F(r°), 
 
 or z-+axr+ by=F(w—al+(y— BY +2°*), 
 
 a relation among the co-ordinates of the generating circle in any 
 position, and therefore the equation to the surface which it de- 
 scribes. 
 
 If the axis of revolution coincide with the axis of z, 
 a=b=0, a=B=0; 1. 2=F(a?+y7°+2), or z=W(r" +-y"), 
 as we have already seen (Art. 20.) 
 
 Ex. To find the equation to the surface generated by the 
 revolution of any straight line about the axis of z. 
 
 Let r=az+a, y = b'z+/3' be its equations ; 
 thn z=co, P +7 4+2=Pr° 
 
 are the equations to the generating circle; therefore eliminating 
 L,Y 2; 
 
143 
 
 (’cta)y + (bc + PY + =7°, 
 or, restoring the values of r and c, 
 
 (a’z +a’ + (Wz + BY. =xr+y', 
 which is the equation to the surface. 
 
 If in this equation we make x=0, or y=0, in order to 
 find the equation to a section through the axis, the resulting 
 equation represents an hyperbola, therefore the surface is an 
 hyperboloid of revolution. Also if in retaining one of the above 
 equations to the directrix, we change the signs of the second 
 member in the other, the equation to the surface is not altered ; 
 therefore the same surface may be generated by two different 
 straight lines, and through any one of its points two straight 
 lines may be drawn so as to entirely coincide with the surface. 
 
 85. To find the differential equation to surfaces of revo- 
 lution. 
 
 One of their distinguishing properties is that the normal 
 always meets the axis of revolution. 
 
 Let «’ = az + a 
 
 y = bz +B 
 
 r—-@f + pz’ — z)=0 
 
 yy —y tq’ — d=0 
 then if these two lines intersect, their equations must be satisfied 
 by the same values of 2’, yf 2’, which will be the co-ordinates 
 of their point of intersection. Hence if we eliminate 2’, y’, 2’, 
 
 there will remain an equation between p and g, which will be 
 
 the differential equation to the surface. Now the equations 
 to the axis may be written 
 
 be the equations to the axis, 
 
 the equations to the normal ; 
 
 x —xr=a(z? —2z)+az+a—xz, y —y=b (2-2) 4+bz+B-y; 
 
 therefore, by substitution for 2’ — a, and y’—y from the equa- 
 tions to the normal, 
 
 (a+p) (2’—z) + az+a—xr=0, (6+ 9) (2’—z)+ 62+ B—y=0 : 
 hence, eliminating z’ — z, 
 
 (bz +B—y)(at+p)—(az+a—x) (6+ 4)=0, 
 
144 
 
 dz dz 
 a —_— y om 7” } — om —s »\ —_— 
 or (62+B—y) — (azt+a—z2z) an y) a (a —x) b=0, (1), 
 
 the required equation; which expresses that a surface is one 
 of revolution about an axis determined in position by the con- 
 stants a, 6, a, 2, independent of the generating curve; its 
 complete integral is of course the equation of Art. 84. 
 
 This equation might also have been obtained by considering 
 that the tangent plane is always perpendicular to a meridianal 
 plane, that is, one which contains the axis of revolution and 
 passes through the point of contact. If the axis of revolution 
 coincide with the axis of z, the equation becomes 
 
 BY ae 
 
 86. A curve surface, whose equation is given, revolves 
 round an immoveable axis to which it is fixed; to find the 
 equation to the surface which touches and envelopes it in every 
 position. 
 
 The same series of points of the moveable surface, namely, 
 those which are at the greatest distance from the axis, are 
 always in contact with the enveloping surface; therefore the 
 required surface will be formed by the revolution of the curve 
 of contact, corresponding to any position of the moveable sur- 
 face, about the given axis. Let .*. F(x, y, z) = 0, be the 
 equation to the moveable surface in its first position, then 
 for all its points in contact with the enveloping surface, the 
 equation F(x, y, z)=0, must satisfy the general equation to 
 surfaces of revolution; if therefore we derive from it, values 
 
 dz dz 
 of cae ce and substitute them in equation (1) of the pre- 
 ceding Article, the resulting equation f(x, y, z) = 0, will be- 
 long to the curve of contact; and its two equations will be 
 
 L,Y, 2). — ey (2, 5.2) — 0. 
 
 Hence, knowing the equations to the directrix and axis of revo- 
 lution, the equation to the required surface may be found by 
 Art. 84. 
 
145 
 
 Ex. Supposing an oblate spheroid to revolve about any 
 diameter, to find the equation to the surface which envelopes 
 it In. every position. 
 
 Let x=az, y=0z be the equations to the diameter, 
 
 x + yo + nz? =m the equation to the spheroid; 
 
 substitute these values in the equation to surfaces of revolution, 
 namely, : 
 
 (y — bz) p—(w—az)q¢+ay—bx=0, 
 1 
 
 : re fh — bx. 1-5) =05 
 
 ind we find (ay — bx) ( a 
 
 ride cag Paka 3 aaa 
 
 ° @ wv ye re = 1 2 . . a 
 J ( are the equations to the directrix, 
 Cia ey , 
 
 and z+ax-+ by =c) 
 erty teary 
 
 the equations to the generating circle ; 
 
 eliminating x, y, 2, we find 
 (r?n® — m?) (a? +B) = (6 Jn 1 — of mi? — 7*)'s 
 therefore, restoring the values of 7 and c, the required equation is 
 jn (ce + y? + 2°) — mt (a? + 6°) 
 oH S(z+a0 + by) Jv —1 infin a Fay 2M 
 
 87. ‘To find the equation to a surface which envelopes 
 a series of surfaces described after a given law. 
 
 Let U=0O, be the equation to a surface containing, besides 
 other constants, a parameter a; then if we give a particular 
 value to a, the form and position of the surface in space will 
 be completely determined; and-if we give to it-all possible 
 consecutive values, we shall obtain an infinite number of cor- 
 responding surfaces, intersecting one another two and two. ‘The 
 surface formed by these successive intersections, and which 
 
 T 
 
146 
 
 consequently touches and envelopes each one of the first series 
 of surfaces, supposing them all to exist together, has been 
 named by Monge the Envelope; and the curve in which any 
 two consecutive surfaces intersect, the Characteristic of the 
 envelope, because it indicates the mode of its generation; thus 
 the characteristic of all surfaces generated by the intersections of 
 planes is a straight line, and that of surfaces generated by the 
 intersections of spheres, a circle. ‘To find the equation to the 
 envelope, we observe that if, after having given the parameter 
 a determined value a in U =O, we give it a new value 
 a +a, differing insensibly from the former, we obtain the 
 equation to a second surface differmg in form and_ position 
 insensibly from the first, and intersecting it in a series of points, 
 the co-ordinates of which satisfy the equations, 
 
 U re 
 =): Tt ee ang ey OF tiie Os Nh iia gsi 
 da da 
 
 av 
 da 
 
 therefore that the equations to the characteristic, or curve of 
 intersection of two consecutive surfaces, corresponding to the 
 
 : dU } 
 value a of the parameter, are U = 0, —— = 0; and since the 
 
 da 
 envelope is formed by the assemblage of the characteristics, 
 its equation will result from eliminating a between the same_ 
 
 or, if the surfaces be consecutive, U=0, = 0; it follows 
 
 equations. 
 
 88. Again, if in the equations to the characteristic, after 
 having given the parameter a particular value a, which deter- 
 mines the position of the characteristic im space, we give it 
 a new and insensibly increased value a + oa, the two resulting 
 equations will be those to a second characteristic differing in- 
 sensibly in form and position from the first, and intersecting 
 it, in general, in a finite number of points, the co-ordinates of 
 which satisfy the two sets of equations, 
 
 Ce 
 
 — =0; 
 da . 
 
 U =, 
 
1U dU @U 
 U +s Sa +++ =0, ini Pm 0p amy 
 
 which latter two equations, by virtue of the first, are equivalent 
 to 
 lat) AT eae 
 da * da? ad 
 Consequently the co-ordinates of the points in which two con- 
 secutive characteristics intersect, satisfy the three equations, 
 
 die! PU 
 
 > da > dd’ 
 
 = 0; 
 
 and from these equations we may either determine x, y, and z, 
 in terms of a, that 1s, the co-ordinates of the poimt in which 
 a given characteristic is intersected by the consecutive, or 
 eliminate a between them, and so find the two equations in 
 x, y, and z, to the curve formed by all the successive points 
 of intersection. ‘This curve will be touched by all the cha- 
 racteristics, and will be the edge of regression of the envelope ; 
 for the portions of the characteristics which are on contrary sides 
 of their points of contact with the curve under consideration, 
 will form two distinct sheets of the envelope, and these sheets 
 will touch one another in that curve and have it for their 
 common limit. 
 
 89. The following are examples of the two last articles. 
 
 Ex. 1. To find the equation to the envelope of all mght 
 cones of a constant volume, whose axes are in the same straight 
 line, and bases in the same plane. 
 
 The equation to the surface of any cone is 
 
 where a=altitude, 6=radius of base, and the origin is in the 
 center of the base. 
 
148 
 
 Let the given volume be that of a hemisphere, diameter 3c ; 
 
 Lr 
 
 9 +¥ ) . 
 b ; 
 
 mab (3c) % 270° a a ( aid 
 
 , ee Of OS ee ee 
 ‘2 y) ra 
 46° 
 
 differentiate with respect to the parameter 6, 
 
 978 818% a 
 
 ”, 0 = — —>z a+y’*, ob == /2xe+y’; 
 ont ae Sty; aM J 
 
 270° Q c* 
 
 See ee eee tee ees the equation required. 
 o@ ty) 3) ty 
 
 a 2b 270° 2 9c 
 
 Also ety = —, and z= —,; ( ~=) es 
 
 Sait y Be 46° 3 4°? 
 
 are the equations to the characteristic, that 1s, determine. the 
 radius, and position of the center of the circle, in which the cone 
 the radius of whose base = 0, is intersected by the consecutive 
 cone of the same volume. 
 
 Ex. 2. If the center of a sphere whose radius =a, describe 
 a curve in the plane of ry, to find the equation to the annular 
 surface which envelopes the sphere in every position, 
 
 Let a and § be the co-ordinates of the center of the sphere 
 in any position, and y = fx the equation to the curve which it 
 describes, therefore B = fa; 
 
 “.(@—a)+ (y—fa)’ + 2 =a? 
 
 is the equation to the surface, and by differentiating with respect 
 to a, 
 ie eee bey, — fa) fra — i 
 
 which are the two equations to the characteristic, whose position 
 and magnitude depend upon a; and if we eliminate a between 
 them, we find the equation to the envelope. The characteristic 
 in this case is manifestly a circle of constant radius whose center 
 is a point in the curve which forms the axis of the surface, 
 and whose plane is perpendicular to the tangent line at that 
 point, as is expressed by the above equations; hence its nature 
 
149 
 
 is entirely independent of the curve whose equation is y=/fu ; 
 it is also the curve of greatest inclination, for its tangent line 
 at every point is perpendicular to the intersection, with the plane 
 of xy, of the plane touching the surface at the same point, 
 therefore its equation is 
 
 dy salle (Art. 26. Cor. 1.) 
 a? eee 
 
 Its remaining equation is that to the envelope, since itis situated 
 thereon, which we obtain by observing that the normal always 
 meets the plane of ay in the curve which forms the axis of the 
 surface, and the length is equal to the radius of the generating 
 sphere ; 
 
 esata er Jitp +9 Se Mea bee: 
 or’ 2 (1 +p? + 9?) =a"; 
 
 which might also have been obtained by eliminating the arbitrary 
 function from the two equations to the characteristic ; hence the 
 integral of the equation z°(1 + p*+ 4°) =a’ is represented by 
 the system of equations to the characteristic. 
 
 Ex. 3. If the vertex of a right cone describe a given curve 
 in the plane of xy, and its axis be always perpendicular to that 
 plane, to find the equation to the surface which touches and 
 envelopes it in every position. 
 
 Let a and ( be the co-ordinates of the vertex, y=fvr the 
 equation to the curve which it describes, therefore BP = fa; 
 a = tangent of the angle which the side of the cone makes 
 with the plane of ry; 
 
 we (ta) +(y-fay = 
 
 &.| No 
 
 is the equation to the surface; differentiate twice successively 
 with respect to a; 
 
 “. (y—fa) fiat x—a=0, (y a ide 1—(f'a)’=0; 
 
 from which equations we may deduce as above the equation to 
 the envelope, and the equations to its characteristic, and edge of 
 regression. 
 
150 
 
 The tangent plane to the envelope at any point, is also 
 the tangent plane to the generating cone, and therefore makes 
 a constant angle (y) with the plane of xy; but 
 
 1 
 —— : 3 
 A) Veni + 9 
 therefore, since a = tan +, the differential equation to the surface is 
 . . . De ot 7 i 
 the integral of which 1s of course represented by the system of 
 equations to the characteristic. 
 
 cos yY = 
 
 Hence, if S denote the area of any portion of the surface, 
 
 d’s 
 nae = ~j/ltpt+@=secy, or S=A, sec y; 
 x 
 
 that is, any portion of the surface bears a constant ratio to 
 its projection on the plane of xy. The characteristic in 
 this case is evidently a side of the generating cone, and 
 therefore perpendicular to the intersection of the tangent plane 
 with the plane of xy; therefore its two equations are 
 
 2 Q Q dy 4 
 p+q =a, dx p’ 
 
 hence the curve of greatest inclination of all surfaces generated 
 in this manner is a straight line, inclined at a constant angle 
 to the plane of ry. 
 
 90. Tf the differential equation to the envelope be known, 
 the equations to the characteristic may be deduced from it. 
 
 Let f(a, y, 2, p, Y =O be the equation to the envelope, 
 then at every one of its points it touches one of the enveloped 
 surfaces; therefore at the point of contact wv, y, z, p,q have 
 the same values for the enveloped surface, that 1s, 
 
 FY % P Y= 0 
 is the equation to the enveloped surface. But of the quantities 
 2, Y, 2, Pp, J, the two latter only depend upon the value of.a, 
 
 which fixes the position of the particular enveloped surface 
 under consideration; and, as we have seen above, the equations 
 
151 
 
 lf ; 
 to the characteristic are f= 0, 2 = 0; therefore in this 
 a 
 case they are 
 dp dp. af dq pe ot 
 =a, i ee — 0, Or * s 
 DO in acne: da oC adam ade 
 
 suppose. But since the characteristic is situated on the en- 
 
 velope, if x, y, z be co-ordinates of any point invit, then 
 
 ee ay 
 
 —, —~, are connected by the equation 
 an 708 
 
 dz a 
 Sin Cian Bes 
 
 dr 
 
 therefore differentiating with respect to a, 
 _ dp , dqdy. 
 i dawmdadcs 
 
 hence, multiplying by P, and subtracting from the former 
 equation, we find 
 
 PL¢ nee - Q) =0; ise Pi nme 
 
 is the equation to the characteristic, a result which we observe 
 to be verified in all the examples given above. 
 
 91. Every partial differential equation of the first order 
 ST (Ps 97> %, Ys 2) = 0, may be considered as belonging to the 
 envelope of surfaces represented by I(x, y,2,a, pa) = 0; 
 that is, as the result of the elimination of a and ma between 
 
 dF 
 the equations £ = 0, iin O; for if we differentiate the 
 a 
 
 first with respect to wx and y separately, regarding a as 
 constant by virtue of the second, we obtain two equations, 
 one between p, x Yy, 2, a, a, and the other between 
 Qs Xv, Y, 2, a, a; between which and f’=O we may 
 eliminate a and ga, and there will reman f=0. This 
 makes the consideration of the characteristic of great import- 
 ance in the integration of partial differential equations, as 
 will be seen in the following examples. 
 
Ex. 1. To integrate the equation Pp + Qq -—-R=0, 
 where P, Q, R are known functions of x, y, z. 
 
 Let x, y, 2 be considered as co-ordinates of a point in 
 the characteristic; then, since 
 
 foe ’ 7 oe 3 ‘ eB Wis 4: 
 dp dq dx 
 Ah dz dy 
 and siice —-~= 4 — 
 dx Tae 
 1 
 i p+Qq=R, or P——R=0 
 
 dz 
 and consequently Me — ae 
 
 which are the three equations to the projections of the cha- 
 racteristic upon the co-ordinate planes. Suppose the two 
 first are integrable and give M=a, N=; im this state 
 these two integrals belong to all the characteristics that can 
 exist on all the possible envelopes comprised in the given 
 differential equation; but if the characteristics are to follow 
 one another according .to a certain law, a@ and $B are not 
 independent, but one is a function of the other; 
 
 “ M=a, N=qga; .«. N=o(M) 
 is the equation to the envelope. 
 In like manner, if we can integrate any two of the equa- 
 tions to the characteristic, either conjointly or separately, and 
 put one result equal to an arbitrary function of the other, 
 
 we obtain the required integral. But if none of the equations 
 to the characteristic 1s immediately integrable, differentiate the 
 
 : d 
 equation leap — Q = 0, (or any other of the equations to 
 iv 
 
 the characteristic, if more convenient) with respect. to a, and 
 the result is f(2, y, z, y, y”, 2)=03 these two equations 
 
 dz 
 combined with Fas — R=0, will enable us to eliminate z, 2, 
 dx 
 
153 
 
 and the result is P(2, y, y’, y')=0, a common differential 
 equation of the second order, the two imtegrals of which will 
 be in 2, y, y',a; 2, ¥, ¥, 93; a and PB being the arbitrary 
 constants ; from which, eliminating y’ by means of the equation 
 Py' — Q=0, we obtain the two integral equations to the cha- 
 racteristic M=a, N=da, and therefore N=@M as before. 
 Hence the integration of every partial differential equation 
 of the first order, which is linear, depends only upon that of 
 a common differential equation of the second order, involving 
 two variables, one of which is the independent variable. 
 
 Ex. 2. Let one of the quantities P, Q, R vanish, R 
 for instance, the equation will be Pp+Qq=0; 
 ply dz 
 
 —Q=0; '-— =Owwist nee 
 or dx 
 
 that is, for the same characteristic, z is constant, and there- 
 fore the characteristic is in a plane perpendicular to the axis 
 of z; substitute, if necessary, for z im the first, and integrate, 
 the result is F(a, 2 Ys Uy ga)=0, ga being the constant 
 introduced by integration; and therefore the complete integral 
 
 is F(z, y, 2, @z) = 
 
 Ex. 8. To find the equation to the surface which cuts 
 at right angles all surfaces comprised in the same total dif- 
 ferential equation. 
 
 Let dz=Pdzr+ Qdy be the equation to the cut surfaces, 
 
 dz=pdx-+qdy the equation to the cutting surface; 
 
 then the equations to the tangent planes applied to them at 
 one of their common points (vy z), are 
 
 Z—z=P'—2)+Qy—y), 2-2=pa'—a)+q(y -y); 
 and since the planes are at right angles, therefore 
 Pp +Qq+1=0 
 
 is the partial differential equation expressing the condition of 
 
 U 
 
154 
 
 the question; the integration of which depends on those of 
 any two of the equations 
 
 “d 
 pHa qeojl P Bigmco, Qa hi =o. 
 dx x dy 
 
 Suppose the cut surfaces to be a series of ellipsoids, re- 
 sulting from giving D all possible values in the equation 
 
 ax + by +c2°=D, 
 
 a, 6, ¢ being the same in all; 
 
 ax b 
 ° P= A ; Q= Poa ede 
 Cz cz 
 and the equations become 
 
 axdz by dz 
 
 BE pee == 3 “J —1=0; 
 
 czar cz dy 
 
 2° 2° ps z 
 
 se — =a, wectees JO : ~= 9 (5) 
 
 is the required equation. 
 
 If a=b=c, the cut surfaces become concentric spheres, 
 ; ‘ z Z ji 
 and the equation to the cutting surface = (2): which 
 
 represents a conical surface, whose vertex is in the origin. 
 
 Ex. 4. Let z—pr-—qy=nJf/etry +2, the geome- 
 
 trical signification of which is, that the portion of the axis of z 
 intercepted between the origin and tangent plane, bears a con- 
 stant ratio to the distance of the point of contact from the 
 origin. 
 
 In this case, 
 
 dz _z-nJfa'ty +2 dy 
 dx x / dx 
 
 adie 
 i" 8 
 x 
 
155 
 
 the latter gives y=ax, and shews that the characteristic lies in 
 a plane through the axis of z; 
 
 HSS a ma > § 
 dx x’ 
 
 and therefore 2”~'(z + e/ v+y+2y)=g (*) ; 
 
 92. In some cases the problem furnishes two relations 
 between p and g; we then obtain a total differential equation, 
 and the resulting integral contains no arbitrary function. 
 
 Ex. 1. To find the nature of a surface which has the 
 property, that the normal at any point passes through the 
 center of gravity of the tnangle whose sides are the co- 
 ordinates x and y of that point. 
 
 By Art. 28. the co-ordinates of the foot of the normal 
 are r-+pz, y+q2;5 
 
 BG ihe. Soe ae me UNE ae 
 Sed AT ed y die TF ia so. p= 
 
 x 2y 
 “. dz=pdxt+qdy= ——dr—-—dy; 
 “Ten Lod, 3z 5 ale 
 therefore 3274+2y°+a2°=C is the required equation. 
 
 Ex. 2. To find the equation to a surface which at once 
 belongs to conical surfaces, and to surfaces of revolution 
 about the axis of z. 
 
156 
 
 In this case p and gq must satisfy the equations 
 py—qr=0, plr—a)t+q(y—b)=z—-c; 
 hence obtaining p and q, and substituting them in 
 dz = pdx + qdy, 
 we find for the equation to the surface 
 dz * adx-+ ydy . 
 z—c a(t—a) + y(y—b)’ 
 
 which does not admit of a single relation between the three 
 variables for its integral; unless a=6=0, when we find 
 
 g—02=C,f/2+y', 
 
 the equation to a right cone, as we might have foreseen. 
 
 93. 'To determine the envelope, when the equation to 
 the series of surfaces contains two mdependent parameters. 
 
 Suppose the equation U=0 to contain two parameters 
 a and 3 independent of one another; then the equation to 
 the surface which differs insensibly from that represented by 
 U=0, will be 
 
 U +5 as = 8a a 7a°P +: 
 and therefore the co-ordinates of the points in which these 
 surfaces intersect must satisfy their two equations, that is, 
 they must satisfy 
 
 dU dU 
 
 U = 0, = ay é Eig) ha 4 
 ati B B + 0, 
 or, 1f the surfaces be consecutive, 
 dU dU 
 U=0, — pan Sos 
 da | dp 2 
 
 where m = the limit of 3,3 but, since m may have any 
 a 
 
157 
 
 value, this latter equation resolves itself into 
 
 we ao, 
 dae? gO Ray 
 
 which, together with U=0, are the three equations for de- 
 termining the curve of intersection of two consecutive surfaces; 
 from whence we may obtain its two equations, involving only 
 one of the parameters, and finally, by eliminating that parameter, 
 we may obtain the equation to the surface generated by the 
 perpetual intersections of the first series of surfaces. 
 
 Ex. 1. To find the equation to the surface touched by 
 a series of planes so drawn, that the product of the per- 
 pendiculars let fall upon any one from two fixed points is 
 invariable. 
 
 Let the line joming the two points be the axis of xz, 
 2c its length, and the origin in its middle point, z= Ax-+ By 
 + C, the equation to one of the planes; then the lengths of the 
 perpendiculars are 
 
 C+ Ac C—Ac 
 ROB PTEE 4 OTL REE ah 
 CA? 2 
 4 St =b’ suppose, or C? =0?(1 +A’? + B’) + A’c’. 
 
 Differentiate the equation to the plane, successively with respect 
 to A and B, considering C as a function of A and B by 
 virtue of the above equation ; 
 
 dC A 
 Oma tart EEO, or r= —(P LANE, 
 
 dC fad 3 Ade 
 ORY Smee Oa or y= —D°— 
 
 CH 
 A® : BAB*!?_ 2 PHY Ie BtE CB 
 Pe 3 b? Dees eeael = OO a 
 = { toe C +C C C 
 2 g 2 Q2 2 2 2 Q 2 
 x plea, ONE Reade a, 
 
 Bye ale Ce 
 
. 
 
 158 
 
 the equation to a prolate spheroid, whose axis is the line 
 joining the two given points. ; 
 
 m7 m 
 
 x ie ake i 
 Ex.2. Let the equation to a surface bea + ie + ee = 
 
 to find the equation to the surface which it touches in all the 
 positions it can assume, subject to the condition a” + 6"-+-c’=k", 
 a constant quantity. 
 
 Differentiate the equation to the surface successively with 
 respect to a and 0, regarding c¢ as a function of a and 6b, by 
 virtue of the equation of condition; 
 
 ym 2” de yo a 2m qv} 4 ym 2” gr 
 ee = Oo ei oe HE Oe or—_=>—_— .—_:: 
 e ; ° — 3 
 git} c’t1da qu t} cmt c” | a” ce” cn? 
 y™ 2” d C y y™ bh” -1 cre gy” b” 
 Ne I BOR LI cere cones or ae e WiPr eas bam aa oe ORs ae 
 fmtt cmt d b bt 1 cut ct 1 hb” m c” 
 y™ gm gm gt bn wh cm gm en Pua cmtn 
 0 SG +555. c” Te ale + jn men? 
 mn mn mn 
 Z\min cl Aiea’ L\ m+n m+n 
 or () mes similarly G) = a, mm C4 =o 
 mn mn mn mn 
 
 2 germ pt gm tn 4 ymtn = Amt”, the equation required. 
 
 If m=n=1, the equation becomes 4/z + a/a + Jy = J ky 
 
 and agrees with Prob. 8. Art. 29; if m=n=2, it is e+4+y=h, 
 which shews that if a series of surfaces of the second'order have 
 the sum of the squares of their axes constant, they are all 
 
 touched by a plane. 
 
 Similarly, if the equation of condition be abc = k’, it may 
 be shewn that the equation to the enveloping surface is 
 
 cues. 
 
159 
 
 Rs) 
 and if m= 1, we find xyz = = for the equation to the surface 
 
 touched by a series of planes which form with the co-ordinate 
 planes a pyramid of constant volume. This surface has also the 
 property that the tangent plane at any point a, y, z, forms with the 
 co-ordinate planes a pyramid of smaller volume than any other 
 plane drawn through the point of contact; and its equation is 
 
 | 
 ri) | RL 
 
 858 
 
 A Ye), 
 it may easily be shewn that this is the equation to the tangent 
 plane of any surface whatever, which forms with the co-ordinate 
 planes the least possible pyramid, x, y, z, being the co-ordinates 
 of the point of contact. 
 
 Ex. 3. To find the equation to the surface touched by a 
 series of planes which retrench from a given right cone, an 
 oblique cone, such that its volume is constant, or such that the 
 transverse axis of its elliptic base is constant. 
 
 If the volume of the oblique cone be constant, then by a 
 property of conic sections the transverse axis of the elliptic base 
 is also constant, suppose it .°. = 2c, and let 3 = semi-vertical 
 angle of the cone; then it will be found that the equation to 
 the touched surface is 
 
 z= cot’ B(«? + y? +c’), 
 the origin being at the vertex; which belongs to an hyperboloid 
 of revolution about the axis of the cone, and to which the sur- 
 face of the cone is an asymptote, the + axes of the generating 
 
 hyperbola being c and ccot 9; also the point of contact is the 
 center of the elliptic section. 
 
 In like manner if a series of planes retrench from a parabo- 
 loid of revolution a segment of constant volume, the surface touched 
 by them is a similar and equal paraboloid about the same axis. 
 The converse of Prob. v. Art, 48. will furnish other examples. 
 
 94. To find the general equation to developable surfaces. 
 
 The characteristic: property of these surfaces is, that they 
 may be extended upon a plane, so as to coincide with it in 
 
166 
 
 every one of their points, without tearing or rumpling. ‘They are 
 generated by the consecutive intersections of a series of planes 
 drawn after a given law; of this generation we have seen a 
 remarkable instance in the surface formed by the intersections 
 of the normal planes of a curve of double curvature, (Art. 63). 
 The law of succession of the planes which generate a develop- 
 able surface, requires that two of the three arbitrary constants 
 which enter into the equation to a plane should be functions 
 of the third, or, which is the same thing, that they should all be 
 functions of the same parameter a; let ..z=xrga +yfat Wa 
 be the equation to the generating plane, in one of its positions 
 depending upon the parameter a; then the equation to the 
 plane which differs insensibly from it in position, will be 
 
 z=u(patda.coa...)+y (fat f'a.oa...)+Watw'a.oa..., 
 
 and the co-ordinates of the points in which they intersect must 
 satisfy these two equations; that is, they must satisfy 
 
 z=xrpatyfatwa, O=r(pat---)+y(fat---)tWat--:; 
 
 or, if the planes be consecutive, the equations to their line of 
 intersection are, 
 
 =rpatyfatpa, O=aPaty fiat Yajee(I): 
 
 and the general equation to developable surfaces will result 
 from eliminating a between these equations. 
 
 95. The characteristic in this case is a straight line of 
 which the equations are (1); and if we differentiate again with — 
 respect to a, which gives 
 
 _— ropa i y fa ai wa, 
 
 the three equations represent the curve formed by the inter- 
 sections of the characteristics, or the edge of regression of the 
 developable surface, by the tangent line of which the surface 
 might be considered as generated. Since two consecutive cha- 
 racteristics are in the same plane, the envelope may be con- 
 sidered as made up of plane elements of indefinite length, but 
 infinitely narrow, which cut one another consecutively in straight 
 
161 
 
 lines.. If now the first of these elements be turned about its 
 line of intersection with the second, till they are in the same 
 plane; and then the system formed by these two be tumed 
 about the line of intersection of the second and third till they 
 are in the same plane, and if this operation be continued 
 through all the plane elements, they will all thus be brought 
 into one plane, and the given surface will be developed without 
 rumpling or tearing. 
 
 Ex. If a series of planes, passing through a fixed point in 
 the axis of z, have their traces on the plane of ary all of the 
 same length; to find the equation to the surface formed by their 
 intersections. 
 
 If c=distance of the fixed point from the origin, a=length 
 of the trace on the plane of wy, and a=the angle which it 
 makes with the axis of x, the equation to the corresponding 
 plane is 
 
 w A Yy ya 
 + et ease ~ 
 c 
 
 acosa  asina 
 differentiate with respect to a, 
 
 x sin y cosa 
 
 me ee 
 acos a a@ sin @ 
 
 == (ey omtan abe (st 
 x 
 
 therefore, substituting in the equation to the plane which may 
 be written 
 
 DL o oO 2 Pe 
 -,/1+tana e's) l1+cotta + nies 1, we find 
 a a 
 
 Od 
 
 eae z . . 
 +y>)* += = 1, the required equation. 
 c 
 
 2 
 3 
 
 ] 
 or = (2 
 
 The equations to the characteristic in this case are 
 
 x Zz Oe ' 
 a ee Te Se tan a. 
 acoas @ € A 
 
162 
 
 96. To find the differential equation to developable sur- 
 faces. ' 
 
 Since the surface is made up of plane elements of indefinite 
 length, a plane applied to it at any point touches it in a series 
 of points, the projection of which on any of the co-ordinate 
 planes is a straight line. Hence, since the equation to the 
 tangent plane is 
 
 z= px t+ gy +2 — pxr— qy; 
 
 the quantities p, g, and z—pxr—qy remain unaltered for all 
 values of x, y and z, subject to the condition y=max-+n3 and 
 therefore the increments of p and gq, and consequently of 
 z—px—gqy, corresponding to any, and therefore to indefinitely 
 small, increments A and k of 2 and y, subject to the condition 
 k==mh, are evanescent; that 1s, 
 
 rh + sk + &c. = 0, or ultimately 7 + ms = 0, 
 EY (Mer pA A Rios ae oad OF ES et va 8) 4 
 
 Pee A : 
 27 = BEN rt —s° = 0, 1s the equation to the surface ; 
 s 
 dy ih dy r 
 and —~— + ~ =0, or ~— + = 
 PES es mi hs t 
 
 together with rt —s°=0, are the equations to the characteristic. 
 
 Since the three quantities p, g, 2-px—qy are constant 
 together, and variable together, it follows that the partial dif- 
 ferential equation of the first order F'(p, q, z — px — gy) =0, 
 belongs in general to a developable surface. 
 
 Cor. The differential equation to developable surfaces 
 may also be obtained by eliminating the arbitrary functions from 
 the equations to the generating line, which are 
 
 z=urhatyfatwWa O=rha+t+ yfatwa; 
 
 for if we differentiate the first separately with respect to x and y, 
 regarding a as constant by virtue of the second, we find p= qa, 
 
163 
 
 q=fa; and since p and q are functions of the samefquantity, 
 they are also functions of one another, or p=7q; 
 
 ¢ 
 d tC } a 
 a = mq orr =7W7q.5, . i 
 ; \ 
 dp , dq ; 
 ee de = t 
 yy Gi ra A ah a 
 ere . 
 fies ABR Obs roe ks AIO : 
 My t 
 
 By making r¢—s*=0, in the expressions for the radii of cur- 
 vature, one of them becomes infinite ; hence developable surfaces 
 have one of their curvatures evanescent. 
 
 97. To find the equation to the developable surface which 
 touches at the same time two given curve surfaces. 
 
 Let the equations to the two given surfaces be 
 FAG ae OE (25.42) 0: 
 or dz=pdr+qdy, dz=p,dr+qdy; 
 also, let z—-a=rgdat+tywa 
 
 be the equation to the generating plane which touches the first 
 surface at a point (v’y'z’), and the second at a point (v”y” 2”). 
 Therefore for the first contact, 
 
 z—a=avhatywpe, p= ha, q=wWa; 
 
 and for the second contact, 
 
 (—a=s"paty' va p=b% Gave 
 
 The first three equations, being combined with F (2, y’, 2’) =0, 
 will serve to eliminate a’, y’, z', and give a relation between 
 a, pa, a; and the second three equations, being combined 
 with F(x’, y”, 2”) =0, will give a second relation between 
 a, da, Wa; and consequently the forms of @a, Wa, are deter- 
 mined; whence the equation to the generating plane is known, 
 
 and therefore the equation to the required developable surface 
 by Art. 94. 
 
 / 
 
 i) 
 
164 
 
 If we suppose an opaque body to be illuminated by a lumi- 
 nous one, the surfaces which circumscribe the umbra and pen- 
 umbra occasioned by the interposition of the opaque body, are 
 two sheets of the developable surface which touches both the 
 bodies; and its lines of contact with the surfaces of the bodies 
 are, one, the curve on the opaque body which separates the 
 illumined and obscure parts, and the other the curve on the 
 Juminous body which separates the illuminating part, from that 
 which can send no rays to the other body. Consequently the 
 question treated above contains the general problem of umbras 
 and penumbras. In like manner we may find the equation to 
 the’developable surface which passes through two given curves 
 of double curvature. 
 
 98. A curve being traced on a developable surface, to find 
 its nature when the surface is made plane; and conversely, 
 a curve being traced upon a plane, to find its nature when the 
 plane is applied to a given surface. 
 
 Cast 1. Let the surface be cylindrical, x, y, z, co- 
 ordinates of any poit im a curve traced upon it; then the 
 equations to the generating line which passes through that point 
 are 
 
 2,-2=a(e—2), y—y=b(4,—2); 
 
 and the equations to the tangent line of the curve are 
 min rad a a ¥ — 1 / 
 2,~-2= 240 — 2) yry=y @—2), 
 : dz .dy A 
 rea? ; : 
 where z, y, denote —-, —+. Let 2 = the angle between these 
 | dn” dx 
 two lines, y =angle of inclination of the generating line to the 
 axis of x, s=length of the curve ; 
 
 pies by +] 
 
 Now when the cylinder is developed, the generating lines pre- 
 serve their parallelism, and may therefore be taken for the ordi- 
 nates; let a line perpendicular to them be the axis of the 
 abscisse, a aud (3 the co-ordinates of the point in question, and 
 ao =length of the plane curve, 
 
 . COsi= 
 
 f RR Me 
 =cosry (az +by TA ae 
 
165 
 
 dp 
 dao’ 
 
 slidp ‘ ; dx 
 then ER Va “. cosy (az + by’ + I). as 
 
 iy ds wr dp (a2h byl 1) 
 peters ety . hi Powe v4 #) . f 
 u a oe = cosy (az yt 
 
 and (S) = (E): » or Ity?+2"=(1 mn Gas) (3 *), 
 
 between which two equations, if x be eliminated, there will 
 remain a differential equation between 6 and a, which will be 
 that to the required plane curve. 
 
 But if the Jatter equation be given, that is, B=fa, we may 
 eliminate a between the two above equations, and there will 
 remain a differential equation between x, y, and z, which with 
 the given equation to the surface, will be the equations to 
 the required curve of double curvature. 
 
 Cor. 1. If the curve on a plane be a straight line, £2 
 is constant 5 
 
 , , d. ’ 
 es COSY NER “OY oe Die CS 7 
 $ 
 
 is the equation for the determining its nature when the plane is 
 applied to a cylinder; the curve is. the helix, and we have 
 for its length, by integration, 
 
 cos ty (az + by +x) = (s+ C) cos 2. 
 
 It is manifestly the shortest line which can join two points on 
 the surface of the cylinder. 
 
 If the generating line of the cylinder be perpendicular to 
 the plane of yz, 
 dx 
 
 a=b=0, cosy=}); cin ta COM 
 ds 
 
 or the tangents of the helix make a constant angle with a plane 
 perpendicular to the axis of the cylinder. 
 
 Cor. 2. If the cylinder on which a curve is traced be per- 
 pendicular to the plane of xy, and we take the arcs of the 
 
166 
 
 curve in which it meets that plane for the abscisse, and the 
 distance of the two curves for the ordinates, then 
 
 as V/ mre A and S=2z; 
 
 dx 
 
 if therefore, y=@ax be the equation to the cylinder, we can 
 either determine a relation between ( and a, from knowing the 
 other equation to the curve of double curvature z=f2; or con- 
 versely, if the equation to the plane curve B=wWa, be given, 
 we can determine z=fx, the equation to the curve of double 
 curvature. 
 
 Ex. The nature of a curve traced on the surface of a 
 right cylinder is such, that its projection on a plane drawn 
 through the axis, produces the circle which forms the base 
 of the cylinder; to find its equation when the cylinder is 
 made plane. 
 
 Oo 9 Q Q 9 . 
 Here y=a°—2r, 2° =a —2, are the equations to the 
 curve of double curvature ; 
 
 da Wi dy\* a a 
 St mngeh st — a +(@ ) aA 
 ( 
 
 dx /a—x? B° 
 since” i= i2 = Nae ee re 
 
 and — = — 
 
 da a “An 
 
 er aie 2? alee Sc 
 
 ao correction being added, since a=O when 3 =a; 
 
 ee B =a cos — 
 a 
 
 is the required equation, as we might have foreseen. 
 
 99. Case 2. Let the developable surface be a cone, 
 a, b, c co-ordinates of its vertex, xv, y, z co-ordinates of a 
 point in a curve traced upon it, r the distance of that point 
 from the vertex. Now, when the cone is developed, the 
 
167 
 
 generating lines become radti vectores, and if @ be the 
 angle which r makes with some fixed radius vector, since 
 the length of any element of the curve remains unaltered, 
 
 fan, ot YP YO: 
 
 and 9° =(«—a) + (y— bY + (2-0); | 
 
 from which, together with the two equations to the curve, 
 we may elimimate 2, y, z, and there will remain a relation 
 between r aud 6, the equation to the plane curve when 
 the surface is developed; or if r= f 0, and the equation 
 to the cone be given, we may eliminate 7 and 0, and so 
 arrive at the remaining equation to the curve of double 
 curvature, resulting from applying the plane, on which a given 
 curve 1s traced, to a given conical surface. 
 
 Cor. 1. All curves which make a constant angle with 
 
 the generating line of the cone, are characterized by the 
 al de 
 equation Tete B EN and become equiangular spirals when 
 ds 
 the surface is developed. 
 
 Cor. 2. All curves which become straight lines when 
 the cone is developed, and which, therefore, are the shortest 
 lines that can connect two points on its surface, are cha- 
 racterized by the equation 7." 
 
 ds ve ee an 
 
 dx dx : 
 taking the vertex for the origin, a being the length of the per- 
 pendicular let fall upon the line from that point, 
 
 d 
 
 dx 
 
 or finally, by developing, 
 (cy’—y) +(x7— 2) +(yy'— 27% =a 4-7” -|- ae); 
 
 rae 
 when y = —, &e. 
 . be 
 
168 
 
 When the surface is a right cone ona circular base, the problem 
 admits of a very simple solution, as will be seen in the following 
 article. 
 
 100. A given curve being traced on the surface of a right 
 cone whose base is a circle, to determine its equation when the 
 surface 1s developed. 
 
 Let P (fig. 53.) be a point of the curve situated-on the 
 generating line CR, Q its projection on a plane perpendicular to 
 the axis; then our object is to find a relation between CP =r, 
 and the angle which CP makes with CA, when the cone is 
 developed ; Vall this angle 0, and let 
 
 CQ =7', QCr=0,m= cosec ACD; 
 
 now, when the cone is developed, G1 is subtended by a circular 
 arc, length AR; 
 
 * "PPA. = Oe Ls or, 0. == mee 
 also CP..sin PCD=CQ, or, r= = 
 if, therefore, f(7”, 0’) =0 be the given equation to the locus of Q, 
 
 1 
 ee m0) =() is the equation to the curve, when the surface 
 m 
 
 of the cone ts develop 
 
 Lida Pi... : 
 In like manner, if a curve be traced in a circular sector, we 
 may determine its equations when the sector is formed into a 
 
 right cone; let a = Z of the sector; therefore, sine 4 vertical 
 
 anele of cone vs big suppose; also let f(r,0)=0 be the 
 air peat 
 
 equation to the plane curve, where @ is measured from the 
 bounding radius of the sector which afterwards becomes the ge- 
 nerating line of the cone situated in the plane of xz, that is, 
 AC; then if 7”, 6’, be polar co-ordinates of Q, 
 
 r , , ‘ 6! 
 = aU it a —) sate} 
 it 
 
 m 
 
 is the equation to the locus of Q. 
 
169 
 
 Ix. 1. Ef a semi-circle be described on the bounding ra- 
 dius of a quadrant of a circle, to find the equation to its projec- 
 tion upon a plane perpendicular to the axis, when the quadrant 
 is formed into a cone, 
 
 T . eee 
 Here m =2a + She =4, also the equation to the sem!-circle 
 
 is * = acos@, therefore the equation to its projection is 
 ren aga) eee 
 
 7 = =~ COS; 
 4, 4, 
 
 the form of this curve is seen (fig. 54); it contains the projec- 
 tions not only of the semi-circles described on the bounding 
 radii of the quadrant, but also of those on the bisecting radius, 
 as seen (fig. 55). 
 
 Again, to determine the equation to the Pee of the 
 chord of the quadrant; its equation 1s 7 ae sec #, therefore 
 the equation to its projection is 
 
 bial © 
 a ae 4 3 
 
 the form of this curve 1s easily traced; the lines on the quadrant 
 are seen (fig. 55). : 
 
 Ex. 2. A right cone being intersected by a sphere whose 
 center is in its surface, to determine the equation to the curve 
 of intersection when the cone is developed. 
 
 Let B be the center of the sphere (fig.53) BC = c, a = 
 its radius, a = 4 vertical 4 of the cone, then’ 
 
 (csna— 2) +(ccosa—x)+y' = 
 
 is the equation to the sphere, and 2° tan’a = x -+- y° is the 
 equation to the cone; therefore, the equation to the projection 
 of their intersection on the plane of wy, is 
 
 x 
 
170 
 
 9 Q Q 2 , cos a 3 o P 
 Ce +7 )(l'+ cot a)—2cxsma— 2¢ fity =a, 
 
 sIn a 
 4 a . . , U he / 
 and its polar equation is, making «=7 cos@, y=r sin@, 
 ’ soos i} 2 2g 
 ctr? cosec’ a — 2cr’ sin acos 0’—@cr cosecacos a=a. 
 / / 
 But m=coseca, r=? coseca, 06 = m0; 
 ’ 3 ? 
 
 therefore, when the cone is developed, the equation to the curve 
 of intersection 1s 
 
 ce? + 7? — Qer(sin? a cos mO + cos? a) — a® =0; 
 ; A iy ee) 2 : 
 or, — —; (cosm@ + m —1)+c¢(—n’) = 0, ifa=nc. 
 m 
 
 2 
 If we make 90 = —~ + dp, cosmO = cos (27 +m) = cosm@, 
 m 
 
 and the form of the equation is not changed ; hence in tracing 
 ree Qa 
 
 the curve, it is only necessary to take @ from O to —., after 
 m 
 
 which the same form recurs. 
 
 Again, if @= a + d,cosm@ = cos(mr + mo) = — cosmo, 
 m 
 hence that part of the curve which is produced by taking @ from 
 
 0 to —, is divided into two similar and equal portions by the 
 m 
 
 ; 7 
 radius corresponding to —, so that in fact we need only take 
 m 
 
 6 from O to ify 
 : m 
 
 The same value of 6 gives two values of 7, and by solving 
 the equation we find 
 
 1 Pie Luge Ne eee 
 r=ec (1—srers m0) +c V0- —; vers m@) —(1—n’); 
 7 m m 
 
 therefore versm@ cannot exceed m*(1— x / 1—n’). 
 
171 
 
 If this quantity be >2, of course all values of 0 are admis- 
 sible, but if it be <2, @ can only be taken from 0 to that 
 value which makes 
 
 versmO@ = m*(1 — Ail —n), 
 Hence there arise two cases. 
 I, When m?(1 — SI — n°) >2; 
 
 in this case, the two branches of the curve cannot intersect ; for, 
 in order that the two values of 7 may be equal, we must have 
 
 vers m@ = m?(1—a/1—n?) >2. 
 
 By differentiating the equation to the curve, we find 
 
 d 5 
 (- —c+ vo vers nO )<— ut ead sinm@0=0; 
 m*” d@ 
 
 m 
 
 let @ = the Z at which the radius vector cuts the curve, then 
 
 1 Ci. 
 “. 7T—C (: — —; vers m0) + —sinm@ tang = O (1). 
 m* m 
 
 Tr wT AES go Set Q 
 Hence, when 92=0, —, —, tan @ 1s infinite, or the radius 
 m m 
 vector cuts the curve at right angles and is a maximum or mi- 
 
 Tv 
 nimum, its corresponding values being, when @=0, or an 
 r— 8r+tce(1—n) =0, or r=c(ltn); 
 
 1% 
 
 vn) 9 
 and when 0 = —, r* — 2cr (1 _ —) +c (1—n’) =0, 
 
 m 
 2\? y 
 ( ~—) = (li 7t: ) 
 me 
 
 2, 
 or r=c(1-—) +6 
 m 
 
 Sed. 
 Hence the form of the curve from 0=0, to 0=—, is that 
 m 
 
 in. (fig. 56), where 
 
172 
 
 BB = 2c iiss — —) —(1—n’), AA’ =2en, 
 
 and the two branches are so connected that for any radius vector 
 CP.CP’=CA.CA’. Ifm bean integer, the whole curve will 
 
 consist of m such portions, the spiral angle having increased 
 
 ; : p ; 
 from O to 27r; if m be a fraction supe the above form will recur 
 b] g > 
 
 p times, and the radius vector will have described an angle 
 = 2q7. BB will vanish if m°(1—,/1—n*)=2, in which 
 
 case the equation to the curve becomes 
 
 : l ; 2° 
 r= Qer (1 —-5 vers m0 ) +e (1 - -;) =0, 
 m m 
 
 and the two portions touch one another. 
 
 Suppose m = 3, n = -—3; therefore the equation becomes, 
 
 < 
 
 cole 
 
 + cos 30) + 5c” =0, 
 
 where c’ =—, and its form is seen (fig. 57). 
 
 If m = 3 and the two branches are in contact, the equation 1s 
 
 r — Qc'r (8 + cos30) + 49c¢? = 0, making ¢ = 
 
 Cts 
 
 II. When m’* (1 —/1— n°) <2, 
 
 In this case when 90=0, we find as before r=c (1+), 
 and when @= 6’, where vers m6! =m? (1—,y 1—n"), we find 
 
 1 ee es 
 pe c( _ —, versm’ ) = crf) —n’, 
 m 
 
 that is, the two values of 7 are equal, and the radius vector 
 becomes a tangent to the curve, therefore tang =0,.as appears 
 
173 
 
 ; on ’ 
 from equation (1); hence from @=0 to 6 =——,, the form of the 
 m 
 curve is that in (fig. 58), and will recur m times, if m be an 
 integer, or p times if m= ie fraction ; in the one case there 
 
 q. 
 will be (m), in the other (p) ovals. 
 
 ' 3 32 tale 
 For instance let m = 3° n= es then the equation is 
 
 : ; 30 - Set asace- sal 
 r —2cr (4cos a 5) + 49c~ = 0, making c =: 
 
 the form of which is seen in (fig. 59), where 2 ACP = 40°, 
 and CP = 7c. 
 
 Cor. If n=1, or the vertex of the cone be in the surface 
 of the sphere, the equation becomes 
 
 ' ] 
 r= 2 (1 — —, vers m0), 
 m 
 
 if m* > 2, r can never =O; hence the form is that of figure (56), 
 without the interior portion. 
 
 If m” <2, that is, if m be between 2 and 1, then from 
 
 > 
 
 e , ; : 
 8=0, to eal the form 1s that of (fig. 60), and if m = P 
 m q 
 
 it will recur (p) times. 
 
 + 5 
 Thus if m = 209% the equation to the curve be 
 
 ~(7+9 by 
 <a EGO Sie 
 ies renee 
 the form is that of (fig. 61). 
 101. A curve of double curvature traced on a surface of 
 
 revolution always cuts the generating curve at the same angle, 
 to find its equations. 
 
174 
 
 Let BP (fig. 49) be the curve, BQ its projection on the 
 plane of the base, AD a section of the surface through its axis ; 
 CO =n thO=2. DE =s, IBCQ te sfromypsa ipgintiinetie 
 curve near to P, draw pr perpendicular to AP, and let a=the 
 constant angle at which BP cuts the meridian, then tana 
 
 d@ EN oe ae UN 
 = ie ee NN =F cotta( =), 
 
 = limit of 
 
 pr 
 
 Pr 
 
 ee AON 
 
 or 1 + (fr) =r’ cot’a (=) A 
 2 
 
 if z=fr be the equation to the generating curve; therefore the 
 equation to the projection of the required curve is 
 
 dr r cota 
 
 or, if p=the perpendicular on the tangent from C, the equation 
 
 fe / t+ (far)? 
 Be sec’a+(f'r) 
 
 co 
 
 1S 
 
 Ex. 1. Let the surface be a sphere; 
 
 dz ? 
 ° oes 2 2 He 
 ee oo = DA eH fe 9 ah Ta55 > 
 dr J/a—r 
 oO 
 " arr? 
 oe Pp — 2 
 a’ cosec’a—r* cot’a 
 
 the equation to the hyperbolic spiral, as we have already seen 
 
 (Art. 79). 
 Ex. 2. If the surface be a right cone, and the origin in the 
 center of its base, 
 dz r 
 
 (c—z)=rcotP — =- cot B; p= ? 
 ely i / 1+ cot’a sin? 
 
 the equation to the equiangular spiral, as we might have fore- 
 seen. 
 
175 
 
 Ex. 3. Let the surface be an oblate spheroid, 
 
 b d br 
 then zg=-r/a—r? Ce a 
 a 
 
 3 ao 33 3 
 dr a JS a—r 
 d@ tana 4 J a®— er” 
 riokds 4, r aQnsriy 
 To integrate this, make 
 2 2.2 g 
 65 8 4 7 Ee Tee 
 ee Qo” —_— 6} 3 ee / I aS a 2Z 2? 
 a-—r v —e 
 
 1 dr v v o dr ] e° 
 carting and ETRE re! 8 Terme at Sr Sin eS Pacey eer omar 
 rdo.v—-1l-wv-é rdv v—-1l we’ 
 dé v 
 therefore since — = tana-, 
 dr r 
 
 dé v dr ; ( 1 e ) 
 — =tana— — = tana = asicor ar or ad IC 
 dv r dv rie ei a ie hs FF A: 
 
 therefore integrating, 
 
 Q+. ’ (i t—]1 —) 
 c=—tana og og 
 o ” e+] ‘1 votes’ 
 
 is the equation to the required projection. 
 
 102. ‘To draw the shortest line between two given points 
 on a surface of revolution. 
 
 Let PC (fig. 62.) be the axis of the surface, which take for 
 
 the axis of z; 4B a portion of the shortest line = s, 2 DCN=0, 
 CN=T, DN=z; 
 
 Ne ee ga0" dz , de 
 2 oe dr® EAS ate dy” tS MES 
 
 because from the equation to the generating curve z=f7'; 
 
 $= fda Jfitap +f 7 SyVeds, 
 
176 
 
 d@ 
 
 putting r=2, O=y, an Re in order to adapt the expression 
 dr 
 
 to the usual formule of the calculus of variations. Hence, 
 
 dV | ‘i 
 dy To EF ap + (fae 
 
 dP Re: 
 but N — Pie + &Xc. = O, for a maximum or minimum ; 
 i 
 
 therefore, in this case, P= a constant, 
 
 xp rd@ 
 or aT aa = = C3 
 Jibar tio ds 
 . ; i 
 “h a0 = , or sin PBAcc ON? 
 Ss uf? ' if 
 
 that is, the shortest line always cuts the meridian at an Z whose 
 sine varies inversely as the distance of the point of intersection 
 from the axis. 
 
 Hence the equation to the projection of the shortest line on 
 the base, is 
 
 i *) = gi) aia aa Gon +r), 
 dO pl + (fr) s fj + (fry 
 
 c ark ds mA 
 or, ——- = - TFetTS Gaus BVO SIT etna tights 
 ** dr sf r—c dr rile F 
 
 c being the distance from the axis, at which the shortest line 
 cuts the meridian perpendicularly. Or, if p be the perpendicular 
 from C on the tangent to the projection of the shortest line, the 
 equation 1s 
 
 rte’ (fry ; 
 
 p=rc 
 
 Ex. 1. Let the generating curve be a parabola and 
 yr =Qaz its equation. 
 
177 
 
 nee at 
 then, p= c a ONY. 
 
 ] 
 a* +c 
 
 and for a right cone, if mz=ra/1—m’, 
 
 then 7 = csecm(0 + a) is the equation to the projection of 
 the shortest line, which is, therefore, an elliptic spiral. 
 
 Ex. 2. Suppose the figure to represent a sphere, and AB 
 to be the arc of a great circle, 
 
 then sin B = — 
 
 if now the meridian revolve round PC, 
 j ] 
 i gia ge Ngee one 
 sn PB CN 
 hence, an are of a great circle is the shortest line that can con- 
 nect two points on the surface of a sphere. 
 
 sin Boc 
 
 Ex. 8. To determine the length of the shortest line on 
 the surface of an oblate spheroid. 
 
 Let CEB (fig. 63.) be the plane of the equator, PE, PN 
 two meridians, AB the shortest line; 4 @ the amplitude of M, 
 so that Cl=r=asng, TM=bcosd; AM=s, 4 ECN=vW; 
 
 2. dv =rdV' +d (a’ cos’ p + b' sin’ d). But r°dy= Cds, 
 ds” C* 2.02 2 2 poe 4 Ke 
 ae 7 ~ Fang) = cos @ + b sin P= 6 +c’ cos ®, 
 if a? —b?=c*. But sn PMA wes let PAM = 90°, and a= 
 
 r 
 
 amplitude of A, therefore C=asina; and since a is the least 
 value of @, let cos @=cos acos@; 
 
 Ge sin’a — cos-asin ee dp 
 a sin sin® sin” dQ7 ’ 
 ¢ , 9 fo) . 
 (= = §? + ¢° cos’acos*0=b* +c? cos a—c’ cos’ asin’@; 
 d@ 
 c? cos’ a b 
 
 let -s——3—3—- = mm’, and .° - Sb +c cosa 
 9 rs 2 Keen ° c s° a= heres 
 b° +c? cos a t Tae Se Ay Ra 
 Z, 
 
178 
 
 ds 
 
 ba b Ca. Saree yy 
 6, Anwy sin @. 
 
 : ' b 
 Take Pm an ellipse whose semi-axes are a = and b, 
 1—m 
 b sin 8 
 and let @ be the amplitude of Q, so that CR= =; then 
 I—m 
 
 arc AM= PQ, the amplitudes of Q and M being connected by 
 the equation cos@=cosacos@. At the point B, d =,90", 
 and therefore @ = 90; hence AB = Pm; also sinPMA 
 
 asina 
 
 = ee =sina at the point B, «. 4 ABE=90—a. 
 
 For the equation to the projection on the plane of the 
 equator, we have 
 
 dw Cds asina 
 
 tee +c’cos?acos’@ 
 ie Ee eS ee LE EL, (eae pe 5 
 d@ rd@ asin “op c/casta cos: 
 
 5] 
 1—cos’ a cos’@ 
 
 which can be integrated only by elliptic transcendants; if the 
 
 eccentricity be small, we may approximate to it by making 
 
 ce 
 
 o 
 —. =e, cos*-acos 0=n; 
 
 b- 
 
 d sina sina l1—n 1—n)(3+n 
 . Ay l+ne _ }1- aes )( fain : 
 d@ 1—n Ite I1-n 2 8 
 
 BNE sina e 3 e*cos’a 
 
 Sf eS, ey © ee cos°8—&c.> ; 
 * d0  \—cos*acos’6 F 8 16 5 
 
 tan@ e 3é€ 
 or, W=tan7*- — Osina(5-— 
 sina 
 
 + = (0++sin 26) sina cos a-- &c. 
 
 The indefinite continuation of the curve will manifestly produce 
 an infinite number of spires, similar and equal, and contained 
 between two parallels equidistant from the equator. 
 
Jas Nede teulp. 352 Strand . 
 
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UNIVERSITY OF ILLINOIS-URBANA 
 
 516.33H99T C001 
 A TREATISE ON ANALYTICAL GEOMETRY OF THR 
 
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