'. II ) RKEIIY *. .;y Of UlfOKNIA NEW MATHEMATICAL WORKS. An Elemeiitaiy Treatise on the Differential otid Integral Calculus. ]{y (J. W. 1[i;mmin(;, M.A., Ffllow of St. John's College, C'anihiidpc, bvo. bds. 9s. Elementaiy Mechanics, accompanied by numerous Examples solved Geometrically. iJy J. 15. Tnu-ui, M.A., Fellow and Mathematical Lec- turer of Clare Hail. 8vo. bdt. 10*. 6rf. A Short and Easy Course of Algebra. Cliicfly designed for tlie use of tlic Junior Classes in Schools, -with a numerous collection of Original Easy Exercises. By the Rev. T. Lund, H.D., late Fellow of St. .John's College. 12mo. bds. 3a. 6d. " His definitions are admirable for their simplicity and clearness." Athenjeum. " In order to ascertain how far the Author's performance comes nj) to fiis design, ire /tate paid particular attention to those places where the learner is most likely to stumble xtpon acknowledged difficulties. , . , In all these toe have much reason to admire the happy art of the Azithor in making crooked things straight, and rough places sfnoofh. The Student must be hopelessly obtuse rcho does not, in folloxcing the guidance of Mr. Lund, obtain increasing light and satisfactimi in every step of his way; and such, too, is the strictly scientific as well as simple nature of the course pursued, that he who makes hitnself master of it, will have laid a firm foundation for an extensive and lofty superstructure of mathematical acquirement,", — The Educator. An Elementaiy Treatise on the Differential Calculus. 15y I. TuDiiuNTEu, M.A., Fellow of St. John's College. About Christmas . This Work is intended for the use of Schools as well as for Students in the Universities. Plane Astronomy. Including Explanations of Celestial Phenomena and Descriptions of Astronomical Instruments. By the Kcv. A. R. Grant, M.A., FelloAv of Trinity College. 8vo. bds. 6s. The Principles of the Solutions of the Senate -House " Riders" Exemplified in the Solution of those proposed in the years 1848 and 1851, By F. J. Jameson, B.A., Caius College. In October, A Treatise on Dynamics. By W. P. Wilson, M.A., Fellow of St. John's College, and Professor of Mathematics iji Queen's College, Belfast. 8vo. bds, 9s. Gd, Cambridge: 3IACMILLAN AND CO. London: GEORGE' BELL. SOLUTIONS OF THE CAMBRIDGE SENATE-HOUSE PROBLEMS FOR FOUR YEARS :-1848-51. N. M. FERRERS, B.A., OF GOXVILLE AND CAIUS COLLEGE, CAMBRIDGE, AND J. STUART JACKSON, B.A., ALSO OF GONVILLE AND CAIUS COLLEGE, arambrtbge: MACMILLAN & Co.; lEontJon: GEOKGE BELL; ©ubiin: HODGES & SMITH; ^(nbnrg^ : EDMOXSTON & DOUGLAS ; (Glasgoto : J. MACLEHOSE. 1851. LOAN STACK CAMBRIDGE: PHINTKD I!Y MF.TCAI.FE AND PALMl.I'. PREFACE. It will, we believe, be universally admitted that there is no easier means of becoming acquainted with any branch of Mathematics, than the study of Examples illustrative of its principles. It is also indispensably necessary that the ingenuity of the Student be thoroughly exercised in attempting to dis- cover for himself the solution of any problem which may be put before him : it is by no means our object, in publishing this book, to save him the trouble of doing so. But we believe that if, after having done his best to master a problem for himself, he is still unsuccessful, he will then derive great benefit from referring to the solution obtained by another person. It is for this pui'pose that we hope the present collection will be foimd of service. Of the intrinsic value of these Problems we could have no doubt, even if we knew less of them, coming as they do from such high authorities. We have as little doubt that they are on all accounts the best problems that we could have chosen for solution for the benefit of the Student, for their general value, their variety, and because they shew what Senate- house Problems are, and arc likely to be in coming years. 418 IV PREFACE. Part 1. contains the solutions of those proposed in tlic first three days of examination : they are of a simpler kind than those proposed in the remaining five days, the solutions of which form Part II. llie solutions of many problems have been kindly timiished by the Moderators by whom they were pro- posed: we take this opportunity of returning om* acknowledg- ments to them and others of our friends who have assisted us in the progress of the work. We shall also feel much obliged to any of our readers who will send us either correc- tions of our solutions or improvements upon them. Some difficulty, it will easily be understood, has been found in bringing all the problems to appear m their right places: any problems, however, which have been omitted in the body of the work, will be found in the Appendix, The problems on pp. 226, 241, should have been placed among the Trigo- nometry of 1850, and the Geometry of Three Dimensions of 1851, respectively. C O IS T E N T S. PART I. Euclid . Algebra Trigonometry Conic Sections Statics . Dynamics Newton Hydrostatics Optics . Astronomv • Page 1 9 19 25 32 37 48 53 59 64 PART II. Euclid . Algebra Plane Trigonometry Spherical Trigonometry Theory of Equations Geometry of Two Dimensions Differential Calculus Integral Calculus 69 74 98 107 112 110 171 183 i CONTUNTS. Page Geometry oH Three Dimensions . . . 1*«^3 Ditferential Equations .... 222 Definite Integrals .... 233 Calculus of Finite Differences .... 240 Statics ..... 245 Dynamics of a Particle .... 2*37 Rigid Dynamics . . . . • 283 Hydrostatics ..... 323 Hydrodynamics ..... 333 Geometrical Optics ..... 339 Astronomy . . • • • 3J5 Disturbed Motion ..... 362 Attractions ..... 36j Physical Ojjtics ..... 375 Calculus of Variations .... 380 Appendix ...... 384 PART I. SOLUTIONS OF THE SENATE-HOUSE PROBLEMS. EUCLID. ERRATA. I'AGK LINE /. ; J ,.' 87, 5 from bottom, for p read n . 95, 10, for (i,,.,n"p ^''^"'■^ "7'---i + "^• — 17, for ... t '• (tig- 08) 7-m(/ (tig. 89). Figure 4f5, /b;- j> 2/ s »e'^t^ ~ i^ y- Again, because EF is a diameter of the circle, therefore the angle FCE is a right angle. But CE bisects the inght angle ACB, therefore ACE is half a right angle, therefore also FCA is half a right angle ; that is, FC bisects the supplement of the right angle ACB. 2. A, B, C, (fig. 2) are three given points in the circum- ference of a circle ; find a point P, such that if AP, BP, CP meet the circumference in D, E, F, the arcs DE, EF, may be equal to given arcs. Join AB, and on it describe a segment of a circle, containing an angle equal to the sum of those subtended at the circiim- SOLUTIONS OF THE SENATE-HOUSE PROBLEMS. EUCLTD. 1848. 1. If the hypotlienuse AB (fig. 1) of a right-angled triangle ABC be bisected in D, and EDF drawn pei-pendicular to AB, and DE, DF cut off each equal to DA, and CE, CF joined ; prove that the last two lines will bisect the angle at C and its supplement respectively. Join CD, then shall CD be equal to half the hvpothenuse AB, that is, to DE or DF ; therefore a circle described from centre D, with radius DC, will pass through B, E, A, F. Let this circle be described, then the angle ECB, at the circum- ference, is equal to half the angle EDB at the centre, that is to half a right angle, and therefore to half the angle ACB ; that is, CE bisects the angle ACB. Again, because EF is a diameter of the circle, therefore the angle FCE is a right angle. But CE bisects the right angle ACB, therefore ACE is half a right angle, therefore also FCA is half a right angle 5 that is, FC bisects the supplement of the right angle ACB. 2. A, B, C, (fig. 2) are three given points in the circum- ference of a circle ; find a point P, such that if AP, BP, CP meet the circmnference in D, E, F, the arcs DE, EF, may be equal to given arcs. Join AB, and on it describe a segment of a circle, containing an angle equal to the sum of those subtended at the circum- H 2 SOLUTION OF SENATE-HOUSE PHOBLEMS. [1849. forence by the arcs AB and DE. Also join BC, and on it describe a segment of a circle containing an angle equal to the sum of those subtended at the circumference by the arcs BC and EF. These segments shall intersect in the required point P. For join AP, BP, CP, and produce them to meet the cir- cumference in D, E, F, respectively. Join AE, then the angle EAD is equal to the difference of the angles APB, AEP, that is, to the angle required to be subtended by the arc DE. Therefore DE is equal to the arc required. Similarly it may be shewn that EF is equal to the arc required, and therefore P is the required point. 1849. 1. Through a point C (fig. 3) in the eirciunference of a circle, two straight lines ACB, DCE, are drawn, cutting the circle in B and E ; prove that the straight line which bisects the angles ACE, DCB, meets the circle in a point equidistant from B and E. Let CP be the line bisecting the angles ACE, BCD ; P the point in which it meets the circle. Join PB, PE, BE ; then because the angles PBE, PCE are in the same segment, there- fore they are equal to one another. Again, because the angles BCP, BEP are opposite angles of a quadrilateral inscribed in a circle, therefore they are together equal to two right angles, that is to ACP and BCP. Therefore, taking away the common angle BCP, ACP is equal to BEP. But ACP is equal to ECP by constniction, therefore from above ACP is equal to EBP : and it has been shewn to be equal to BEP, therefore the angles EBP, BEP are equal to one another, therefore PE is equal to PB. That is, the point P in which the bisecting line CP meets the circle is equidistant from B and E. 2. Two circles intersect in A and B (fig. 4). At A, the tangents AC, AD are drawn to each circle and tenninated by the circmnference of the other. If BC, BD be joined, shew that AB, or AB produced if necessary, bisects the angle CBD. 1849.] EUCLID. 3 Produce CA, DA, to E, F. Then the angle CAF is equal to the angle DAE : but the angle CAF is equal to the angle ABC in the alternate segment, also the angle DAE is equal to the angle ABD in the alternate segment. Therefore the angles ABC, ABD are equal to one another, and AB, produced if necessary, bisects the angle CBD. 3. Draw a line to touch one given circle, so that the part of it contained by another given circle may be equal to a giveii straight line, not greater than the diameter of this latter circle. Let ABC, DEF (fig. 5) be two given circles ; it is required to draw a straight line touching the circle ABC, so that the part of it contained by DEF may be equal to a given straight line, not greater than the diameter of DEF. In the circle DEF place the straight line DE, equal to the given straight line. Find G the centre of this circle, and with G as centre describe a circle touching DE. Draw AFH a common tangent to this latter circle and ABC, cutting DEF in F, H, this shall be the line required. For since FH, DE each touch a circle whose centre is G, therefore they are equidistant from G, the centre of the circle DEF. Therefore FH is equal to DE. Hence AFH is drawn touching the cii-cle ABC, and the part of it contained by DEF is equal to the given straight line. 4. A quadrilateral figure possesses the following property : any point being taken, and four triangles fonned by joining this point with the angular points of the figure, the centres of gravity of these triangles lie in the circumference of a circle ; prove that the diagonals of this quadrilateral arc at right angles to each other. Let ABCD (fig. 6) be the quadrilateral, P any point within it : E, F, H, K, the middle points of the sides. Join PE, PF, PH, PK. And in them take PG„ PG,, PG„ PG,, respectively equal to two-thirds of PE, PF, PH, PK. Gfi.jGjOc^ shall be the centres of gravity of the triangles PAB, PBC, PCD, PDA. Join EF, FH, HK, KE, these lines are evidently parallel to b2 4 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. ^^.^\i C^^Gs, G3G,, G,G, ; and therefore E, F, H, K, lie in the cireumference of the same circle. But EF, HK are each parallel to the diagonal AC, therefore also to each other. Similarly FH, EK arc parallel to each other, therefore EFHK is a parallelogram. And since it is inscribable in a circle, each of its angles is a right angle. Therefore also the diagonals AC, BD, which are respectively parallel to the sides of the parallelogram, are at right angles to each other. 1850. 1. If ABCD (fig. 7) be a parallelogram, and P, Q two points in a line parallel to AB, and if PA, QB meet in E, and PD, QC in S, prove that RS is parallel to AD. Because CD is parallel to QP, therefore SD is to SP as CD to PQ. And because AB is parallel to PQ, therefore RA is to RP as AB to PQ. But AB is equal to CD, therefore RA is to RP as SD to SP, therefore RP is to AP as SP to DP, there- fore RS is parallel to AD. 2. Two sides of a triangle, whose perimeter is constant, are given in position ; shew that the third side always touches a certain circle. Let ABC (fig. 8) represent the triangle ; AB, AC being the sides given in position. Describe a circle DEF touching BC and AB, AC produced. The side BC shall always touch the circle DEF. For since BD, BF both touch the same circle, therefore BD is equal to BF. Hence AD is equal to AB, BF together. Similarly AE is equal to AC, CF together. Therefore AD, AE together are equal to AB, AC, BC together ; that is, to the perimeter of the triangle ABC, which is constant. But since AD, AE touch the same circle, therefore AD is equal to AE, and their sum has been shewn to be con- stant; therefore AD, AE are each constant, that is, the circle touching BC, and AB, AC produced, touches AB, AC in fixed points ; that is, it is a fixed circle. Therefore BC always touches a fixed circle. 1851.] EUCLID. 5 1851. 1. In AB, the diameter of a circle, take two points C, D, equally distant from the centre, and from any point E in the circmnference draw EC, ED ; shew that EC' + ED'' = AC + AD^ Take O (fig. 9) the centre of the circle, and join EO, and di'aw EF pei'pendicular to AB. Then because CD is bisected in O, and produced to A, .-. AC' + AD' = 2 (AC + OC) [Em. ii. 10). Again, because EC is opposite to an acute angle O of a triangle ECO, therefore EC + 20C.0F = EO' + OC [Euc. ii. 13). And because ED is opposite to the obtuse angle O of a triangle EOD, therefore ED'' = EO' + OD' + 20D.0F [Euc. ii. 12), = EO' + OC + 20C.0F ; .-. EC^ + ED' = 2 (EC + OC), = 2(A0' + 0C), = AC 4- AD' from above. 2. If through the fixed points P, Q, (fig. 10) parallel lines be drawn meetmg two fixed parallel lines in the points M, N ; then the line through the points M, N, passes through a fixed point. Join PQ, and let it meet MN in O, and the given pair of parallels in Ii, S, O shall be a fixed point. For since QN is parallel to PM, therefore QO is to PQ as ON to NM. And since NR is parallel to MS, therefore OR is to RS as ON to NM. Therefore OR is to RS as OQ to QP, or OR is to OQ as RS to QP ; that is, QR is divided in a con- stant ratio in O, and therefore O is a fixed point. 3. In a given circle it is required to inscribe a triangle, similar and similarly situated to a given triangle. 6 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Let ABC (rig. 11) be a given triangle, DEF a given circle; it is required to inscribe in DEF a triangle, similar and similarly situated to ABC. At the point A in the straight line AB, make the angle BAG equal to the angle ACB. Find H, the centre of the circle DEF, and draw UK parallel to AG, HD pei*pcndicular to HK. Through D draw DE, DF parallel respectively to AB, AC, DEF shall be the triangle required. For di'aw DL parallel to HK or AG, and therefore touching the circle at D. Then the angle LDE is equal to the angle GAB. But GAB is by constniction equal to ACB, and LDE is equal to DFE in the alternate segment. Therefore the angle DFE is equal to ACB. Similarly the angle DEF is equal to the angle ABC. Therefore the remaining angle EDF is equal to BAC, so that the triangles ABC, DEF are similar. And since the sides DE, DF are parallel respectively to AB, AC, therefore EF is parallel to BC, and they are similarly situated. 4. Describe a circle which shall pass through two points, and cut off from a given straight line a chord of given length. Let A, B (fig. 12) be two given points, CX a given line, it is required to describe a circle passing tlu'ough A, B, and cutting off from CX a chord of given length. Join BA, and produce it to meet CX in C. Bisect AB in E, and draw EF perpendicular to AB. With A as centre, and radius equal to half the required chord, describe a circle FGH, cutting EF in F. With F as centre, and FA as radius, describe a circle BAG. Join CF, and let it cut the circle BAG in K, L. From CX cut off CM, equal to CK. The circle described through A, B, ]\I shall be the circle required ; that is, if N be the second point in which it meets CX, MN shall be equal to the required chord. For the rectangle CM.CN is equal to CA.CB by the pro- perty of the circle ABM. And the rectangle CA.CB Is equal to CK.CL by the property of the circle KAB. Therefore the rectangle CK.CL is equal to CM.CN. 1851.] EUCLID. 7 But CK is equal to CM, therefore CL is equal to CN, and therefore KL is equal to MN, and KL is equal to the required chord, therefore MN is so, and the circle ABNM is the circle required. 5. Give a constniction* for finding the common tangents of two circles, and shew that if through the intersection O of two of the common tangents which meet in the line joining the centres of the two circles, there be drawn a transversal meeting the circles in A, A', and B, B', respectively, then (the points denoted by B, B' being properly chosen) OA.OB' = OA'.OB is independent of the position of the transversal. Let ABC, DEF (fig. 13) be two circles, whereof DEF is the greater; find G, H, their centres, and with H as centre, and radius equal to the differencef of the radii of the given circles, describe a circle. Through G draw GK, a tangent to this circle ; di-aw GA perpendicular to GK, and AD perpendicular to GA, meeting DEF in D : AD shall be a common tangent. Let C, D (fig. 14) be the points of contact of one of the coimnon tangents. Then we easily see that OA:OA' :: OB: OB', .-. OA.OB' = OA'.OB ; and also OA.OB' : OA.OA :: OB.OB' : OA'.OB, or OA.OB' : OC^: OD^ : OA'.OB, ::0D^ OA.OB', .-. OA.OB' = OA'.OB = OC.OD, which is independent of the position of the transversal. 6. Shew that a triangle made to revolve in the same direc- tion about its three angular points in a proper order through angles double of the angles of the triangle at the same angular points, will return to its original position. * A demonstration of this construction is not required. t We might also take a radius equal to the sum of the radii of the given circles, in which case the common tangent would touch the two circles on opposite sides of the line joining their centres. 8 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1851. Let ABC (fig. 15) denote the triangle in its first position. At the point A, in the straight line AB, make the angle BAG' equal to the angle BAG, and make AG' equal to AG. Again, make the angle GAB' equal to the angle BAG', and AB' equal to AB. Join B'G' ; then in the triangles GAB, G'AB', the sides GA, AB are respectively equal to G'A, AB', and the angle GAB is equal to the angle G'AB' ; therefore the triangles are equal in all respects. And the angle B'AB is double of the angle GAB, therefore G'AB' is the position of the triangle after revolving round A through an angle equal to 2. GAB. Again, join BG' ; make the angle BG'A' equal to the angle B'G'A or BGA, and G'A' equal to G'A. Join A'B ; then in the triangles AG'B', A'G'B, the sides AG', G'B' are respectively equal to A'G', G'B, and the angle AG'B' is equal to A'G'B. Therefore the triangle A'BG' is equal in all respects to AB'G', and therefore to ABG. And the angles BG'A', AG'B are together double of B'G'A or BGA, therefore A'BG' is the posi- tion of the triangle after revolving round the angles A, G. Again, since the angles G'BA', G'BA, GBA, are all equal, GBG' is double of A'BG' or ABG. Therefore the triangle, after revolving romid its three angular points in succession, through angles double of the angles of the triangle at those points, returns to its initial position ABG. ( 9 ) ALGEBRA. 1848. 1. A ship sails with a supply of biscuit for 60 days, at a daily allowance of 1 lb. a-head : after being at sea 20 days she encounters a storm, in which 5 men are washed overboard, and damage sustained that will cause a delay of 24 days, and it is found that each man's allowance must be reduced to f lb. Find the original number of the crew. Let X = the original number of the crew. Then 60a; = number of lbs. of biscuit with which they started. 4:0a; = remaining after being at sea 20 days. And the remainder of the crew = a; — 5, who have to remain at sea for 64 days, under a daily allowance of f lb per man. .-. f 64 (a; - 5) = 40a;, .-. 64a; - 320 = 56a;, .-. 8a; = 320, and X — 40, the original number of the crew. 2. If a, i, and x be positive, and a > b, prove that X + a X -\- b > < (a;" + a^)i {x' + h' ,,, , x -\- a X + b W e have t-^. ^. > < yjTi , according as a- > < [ab) [x' + ay (a;* + b'J ' x^ + 2aa; + a" x^ + 2bx + h as = :, — > < X* + d' x^ + W ' 2aa; ibx x' + «' x' + /'' ' 10 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. a h . . .. as -5 r. > < -5 Ti , Since x is positive, as [a — h) x'' > < d'h — a//, as (a — h) x^ > < ah [a — Z>), as x^ > < ai, since a is greater than b, or as X > < («&)*. 3. If a, ^, c, be in haiinonic progression, and ti be a positive integer, shew that a" + c" > 25". Suppose a, 5, c, all positive.* Then we have - , T 1 - •) in aritlinietic progression, a b c 2ac but a + c > 2\/«c, 2ac ^ , .*. -^— > 2v«c, or V^c > 5, but a" + c" > 2a*"c*", a fortiori a' + c" > 2V. 1849. 1. Reduce to its simplest fonn the expression (1 - gQ (1 - b'') (1 - d^ + [n— 1) «J, > al ; .'. a + ^-^ h > {aI)K Hence, in all cases, P > [al)-". (/3) Let r be the common ratio in the geometrical pro- gression, S the sum of its tenns ; then b = a — — , r — 1 ' and I = ar"'^^ .'. a + l = a{l+ r"-'). Now ar"^ + ar"-"-' - (a + 7) = « ('•'" - 1) - « (^''"' - »•""""'), = « (r - 1) (1 - r ), which is negative for all positive values of m. Hence the smn of the first and last tenns is greater than the sum of any other two tenns equidistant from the mean. The property enunciated follows at once from this when n is even. If n be odd, the middle tenn is and ri^"-'> < ^^f^, ,,„_!) a + I 2 Hence, in all cases, S < (a + /) ^n. 14 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. 1851. 3.r — 2 1. If , -xV 7:^r} ttn be expanded in a series ascend- [x— 1) [x — 2) [x — 6) ^ lug by powers of a:, find the coefficient of x\ This is best effected by resolving the given expression mto its partial fractions, and expanding each fraction separately. For this piu'}30se, assume 3a; -2 A B G {x-l) {x-2)[x-^)~ x-l'^ x-2'^ x-^"* A, Bj C being independent of x. ThenSx-2=A{x-2){x-3) + B{x-3){x-l) + C{x-l){x-2) Hence, putting a; =1, identicalhj. 3.1 - 2 = ^(1-2) (1-3), or 1 = 2A (1). Similarly, putting a; = 2, 3.2 - 2 = 5(2-3) (2- 1), or 4 = - B (2), and putting a^ = 3, 3.3 - 2 = 0(3-1) (3-2), or 7 =2C (3); Sx-2 _ 1 4_ 7 •'' (a;-l)(a;-2)(a;-3) ~ 2 (a; - 1) a; - 2 "^ 2 (a; - 3) ' 11 2 7 1 2"^ u J. — J 2 1-a;l-^a; 6 1-ia;' = -i(l+a;+ ... +a;"+...) + 2{l+^x-}- ... + {^xY + -l{l + ^x + ... + i^xr + ...], in which the coefficient of a;" _ _ 1 2 _ 7 1 ~~2'^2^~6 3""' _ J 1. _ 7 1 - 2«-i 2 2 3^" the required coefficient. 1851.] ALGEBRA. 15 2. Find the siun of the different numbers which can be formed with m digits a, n digits /3, &c., the entire series of w + » + &c. digits being employed in the formation of each number. The total number of numbers which can thus be fonned, is equal to the number of permutations of m -^-n -\- things, whereof wi are of one kind, n of another..., taken all together, _ 1.2...(m + /i + ...) ~ (1.2. ..w) (1.2...??)... ' Now, the number of times a will be found in any assigned place : the nmnber of times j3 will be found there :...'.: m '. n : ... \ therefore the number of times a will be found there m 1.2...(w + w + ...) 1.2...(7n + »i+...-l) — = m ??i + « + ... [1.2... m) [1.2... n)... (1.2...w) (1.2...w)... ' Similar expressions holding for the number of times y3, 7,... ■will ■ be found there : we have, if S be the sum of the digits in any assigned place, [ma + n^+ ...) \.2...[m + n-\- ... - 1) _ [1.2... m) [1.2... n)... ' therefore if 2 denote the sum of all the numbers, 2 = ;S(1 + 10 + 10''' + ... + iO"'+"+--^), = ^ 9 ' _ 10»'+"^--_l (»;a + 7?/3+...) 1.2...(7>? + »+•..-!) ~ 9 [1.2... m) [1.2... n)... ' the sum required. 3. The difference between the arithmetic and geometric means of two niunbers is less than one-eighth of the squared difference of the numbers divided by the less number, but greater than one-eighth of such squared difference divided by the greater nimaber. If a*, y be any two nimibers, a-^, y, their arithmetic and geometric means, a:^, y^ the anthmetic and geo- metric means of iCj, 3/,, and so on, find major and minor limits for the difference ic„ - ?/„. 16 SOLUTIONS OF SENATE-HOUSE PKOliLEMS. [iHol. (a) Taking the notation of the Litter part of the problem, we have x-2 (a-M)i + y (x* - iAY • • ^, ^1 - 2 ~ 2 * Now, one-eighth of the squared difference of the numbers K-yJ 8 4 And [x^-\-y'^y lies between [2y^f and (2a;*)'\ or > 4?/, < 4a? ; 4 > 3/ 8 ' 1 [x-yY X 8 (/3) From above it appears that „ _ „, ^ (^«-i ~ 3/«-i) ^ (^«-i ~ 3/»-i) . therefore, a fortiori^ X — V < ^ (^"-2 ^/n-a) 2 ' < > < {^-yf 1 (^«-2 3/«-2) 8^«-, . (8^.-2)^ ' (^-2/f %„-.(83/„-2)^"(%)''"' 8a.,^_,(8.0^..(8a.)^"-^ 4. If all the sums of two letters that can be formed with any n letters be multiplied together, then in each term of the product, the sum of any r of the indices cannot exceed the number rn — \r[r-\-\). 1851.] ALGEBRA. 17 Let rtj, a.^, f/^^, be the letters. Then the product will be of the form Now, any one of the letters, as «|, only appears n—\ times in this product, that is, once in each of the factors a^ + a.^ . . . a, + r/,^, therefore in no term can its index be greater than n—\. And in such terra, the index of a^ cannot be greater than w— 2, for a,^ can only enter n—2 times as a factor of such a tenn, the factor a, + a.^ being excluded. Similarly, in the term involving a"~^a^~'^ the index of a^ cannot be > 7? — 3, >i-l n-2 n-r+l ttj rtj ...(i,_^ a,. > ?« — r, therefore the sum of any r of the indices cannot be greater than (m-1) + (n-2) +... + [n-r), = rn - -^r (/•+!), the required limit, which the sum of the r indices cannot exceed. 5. Eliminate x from the equations [x — a) [x — h) = [x — c) (x — d) = [x — e) [x —f)^ and from the same equations with the additional relation e =/", find a quadratic equation for determining the quantity e or f. Shew also that if ?«', vi" be the values of e or f^ then m" — m is a harmonic mean between a — in\ b — m\ and between c — m\ d — m. (a) Since [x — a] [x — h) = [x — c) [x — d) = [x — e) [x—f), we get x^ — [a + b) X + ah = x'^ — [c + d) x + cd = x^ — {e +f) x + ef] .'. ie +f— a — b) X = ef — ab^ [e +f — c — d) X = ef — cd^ ... [ef- cd) [e +f- a-b) = [ef- ab) {e +/'- r - d). ..{!), 18 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. an equation from which x is eliminated, and which may be put in the more symmetrical form ab{c + d-e-f)+ccl[e+f-a-h) + ef{a + h-c-(l)=0...{2). (/9) If (' =/, equation (1) becomes [e' - cd) {2e - a - b) = {e' -ah){2e-c- d), ... i^c-\-d-a-h) i + 2[ah - cd) e -\- {a + b) cd - [c -\- d) ab = 0.. .(3), the quadratic for the deteniiination of e or/. (7) If m'^ m" be the roots of equation (3), ab — cd m + m" = 2 mm a + b — c — d'' ab {c + d) — cd [a + b) ^ a + b — c — d ' {ab — cd) {a + b) .'. 2 (ah + m'm") =2 , , , ^ ' a + b - c -d ^ = (a + b) [m + m"). Now 2 {ab->rm'm') — {a + b) {m' + m") = [b—m') {a—m") — {a — m') {m"— b), whence {a — on) {m" — b) = {b — m) {a — m")^ .'. a — m' '. b — m! '.'. a — m" : m" — b^ or a — m' : b — m :: a — m — {m" — m') : m" — m — {b — m')^ whence m" — in' is a harmonic mean between a — m\ b — on. Similarly, it is a harmonic mean between c — on, d — on. 19 TRIGONOMETRY. 1848. 1. The angles of a quadrilateral inscribed in a circle taken in order, when multiplied by 1, 2, 2, 3, respectively, are in Arithmetical Progression; find their values. Let 6^ (j) be two adjacent angles of the quadrilateral, then TT — 6, TT — (j), will be the angles respectively opposite to them ; and, by the conditions of the problem, 0, 2^, 2(7r — 6), 3(7r — 1, are in Arithmetical Progression. .-. 2(f)- e = 2(7r-^) - 2(f) = S[7r-(f)) - 2(7r-^), .-. A(f) + e = 27r, 4^ - (/) = TT, .-. 17^= Gtt, 17<^ = Ttt, 17 ' ^ 17 ' a llTT , IOtt 17 ' ^ 17 ' the required values of the angles of the quadrilateral. 2. Prove that sin 3^ sin'^ + cos3^ cos'^ = cos^2^. We have cos 3^ = 4 cos"^ — 3 cos ^, sin 3^ = 3 sin^ - 4 sin'^; .-. sin 3^ sin'^ + cos 3^ cos'^ = 3 (sin*^ - cos'<9) + 4 (cos«^ - sin"^), = 3 (cos^^ + sin''^) (sin'(9 - cos*^) + 4 (cos«^ - sin"^), = cos'^ - 3 cos'^ sin'-'^ + 3 cos'^ siu'(9 - sin"!?, = (cos'^-sm''^)=', = cos" 2^, the required result. C2 20 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1848. 3. Having given the three right lines di'awn from any point to the three angular points of an equilateral triangle, determine a side of the triangle. Let ABC (fig. 16) be the triangle, the point from which the lines are dra^vn, OA = a, OB =5, = c ; also let the angle BA 0=0^ CA = (f>^ and let a side of the triangle = x. Then, by the triangle BA 0, x' + d^ - 2ax cos6 = If (1). By the triangle CA 0, x^ + d^ — 2ax cos(f) = c^ (2). Also 6 + (f) = ^TT. Adding (1) and (2), we get, observing that cos ^ + cos ^ ^ 0-\- cf> e- = 2 cos — — ^ cos —!- , 2(a;^ + a') - ^ax cos^tt cos — -^ = W + c' (3). Subtracting (2) from (1), and observing that cos ^ — cos ^ = 2 sm — - sm — ^ — - , Q JL Aax sin^TT sin — — -^ = b^ — c^ (4). By (3) and (4), =^ ^^ COS^ ^TT sin""' ^TT ,,,'^ ji+i^\'^-:±^ ^ac^ff , ... [y^ + c^ _ 2d'y - ix\W + 6' - 2d' + ?>a^) -h 3 (// - d')' + ^x' = 0, ... a;* _ [a' + P + c') x' + a' + b* + c*- {bV + c'd' + d'F) = 0, 2 d' + b^ + d' {{d'+b^+Cy , 4 , ,4 , 4N , 72 2, 2 2, 271! .-. x'= ± < ^ — (a* -I- b* + c) + Fc+d'ar^-a'b' = '"'"^t'"^''' ± % [2 (?>V + cV + d'h') - [a' + b' + c*)}4, which determines a side of the triangle. 1849.] TRIGONOMETRY. 21 1849. 1. If — a, 0, + a, be three angles whose cosines are in Harmonical Progression, prove that cos (f> = 2^ cos ^a. Since cos(^ — a), cos cos {(}> + a) cos ((jb — a) ' 2 cos<^ cos a cos (0 — a) cos (^ + a) ' /. cos^<^ cosa = cos(^ — a) cos((^ + a), = ^ (cos 2(f) + cos 2a), = cos^^ — sin'^a, 2 , sin'^'a .*. cos © = , 1 — cosa 4 sin'"* ^a cos'"* ^a ~ 2 sin' ^a ' = 2 cos"''^a, .*. cos<^ = 2- cos^a, the required relation. 2. A person wishing to ascertain his distance from an in- accessible object, finds three points in the horizontal plane at which the angular elevation of the summit of the object is the same. Shew how the distance may be found. Let (fig. 17) be the foot of the object; A^ B^ C the three points at which the angular elevation of the summit of the object is the same ; then they must all be at the same distance from 0. Let x be this common distance. Let the angle AOB = d^ the angle AOC = /^ w sin a cos a 1. It tanp = ;— rr- , 1 — /i sin a shew that tan(a-/3) = (1 — w) tana. ■iiT 1 / ^N tana — tan/3 We have tan (a - /3) = :; : : — ^ , ^ ^' I + tana tan/3 ' 71 sin a cos a tan a — ; ;— 5— 1 — n sm a ?? sin'' a 1 + 1 — n sni a sin a = tan a — n n sm a cos a, cos a sin a — «(sin a + cos a) sin a cos a = (1 — w) tan a. 2. Two triangles stand on the same base, determine in terms of the base and of the tangents of the angles at the base, the distance between the vertices of the triangles. Let ABG^ ABC (fig. 19) be the two triangles. Let BC the base = rt, and let the angles ABC^ A CB = B^ C, respectively, and the angles A'BC, A' CB = j5', C Join AA', and let AA' = ?•, then it is required to find the magnitude of r. Draw ABj AD' pei-pendicular to the base. Then r'= [AD-AD'f + DD'% = {AD - AD'f + {AD cot B - AD cot By. Now a = AD (cot B + cot (7), also = AD (cot B' + cotC) ; 1 1 r' = a' cotB + cotC cot 5' + cotO'; cot-B' cot 5 , / coti^' coti^ Y "^ " Vcot B' + cot C cot B + cot 67 24 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. ( / tan ^ tan C tan B' tan C" (Vtan5 + tanC tani? + tanO tanC tanC N^)* ,tani?' + tanC" tan 5 + tan (7/ j ' an expression of the required form. 25 CONIC SECTIONS. 1848. 1. Given the lengths of the axes of an ellipse, and the positions of one focus, and of one point in the curve : give a geometrical construction for finding the centre. Let MN (fig. 20) be a line equal in length to the axis-minor. With N as centre and a radius equal to the axis-major, describe an arc of a circle. From 31 draw MO perpendicular to MN^ and cutting the arc in 0, MO will be equal to the distance between the foci of the ellipse. Produce 8P to Q (fig. 21) making ;S^^ equal to NO. With P as centre, and PQ as radius, describe an arc of a circle, and Avith S as centre, and radius equal to MO^ describe another arc ; H the point of intersection of these arcs will be the other focus, for /SlP, PH are together equal to the axis-major, and SII is equal to the distance between the foci. If therefore we bisect SH in (7, C will be the centre. Since the arcs described from 6^, P as centres will in general intersect in two points, it appears that there are two positions which the centre may have. 2. P is any point in an ellipse (fig. 22), A A' its axis-major, NP an ordinate to the point P; to any point Q in the curve draw AQ^ A' Q^ meeting NP\n R and S-^ shew that NR.NS = NP\ Draw the ordinate QM, then by similar triangles ANP^ AMQ, XR : NA :: MQ : MA, anS', and passing through By and draw the diameter SB. Join SBy BB; then the angle SBB^ being in a semicircle, is a right angle, also the angle SCB is a right angle. And the angle BSP is equal to the angle SBCj therefore the triangles BSBy SBC are similar. Hence BC: SB:: SB: SB, or SB is a mean proportional between SB and BC. But SB is equal to the semiaxis-major ; therefore the semiaxis-major is a mean proportional between the diameter of the circle and the semiaxis-minor. 3. If ABj CBj two lines in an ellipse, not parallel to one another, make equal angles with either axis ; the lines A C, BD and ADy BC will also make equal angles with either axis. Let A'B'C'D' (fig. 26) be the points of intersection of the peq^endiculars to the axis-major through the points ABCD with the auxiliaiy circle A'B'C'D'. Then it is evident that a line joining any two of the above points as A'B' will intersect the axis-major in the same point as AB does, and any two lines joining the above points as A'B'.^ CD' will be equally inclined to the axis-major, and therefore to either axis, if AB., CD arc so. 28 SOLUTloX.S OF SENATE-HOUSE PROBLEMS. [1850. and vice versa: hence we have to prove that If A'B\ CD' are equally inclmed to the axis-major LFE^ the lines A'C'^B'D' and A'D\ B' C are so. Now lAGL = L A EL + L B'A C\ and z D'HL = l UFL + l B'D' C ; also /:AEL = I UFL, and lB'AC = aB'D'C; therefore lAGL = L D'HL = L B'HG, or AC\ B'D' are equally inclined to LFE. Again, z ALH = L LD'H + z LED', and z B'KG = lKC G ^ lKGC ', also lLD'H=lKC'G, and lLED = lKGC] therefore z ALE = L B'KG, or AD' and 5 '6" are also equally inclined to LFE\ therefore also AC, BD and AD, BC are equally inclined to either axis. Q. E. D. 1850. 1. If from any point P of a circle, PC be drawn to the centre (7, and a chord PQ be drawn parallel to the diameter A CB, and bisected in R, shew that the locus of the intersection of CP and AR is a parabola. Let (fig. 27) be the intersection of CP and AR. Draw AM, CN pei-pendicular to AB, ONM parallel to AB. CN will pass through R. Then CO-.CP'.'. AO:AR, :: OM:MN. But CP=AC=MN, therefore CO = OM, and the locus of is a parabola, of which C is the focus, AM the directrix. 1850.] CONIC SECTIONS. 29 2. From the point P in the ellipse APB (fig. 28), lines are drawn to A^ B, the extremities of the axis-major, and from A, B^ Hues arc drawn perpendicular to AP^ BP; shew that the locus of their intersection will be another ellipse, and find its axes. Let Q be the intersection of the lines pei'pendicular to AP^ BP. Draw P2f, QN perpendicular to the axis-major, then the triangles PBM, BQN ^yl\\ be similar, therefore PM:BM::BN: QN', similarly P3I : AM :: AN : QN, therefore PM' : AM.BM : : AN.BN : ^.V''. But if Z* (7 be the semiaxis-minor of the original ellipse, PM' : AM.BM:: bC : AC\ therefore AN.BN : QN' ::hC'' : AC] therefore the locus of Q is an ellipse whose axes are to one another as IC : AC. And if we draw AB\ BB' perpendieidar to Ah^ Bb^ we have A (1^ ^■^=#' bC which is one axis, the other is equal to -^-^ B'C or A C. 3. If two elUpses having the same major axes, can be inscribed in a parallelogram, the foci of the ellipses will lie in the comers of an equiangular parallelogram. For it is evident that the centres of the ellipses must lie at the point of intersection of the diagonals of the parallelogram, that is, must be coincident, and their major axes are equal; there- fore they will have a common auxiliary circle. The lines joining the points of intersection of this circle and the parallelogram, will, if the right points are joined, be pei'pen- dicular to the sides of the parallelogram, and each of them will contain two foci: hence the four foci will be at their points of intersection, that is, at the comers of a parallelogram, equiangular with the circumscribing parallelogram. 30 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [IBol- 4. If from the extremities of any diameter AB (fig. 29) of an equilateral liyperbola, lines be drawn to any point P in the curve, they will be equally inclined to the asymptotes. From A and P draw the perpendiculars A C, AF, PE, PF, on the asymptotes, AF and PF intersecting in Fj from B draw BD perpendicular to the asjTuptote Oq. Then, since PQ = Bq., and the triangles QFP^ BDq are similar, they are also equal ; therefore PE = Dq and QE = BD, therefore AF=^ CO + EP=OD + Dq = Oq, and PF = OE + AC = OE + BB = OE -\- EQ = OQ', therefore the right-angled triangles APF, qQO are equal in all respects, and the chords PA, PB equally inclined to the asymptotes. 1851. 1. Given a pair of conjugate diameters of a conic section, find geometrically the position of the principle diameters, (1) in the case of the hyperbola, (2) in that of the ellipse. Let PP', DU (fig. 30) be the given conjugate diameters of an hj-perbola intersecting in C. Join PD, PB'-, bisect them in E and F, draw GE, OF', bisect the angle ECF by the line A' CA, and through C draw BCB' pei'peudicular to ACA' ', these will be the principal di- ameters required. For, by the property of the hyperbola, CE, CF are the asjnnptotes, and AC A', BCB', to which they are equaUy inclined, are the principal diameters. 2.* The solution of this part of the problem depends upon the property of the ellipse, that if P (fig. 31) be any point in the ellipse, and CR, CN, two lines at right angles to each other, cut the straight line PRN in the pomts R, N, such that PN is equal to the semiaxis-major, and PR to the semiaxis-minor, CR and CN will be the directions of the principal axes. * For this solution the authors are indebted to the kindness of the Mode- rator, Mr. Gaskin. 1851.] CONIC SECTIONS. 31 Let CP, CD be the given semi-conjugate diameters; draw Pi^ pei-pcndlcular to CD'^ make FK equal to CP; upon CK as diameter describe the circle CFK\ tlii'ough its centre draw PRN'^ join CP, CN'. these will be the directions of the principal axes. For PF. CD = PF.PK = PR.PN = A C.BC, and CP' + CD' = CP'' + PK' = 2 CO' + 2P0' = 2 OE' + 2 0P\ = PR + P^ {Euc. II. 10) = J.0"^ + PC'^ therefore PN = AC, PR = BC, and OP, CW are the du*ections of the principal axes. { 32 ) STATICS. 1849. 1. Two forces F and F'^ acting In the diagonals of a paral- lelogram, keep it at rest in such a position that one of its edges is horizontal; shew that i^seca = i^'seca' = TFcosec(a + a'), where W is the weight of the parallelogram, a and a! the angles between its diagonals and the horizontal side. Let AB (fig. 32) be the horizontal side of the parallelogram. In order to preserve equilibrium, the directions of the forces F^ F' must meet in G^ the centre of gravity. Hence, by the triangle of force, F _ F' _ W smBG W ~ &\nA G W ~ sin^ GB ' F F' W or cosa cosa sm(a + a) therefore i^seca = i^'seca' = TFcosec(a -I- a'). 2. A cubical box is half-filled with water, and placed upon a rough rectangular board ; if the board be slowly inclined to the horizon, determine whether the box will slide dowTi or topple over. Let fi = the coefficient of friction. Then the box would begin to slide when the inclination of the board to the horizon = tan~^yLt. It would begin to topple when the inclination = jtt. Therefore it will begin to slide or topple over, according as /Lt < or > 1. 1850. 1, A heavy body is supported in a given position by means of a string which is fastened to two given points in the body. 1850.] sTATifs. •};^ and then passes over a sjnooth peg: find the k-ngtli ot" the strmg. Let G (fig. 33) be the given position of the centre of gravity, A and B those of the points of support. The position of tlie peg P is determined by the conditions that it must lie in the vertical tlu'ough G^ and that the angles APG, BPG must bi; equal, each = 6 suppose. luQt AG = a, BG = b, lA GP = a, L BGP = /3 ; then PG miPAG sinf^ + a) AG sin APG sin6> = cos a + sin a cot^ PG similarly -^^ — cosyS + sin^cot^, , ^ cosa + sina cot^ BG b thereiore y^ -. — ^ 7: = -j-f, = - , cosp + smp cota AG a whence 6 is known, and length of the string = AP^BP, sin a sin /3 , . , = -. — 7i a + -. — - b is known, sma sma 2. Two spheres are supported by strings attached to a given point, and rest against one another: find the tensions of the strings. Let -4, B^ (fig. 34) be the centres of the spheres, and C the peg. Then, if the spheres are smooth, the strings must lie in the lines CA^ CB; hence the parts of the triangle ABC ai-e known. To determine its position. Let G be the centre of gravity of the spheres, CG must be vertical. Let W^, W^ be the weights of the spheres A and B, therefore AG : BG :: W^ : W^-, and if LACG = e, smjC- 6) _ BG sin^ _ TF, sini? ^ siiT^ "" ATCf&mA ~ it; sin ^ ' 1 /• • /> /I n ^^^, sin 5 therefore smC cot a — cosO = ttt • — 7 1 T^.^sm^ ' whence 6 is known. 34 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. Let T,, T^ be the tensions of the strings which support A and B respectively; therefore resolving the forces on the sphere A perpendicular to AB^ T, sin^ - PF; sin(^ + ^) = 0, _ sin( ^ + ^) ,.^ . or sin -4 and similarly 1\ = t^^ ^ TT,, _ sin(^ + 6>) ~ sin 5 «' whence T^ and T^ are known. 3. A cone of given weight W (fig. 35) is placed with its base on a smooth inclined plane, and supported by a weight W\ w^hich hangs by a string fastened to the vertex of the cone, and passing over a pully in the inclined plane at the same height as the vertex. Find the angle of the cone when the ratio of the weights is such that a small mcrease of W would cause the cone to turn about the highest point of the base, as well as slide. Let a = the angle of the plane, 6 = the half-angle of the cone. Since the resolved parts, along the plane, of the tension W of the string and the weight W just balance, we have PTsina = T'T'cosa (1) ; and' because the moments of the same forces about B are also equal, Wmia.lAC+ Wcosa.BC = W cosa.AC - W sma.BC, W cosa.f^O = ( W ^ + W sin a") BC, from (1), or BC = f sina cosa A C, or tan^ = # sin 2a. 1851.] STATICS. 35 1851. 1. A cone whose semi-vertical angle is tan"* -j is enclosed in the circumscribing spherical surface, shew that it will rest in any position. Let ABC (fig. 35) represent a section of the cone made by a plane through its axis. Divide the axis AD in (r, so that GD = iADj then G will be the centre of gravity of the cone. Join BGj then BG' = BD"' + DG\ But i'AnBAD = ^^, therefore BD' = \AD\ and DG' = ^AD\ therefore BG' = ^AD\ and BG = IAD, ^AG; therefore G is the centre of the circiunscribing sphere. Hence it appears that the height of the centre of gravity of the cone will be the same in whatever position it be placed, therefore it will rest in any position. 2. A string ABCDEP (fig. 37) is attached to the centre A^ of a pully Avhose radius is ?•, it then passes over a fixed point B^ and under the pully, which it touches in the points C and D ; it afterwards passes over a fixed point E^ and has a weight P attached to its extremity ; BE is horizontal and = — , and DE is vertical : shew that if the system be in equilibrimn the weiglit bP of the pully is — , and find the distance AB. Let W be the weight of the pully, and let ^, ^ denote the respective inclinations of AB^ BC to the horizon. The tension of the string will be thi'oughout = P; hence resolving horizontally d2 36 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. and vertically, Pcos^ - Pcos<^ = (1), P(l + sln^ + sin) cos^}, therefore |sm(^ + 4>) = coaO (3). By (1) e = = ^^^^-^^^^^0 = 1500 ' (44 .'. tana = 1500 X 32.2 ' ^ (44F 48300 ' - 4^4 ~ 12075 ' which gives the inclination to the vertical. 3. A nmnber of balls of given elasticity A, B, C are placed in a line ; A is projected with a given velocity so as to impinge on B] B then impinges on (7, and so on: find the masses of the balls B^ C , in order that each of the balls A, B, C may be at rest after impinging on the next; and find the velocity of the «*'' ball after its impact with the (n - 1)'". Let m : 1 be the ratio of the mass of w* ball to that of the {n— 1)'^', then the ratio of the velocity of the {n— ly^ ball after impact to its velocity before, would be, if the balls were inelastic, _ 1 ~ 1 + m ' 1848.] DYNAMICS. 39 Since they are elastic, the ratio is = 1 - (1+ .) f 1 - ^ = 1 - (1 + e) 1 + mj ' m 1 + m and since the (w — 1)"* ball is thus brought to rest, this must = 0, .-. 1 + - = 1 + e, m and ni = - , e so that the masses of the balls from a geometrical progression, whose common ratio is -, and the ratio of the velocity of the I -\- e n*^ ball after impact to that of the (n— lY^ before = ;; = e : therefore if V be the initial velocity of A, velocity of n^^ ball after impact = e"~' V. 4. An imperfectly elastic ball is projected in a given direction within a fixed horizontal hoop, so as to go on rebounding from the surface of the hoop ; find the limit to which the velocity of the ball will approach, and shew that it will attain this limit at the end of a finite time. Let e be the modulus of elasticity, F, V^ F^ the velocities of the ball before the first, second, (n— 1)'^ impacts, 6, ^,...^„ the successive angles of incidence. Then Fj cos 0^ = e Fcos ^, F,8in^, = Fsin^, .-. F/ = F^(sin'^^ + e"'cos^^), = sin''^(l +e:'coi'e)V': shuilarly F/ = sin'^6», {I + e' cof''^,) V;\ 1 1 But sin'^. = 1 +cof^, 1 + rVot'^' 40 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1B49. therefore T^ = sin''6> (1 + c? cof^^J V% and similarly it may be shewn that T7 = sin-^^(1 + e'^"cof^^)F''; hence when n is indefinitely increased, V = Fsin(9, the limit to which the velocity of the ball approaches. Now the distances between the successive points of incidence are 2rcos^ , 2?'Cos^.^ r being the radius of the circle; there- fore the times of describing these spaces are 2?-cos^, 2rcos^„ . , — 1^- ' , — T^— ^ respectively, 1 2 cot 6', 1 , cot^„ 1 = 2r :i + cof^j* TV (i + cof^^j* f; ('cot^ (1 + e^cot-^ji 1 H-e'^cot'6')4 sin (9 F' e'cot^ (l+e*cot'^)* 1 2r :i +e*cot'6')4 sin^ F" ' 2r cos^ 2r jjcos^ "" F^ii^' F^ dK^ ' therefore the ball will attain its terminal velocity, after the time 2r cos^ / 2 s \ 2r cos^ e V sin'^ I -e' 1849. 1. A body is projected from a given point in a horizontal direction with a given velocity, and moves upon an inclined plane passing through the point. If the inclination of the plane vary, find the locus of the directrix of the parabola which the body describes. Let a be the inclination of the plane to the horizon ; F the velocity of projection ; / the latus-rectum of the parabola de- 1849.] DYNAMICS. 41 scribed ; therefore j^r sina ' and I sill a = . 9 But J? sin a is the height of the directrix above the given point of projection ; therefore this height is constant, and the locus of the directrix is a horizontal plane at a distance — above 2g the given point. 2. An imperfectly elastic ball A lies on a billiard-table, deteniiine the direction In which an equal ball B must strike it in order that they may impinge upon a side of the table at equal given angles. The impact must be oblique and the impulse take place in the direction in which A is to go off. This direction makes with the side of the table the given angle a : let ^ be the angle which ^'s direction before impact makes this direction. V = B^s velocity before impact, e = the modulus of elasticity. 5's velocity V sin pei'pendicular to the direction of the Impulse will be unaltered by it: if there were no elasticity, its velocity in direction of the impulse after impact would be ^Fcos^, since the balls are equal, and the impulse ^MJ^cosd: hence the actual impulse will be ^(1+e) MVcosd, and the actual velocity in its direction after impact P^cos^ — ^{1 + e) Fcos ^ or ^(1 — e) Fcos^. Let = 2a, .-. tan^ = ^(1 -e) tan 2a, whence 6 is known. 42 .SOLUTIUXS OF SENATE-HOUSE PliOBLEMS, [1849. 3. A bead running upon a fine thread, the extremities of which are fixed, describes an ellipse in a plane passing through the extremities, under the action of no external force; prove that the tension of the thread for any given position of the bead is inversely proportional to the square of the conjugate diameter. Let the bead be at the point P of the ellipse. Since the tension of the string is the same tkroughout, the resultant force on the bead will bisect the angle SPH^ and therefore be nonnal to the elliptic path. Consequently, as no force acts upon the bead in the direction of its motion, its velocity will be uniform. Now, considering the bead as moving, for the instant, in the circle of curvature at the point P, nomial J, vel.' 1 .,,... torce Gc — 3 — 7: cc — j — ^ , smce the velocity is mii- rad. ot cui'v. rad. 01 curv. •' form : but radius of curvature gc CD\ therefore normal force 1 Now, adopting the usual notation, tension of the string : nomial force -.-.PE: PFr. CD.AGxCD.PF, '.iGB-.BC, therefore tension of the string cc -^^ . 4. The centres of two equal spheres (elasticity e, radius r,) move in opposite directions in a circle (radius B) about a centre of force vaiying inversely as the square of the distance ; deter- mine the motion of the spheres after they have impinged, sup- posing that e = ^ 5 ; and prove that the latus-rectum of the conic section described after the second impact will be 2e^i?. Let (fig. 40) be the centre of force ; (7, C the centres of the spheres. Draw OPQ perpendicular to CC\ such that CQ is perpendicular to 00, and consequently C Q to OC. Then if CQ represent in magnitude and direction the velocity of the sphere before impact, CPj PQ will represent its resolved parts in directions CP^ PQ. Now draw QB perpendicular to 0'^, 1850.] DYNAMICS. 43 meeting CP in B: the triangle QPR is evidently similar to G'PQ^ and therefore to CPQ. Hence BP:PQ::PQ: CP, .: PPiCP:: PQ' : CP' :: CP' : OP' :: r' : B' - r' :: e : I. Consequently BP represents in magnitude and direction the resolved part, pei'pendicular to OQ, of (7's velocity after impact. The velocity PQ remains unaltered by the impact ; therefore the diagonal of the parallelogram PBQ drawn through P will represent in magnitude, and be parallel to the direction, of the whole velocity of C after impact. Now this diagonal makes with PQ an angle equal to BQP or COP or COP, and is therefore parallel to OC. Hence after impact the centres of the spheres will move directly from the centre in the lines OC, OC. They will evidently return to the same positions C and C, and there impinge a second time. For the velocity of C after the second impact it is sufficient to obser\'e that the velocity along OP will be unchanged, while that perpendicular to OP will be again diminished in the ratio of ?: 1. Let PB' = e.PB. Through C draw CS equal and parallel to QB' ; join OS. Therefore the latera-recta of the first and third orbits will be to one another as (triangle OCQY : (triangle OSCf, since these triangles represent upon equal scales, half the product velocity x pei*pendicular on the tangent ; and we may shew that (triangle OCQf : (triangle OSC)' :: 1 : e* ; and the latus-rectum of the first or circular orbit is 2B. There- fore that of the third is 2e^B. 1850. 1. Shew that it is possible to project a ball on a smooth billiard-table from a given point in an infinite number of directions, so as, after striking all the sides in order once or oftener, to hit another given point ; but that this number is limited if it have to return to the point from which it was projected. Let P (fig. 41) be the point of the table from which the ball is projected, PQBSTU its course once round the table. 7?.S' 44 S(.>LIT1U.NS OF SENATE-HUUSE I'liUBLEMS. [1850. may be sliewn to be parallel to QP-^ and if the elasticity be perfect, equidistant with it from the line AD drawn through the comer A of the table parallel to either of them. For the angle SUB = angle QRA = 90° - BQA^^OT - PQD, therefore RS is parallel to QP. Also if QR intersect AD in F, RF'.QF:: RA sin RAF: QA smQAF, :: RA s'mBRS : QA sinDQP, :: RA sm ARF: QA sinAQF. But RA sm ARF = QA sin A QF, therefore RF = QF^ and RSj QP are equidistant from AD. Similarly, RS and TU are equidistant from CE. Through P draw VDEPU pei'pendicular to the parallel lines. Then VD = DP and VE = EU, therefore PU= DE+ EU- DP=DE+ EV - DV= 2DE. The same equation, PU = 2DE, holds whether P and U be on the same side of D and E or on opposite sides of either or both. Hence it is evident that by choosing the direction PQ rightly we may make the ball hit the second given point, through ■which the line TU will pass, after striliing all the sides once : and by lessening DE or projecting the ball more nearly in the direction of the diagonal CA, we may make it strike the second point after striking all the sides twice, when PU will = ADEy and so on ; there being thus an infinite number of directions of projection each more nearly parallel to the diagonal CA than the preceding, which will cause the ball to hit the second given point after striking all the sides once, twice, &c., respectively. If, however, the ball have to return to the point of projection, we must have DE = 0, or the direction of projection parallel to either diagonal; there being thus two directions and their op- positcs, or four directions in all, which will bring the ball back to its point of projection. Through this point it will pass after making each round of the table. DYNAMICS. 45 1851. 1. A body of given elasticity e is projected along a hori- zontal plane from the middle point of one of the sides of an isosceles right-angled triangle, so as, after reflexion at the hypothenuse and remaining side, to return to the same point ; shew that the cotangents of the angles of reflexion are e + 1 and e + 2, respectively. Let ABC (fig. 42) be the triangle, right-angled at -4 ; i> the middle point of AB the point of projection : AD — DB = a. Let e be the modulus of elasticity. Draw DEF perpendicular to BC^ making EF = e.ED: draw FGH perpendicular to ylC, making GH = e.GF: di'aw HD, LF, KB: DEL will be the path of the body. The angle of reflexion at -ff" = 90° — LEG = 6 suppose, L= LEG = (1). 58 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Also ic + weight of the air iii the eyliiider = weight of the fluid displaced, .-. w+gp'V = fjpj^, V (2), where Via the volume and // the height of the cylinder. From equation (2) z is known, and thence y from (1). 1851. 1. A hollow cone floats m a fluid with its vertex upwards and axis vertical ; determine the density of the air in the hollow cone. Let 2^ be the pressiu'e of the air in the cone, IT that of the atmosphere, tv the weight of the cone, h its height, z the height of the cone of compressed air, y the depth of its base below the surface of the fluid, /?, p' the densities of the fluid and uncom- pressed air. Then, equating the pressures at the common smface of the air and fluid, gpy +'n = pressure of the compressed air, =i> = -n (1). Also 10 + weight of the air in the cylinder = weight of the fluid displaced, ,TT z^ — (z — vY _- , , or to + gpV=gp Jf-^ ^ (2), where V is the volmne of the cone. From equations (1) and (2) z and y are known, and thence j-j, and the required density. ( '^9 ) OPTICS. 1848. 1. If Q^ q (fig. 47) be two points in the radius of a spherical reflecting surface whose centre is E^ such that EQ : Eq :: sine of the angle of incidence : sine of the angle of refraction, de- termine geometrically the position of the point P, so that a ray- proceeding from Q and incident upon the surface at P may after refraction proceed from q. Bisect Qq in m, and thi'ough m draw mF perpendicular to EQ meeting the circle in P; Pwill be the point required. For if we join FE^ Fq^ FQ, we have sin^P^ : sin FQE:: QE : FE, and sm^P^ : sinP<^^ :: qE : FE. Now sin FQE = sin FqE, •. • z FQE = LFqQ, .-. smEFQ : sinJ;P^ :: EQ : Eq :: fM : I by the question, therefore the ray QF after refraction at P will proceed as if from q. 2. If a ray of light, after being reflected any number of times in one plane, at any nimiber of plane sm'faces, retmii on its fonner course, prove that the same will be true of any ray parallel to the foi'raer which is reflected at the same surfaces in the same order, provided the number of reflections be even. Let FQR8 (fig. 48) be the course of any ray which starting from P, after reflection at Q^ R and 8^ amvcs again at P, and is there reflected in the direction PQ of the original pro- pagation. Let F'Q'JR'S' be the course of another ray starting from P' in a dii'ection F' Q' parallel to FQ', we have to shew that after reflection at 5", this ray will proceed to P, and there be reflected in the direction P' Q'. Join S'F'. Then since the angle Q'F'A = QPA, and QPA=SFF, there- fore the triangle PpP' is isosceles, and the ptTpciidiiiilar frmn /' 60 SOLUTIONS OF SENATE-HOUSE PROBLEMS, [1849. on F Q = that from P' on PS. Similarly the perpendicular from Q on QR = that from Q on Q'P' = that from P on P Q since PQ is parallel to P' Q ; therefore the pei'pendicular from Q on QR = that from P on PS. By similar reasoning it may be shewn that the pei'pendicular from S' on PS = that from R' on QR = that from Q' on ^i? since Q'R' is parallel to ^^ = that from P on PS. Hence PS' is parallel to PS'^ therefore S'P' is the direction in which R' S' will be reflected from S\ and P Q' is that in which S'P will be reflected from P'. The same proof may be extended to any even nmnber of reflections. If the number of reflections were not even we might still shew that P', S' were equidistant from PS^ but they would be on opposite sides of it, as P', R' are of PP, and the pro- positions would not be true in that case. 1849. If the angle of a hollow cone, polished internally, be any submultiple of 180°, a cylindrical pencil of rays incident parallel to the axis will, after a certain number of reflections, be a cylindrical pencil parallel to the axis, and of the same diameter as the incident pencil. Let fig. 49 represent a section of the cone and the light by a plane through the axis CD of the cone, and let tn^m_^ be the successive points when the ray PQ^Q^Q^... cuts the axis CD. 180° (1). Let the angle ACB be an even submultiple of 180°=— — 90° ^'' suppose, or — . Now the angle Q^m^D = ACD + CQ^m^ = ACD + A Q^P, = 2ACD = ACB, and the angle Q^^n.^D = BCD + CQ,^in^ = BCD + Q^Q^B, = BCD + BCD + Q.m^D, = 2ACB- similarly Q^m^D = 3 A CB, ~ ) and Q„<»,D = uACB = 90°, 1850,] OPTICS. 61 or after the »"' reflection the ray will be pci*pendicular to the axis CD^ and will proceed in a path exactly similar to that already described, finally emerging in a direction parallel to QyP^ and at the same distance as Q^P from the axis CD^ but on the opposite side of it. 180° (2). Let the angle A GB be an odd submultiple of 180°= ^ > ^ ^ 2n+l suppose. Now the angle Q.Q^B = BCD + Q,m,D^ = IACB -\- ACBhj the above, and the angle Q.^Q^A = A CD + Q.^mJ)^ = \ACB + 2ACB, = IACB; similarly Q.^Q^B = IACB, and Q,^Q,^,^A = ^-^ACB = dO'', or after w reflection the ray will be perpendicular to the side CA or CBj at which it has next to be reflected, and will there- fore after that reflection return by the same path as it came by, and will emerge in the direction Q^P. Hence, whether ACB be an even or odd submultiple of 180°, the emergent rays will form a cylinder equal in diameter to the cylinder of incident rays, and having its axis coincident with the axis of that cylinder, if the angle A CB be an odd submultiple of 180°; or if the angle ACB be an even submultiple of 180", the axes of the emergent and incident pencils will lie in the same plane with the axis of the cone at equal distances on opposite sides of it. 1850. 1. If a luminous point be seen after reflection at a plane min'or by an eye in a given position, there is a certain space within which the image of the point can never be situated, how- ever the position of the plane mirror be changed ; find this space. 62 SOLUTIONS OF SKXATK-Ilorsi: I'KOBl.EMS. [iSol. It is easily seen that the distance from the eye of the image foiTned by the mirror equals the actual length of the ray by which the point is seen. This can never be less than the direct distance of the point from the eye ; hence the image can never be situated within the sphere which has the dii'ect distance between the point and the eye for radius. 2. If a be the angle which every diameter of a circular disc subtends at a luminous point, shew that the ratio of the light Avhich falls on the disc to the whole light emitted is as sin'^^a : 1. About the Imninous point as centre describe a sphere with radius unity : also with the luminous point for vertex and the circular disc as base describe a right cone. Then the light received on the circular disc : whole light emitted :: the portion of the surface of the sphere Intercepted by the cone : whole surface of the sphere. Now by a known property of the sphere, the surface of any portion of the sphere cut off by any plane is proportional to the difference of the radius of the sphere and the distance of the cutting plane fi'om the centre. Hence the surface intercepted by the above cone : whole surface of the sphere :: 1 — cos|a : 2 :: sin^'ja : 1, which is therefore the ratio of the light received on the circular disc to the whole light emitted. 1851. A sphere composed of two hemispheres of different refrac- tive powers is placed in the path of a pencil of light in such a maimer that the axis of the pencil is perpendicular to the plane of jimctlon and passes through the centre : determine the' geometrical focus of the refracted pencil. Let r be the radius of the sphere, u the distance of the focus of incident rays from the centre; v^v^v^ the distances of the geometrical foci after the successive refractions, positive lines being measured m the direction opposite to that of the incident light ; fi^fx,, the refractive indices of the two hemispheres. 1851.] OPTICS. 63 Then i = - ^Vzi + ^. (1), 1 /i, 1 - =^ - (2 , (3) + (2)xl + (l)x^,, i=-^.i^>,-i)+^,>,-i)}i+^:i, which gives \\ the distance from the centre of the sphere of the geometrical focus after refraction. ( <>4 ) ASTRONOMY. 1849. 1. Tlierc arc two walls of equal known height at right angles to each other, and running in known directions ; shew how to find the sun's altitude and azimuth by observing the breadth of the shadows of the two walls at any given time. And prove that the sum of the squares of the breadths of the shadows will be the same whatever be the direction of the Avails. Let a, h be the observed breadths of the shadows, h the known height of the walls ; 6 the angle between the base of the wall, the breadth of whose shadow is a, and the line joining the shadow of the top of the line of intersection of the walls with the bottom of that line, ^ the sun's altitude. Then (y = tan 7- , and © = tan j-^ ^,-^, , i ' ^ (a^ + ly are known. Let a be the angle between the wall whose breadth is a and the plane of the meridian ; then a -- ^ is the angle between the plane of the meridian and the vertical plane through the sun, or the sun's azimuth. Hence both the altitude and azimuth are known. Also d^ +h^ = the square of the length of the shadow of the Ime of intersection of the walls ; and the height of this line is the same whatever be the direction of the walls, or a'' + // is independent of that direction. 2. If the same two stars rise together at two places, the places will have the same latitude. And if they rise together at one place and set together at the other, the places will have equal latitudes, but one north and the other south. From the bisection of SS' the great circle passing through the two stars S^ S' draw a quadrant of a great circle perpen- dicular to SS' towards the north pole terminating in the 1850.] ASTRONOMY. (55 point r, and another towards the south pole terminating in T. First, suppose the great circle containing these quadrants to have its point which is nearest to P the north pole, to tlic left of P (drawing the stars on the convex part of the sphere). Then, in order that the stars /S', S' may rise together at any place, its zenith must at some time in the 24 hours come to T\ its co-latitude therefore must be TP. Hence if S^ 8' rise together at two places, the co-latitude of each must be 2!P; hence the latitudes of the places are the same. If the point of the circle nearest to P lie to the right of P, the zenith of the place must pass through T' in their daily path and therefore have the same latitude, viz. 90' — T'P\ S., P' being the south pole. If the same stars rise together at one place and set together at another, the zenith of one must pass tlii'ough T and that of the other through T m their daily paths; hence they will still have equal latitudes, but one will be north and the other south. 1850. 1. Prove that all stars which rise at the same Instant at a place within certain limits of latitude, will, after a certain interval, lie in a vertical great circle ; and detenuine those limits. This Avill happen when the zenith of the place comes to that great circle of the heavens which at the time of the stars' rising was the horizon of the place. Hence it can only happen for those places for which the altitude of the pole is less than the co-latitude : but the altitude of the pole is the latitude, hence if / be the latitude, I must be less than 90° — / or / less than A,'/. 2. Shew how to find the days of the year on which the light of the sun reflected by a given window which has a south aspect will be thrown into some one of the lower windows of an opposite range of buildings. Corresponding to each window opposite, let that point of the heavens be detennined, which lies in the same plane as that window and the horizontal line through the reflecting window pointing to the south, and at the same angidar distance from this r 66 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1851. line as the opposite window in question, and on tlie opposite side of it. Let the north polar distance of this point or its declination be then observed ; the reflected light will enter the window in question on those days when the sun has this de- clination. Suuilarly, the days when the reflected light will enter the other windows may be detenniued. 1851. Altitudes of the same heavenly body are observed from the deck of a ship and from the top of the mast the height of which from the deck is known : find the dip of the horizon and the tnie altitude. Let AB = x, BC= h (fig. 50) be the height of the deck from the sea, and of the mast respectively ; OA = r the radius of the earth. The difference (a) of the observed altitudes is the angle^i^Oori>(9^. Now cosCOE = 5 , and cosCOD = , r + h + x^ r + x' .-. DOE = cos"^ 5 cos ^ = a, r + li + X r + X an equation for the deteinnination of x. T Then the dip of the horizon = OBD = sin~^ is known, and subtracted from the altitude observed at B gives the true altitude. From the above equation we may determine x with sufficient accuracy thus : r + X \ {r + h + xY) r + h + x \ {r+x \2r(h + x)}^- (2rx)^ or - — ^ -^^ — = sma, r r ' omitting h and x in comparison of r ; 2(h + x) .2 « • /2a;\i 2x -^ ' = sm^a - 2 sma — -\ , r \ r J r ^ f2x\^- , . h I I = 4 sin « — — h , , — 9 sma coseca, \r J ^ r ' or X = l^ sma — - coseca I PART II. F-2 ( 69 ) PART 11. EUCLID. 1848. 1. AB, CD, (fig. 51) are any two chords of a circle passing through a fixed point O, EF any chord parallel to AB ; join CE, DF meeting AB in the points G and H, and DE, CF meeting AB in the points K and L : shew that the rectangle OG.GH = OK.OL. The triangles OCG, OHD have the common angle 0, and ^OCD = 180° - EFD = EFH, = OHD, since EF is parallel to AB ; hence the triangles OCG, OHD are similar, therefore OC:OG::OH:OD, or OG.OH = OC.OD. Again, L OCD = OEF = DKO, since EF is parallel to AB ; hence the triangles OLC, ODK are similar, therefore OC:OL::OK:OD, or OL.OK = OC.OD, .-. OG.OH = OL.OK. 2. In a given circle inscribe a rectangle equal to a given rectilineal figm'c. Let AB (fig. 52) be a diameter of the given circle ABC. Draw a square which shall be equal to the rectilineal figure 70 SULI'TIONS OF SEXATE-HOUSE PROBLEMS. [1848. {Eu<\ II. 14,; through D any point of AB draw BE pei*pen- dicular to AB, a third projxjrtional to AB and the side of the square. Through E draw EC parallel to AB, meeting the circle in C ; join AC, BC, and complete the parallelogram ACBF : it shall be the rectangle required. For C and F are each right angles, being the angles in a semicircle, therefore ACBF is a rectangle. Also its area equals AB.DE which equals, by constniction, the square which equals the given rectilineal figure ; therefore it is the rectangle required. 3. Through a given point A (fig. 53) describe a circle which shall touch a given circle BCD, and intersect another given circle LEF in a chord passing through a given point G. From G draw any line GEF Intersecting the circle LEF In the points E, F ; join GA, and in GA produced if necessaiy, take the point H, such that GA.GH = GE.GF. Through the points A, H describe any circle cutting the circle BCD In the points B, C ; join BC, and produce It to meet GA In K. From K draw KD a tangent to the circle BCD. About the triangle AHD describe a circle, it shall be the circle required. And first It shall touch the circle BCD : for since KD touches the circle BCD, KD^ = KB.KC = KA.KH, since one circle has been made to pass through A, H, C, and B ; and therefore KD touches the cu'cle m question as well as the circle BCD, therefore the two circles touch. Also the chord in which it intersects the circle LEF will pass through G. For suppose L to be one of the points In which It intersects the circle LEF ; join GL and produce It to meet the two circles in M, M'. Then GL.GM = GE.GF = GA.GH, by construction; also GL.GM' = GA.GH, since H, A, L, M', lie in the circumference of the same circle, therefore GM = GM' or the points M and M' coincide, and GLM Is the chord in whicli the circles intersect, and the chord passes through G as required. 1851.] EUCLID. 71 Hence tlic circle drawn as above described fulfils the required conditions, and is therefore the circle sought. 1849. Thi-ee circles are described, each of which touches one side of a triangle ABC (fig. 54), and the other two sides produced. If D be the point of contact of the side BC, E that of CxV, and F that of AB, shew that AE = BD, BF = CE, and CD = AF. Let AB, AC touch the circle GDH in G, H ; then BG = BD, and CH = CD, also AG = AH, .•. AB + BD = AC + CD = semiperimeter of the triangle, similarly, BA + AE = semiperimeter of the triangle : .-. BA + AE = AB + BD, and AE = BD ; and similarly, BF = CE and CD = AF. 1851. 1. Let T (fig. 55) be a point without a circle, whose centre is C ; from T draw two tangents TP, TQ ; also through T draw any line meeting the circle in V, and PQ in B, and draw CS perpendicular to TV; then SR.ST = SY\ Join CT, intersecting PQ in U at right angles ; draw CP; it will be perpendicular to PT. Since the triangles CTS, RTU are similar, CT:TS::RT: TU, .-. TS.TR = TC.TU, or ST'^ - ST.SR = CT' - CT.CU, = ST^ + CS^ - CT.CU, .-. ST.SR = CT.CU -CS^: but CPT, PUC are both right angles, .-. CT.CU = CP'' = CV^ = SV^ + cs% .-. ST.SR = SV^ 72 [SOLUTION'S OF SEXATE-HOUSK PROBLEMS, [1851. 2. It" a ciix'lc be described round the poiut of intersectiou of the diameters of a parallelogram as a centre, shew that the sum of tlie squares of the lines drawn from any point in its cir- cmnference to the four angular points of the parallelogram is constant. Join P any point in the circle with A, B, C, D (fig. 56) the angular points of the parallelogram, and with O the centre of the circle. Then, since OB = OD, the square of BP is greater than the squares of OP and OB by the rectangle by which the squares of OP and OD are greater than the square of DP {Euc. IL 12, 13) ; hence Bp2 ^ j)p. ^ Qg. ^ Qj)2 _^ gOP" is constant : similarly, AP^ + CP-^ = AO'^ + OC' 4- 20P' is constant, therefore also AP"' + BP'^ + CP' + DP"' is constant. 3. (a). Let B (fig. 57) be any point in the circumference of a circle whose centre is A ; in AB take two points C and D, such that ACAD = AB'; bisect DC in E, and draw EF at right angles to AE ; in EF take any point G, then will the tangent drawn from G to the circle be equal to GO. Draw GF the tangent to the circle, join CG ; then AG^ = GF^ + AF' = GF' + AB', also AG' = GE^ + AE' = GE' + CE' + ACAD {Euc. ii. 6), = CG^ + AB^ by construction ; ... GF' + AB^ = CG' + AB', and GF = CG. (/3). Describe a circle which shall pass through a given point, touch a given straight line, and cut orthogonally a given circle. Let D be the given point, BFL the given circle, KH the given line intersecting AD in H. In AD take AC : AB :: AB : AD, and HK a mean pro- portional to HD and HC ; the circle through CD and K shall be the required ciicle. 1851.] EUCLID. 73 For the centre of this circle will be at some point G of EG, and since GF = GC, will pass through F, cutting the circle BFL orthogonally at that point. Also since HK is a mean proportional to HD, HC, or HK"^ = HD.HC, HK will touch the circle through the points C, D, K ; hence that circle fulfils all the required conditions and is the circle sought. ( 74 ) ALGKBRA. 1848. 1. A walks to Trumpington and back by Granchester In 1^ hour, starting between 2 o'clock and 2f ; B walks the same distance in the same direction in labour, starting between 2 and 2^ : find the chance that A overtakes B before he gets home. Unless A starts after B he cannot overtake him. Now if he starts between 2 and 2^, it is an even chance whether he starts first or not ; otherwise he cannot. Hence ^'s chance of starting before B = ^.| = ^. Again, A gets home between 3^ and 4, B between 3^ and 3f ; therefore A has an even chance of getting home first ; therefore also his chance of getting home last = ^, and chance of his not overtaking B = chance of his starting first + chance of his getting home last = ^ + ^ = 1; therefore chance of A overtaking -C = 1 — f , the required chance. 2. A paralleloplped is cut by three systems of parallel planes given in number, parallel to the three pairs of opposite faces respectively : find the total number of parallelopipeds formed in every way. Let in, w, j9, be the given number of intersecting planes parallel to the three sides respectively; we thus have m + 2, « -f 2, ^9 + 2 parallel planes in -three several directions. Now, out of the first set of parallel planes we may make ^ '— — — - sets of two each. Similarly, out of the other two sets we may make — ^-^ — — , — ~- sets respectively. 1848.] ALGEBRA. 75 Now each paralleloplped is formed by taking one out of each of the above sets of two parallel planes, therefore the total number of parallelepipeds will be {m + 2){m+l) {n+2){n + l) (p + 2)(p+l) 2 ' 2 • 2 ' _ {m + 1) {n + l){p + 1) [m + 2) (n + 2) (^ + 2) 8 ' the required number. 3. If {^-a){i/-ma) = {n^-mo(.){x-a) (1), and O' - a') (y - m/3') = {m^' - na) (cc - /3') (2) , shew that [—^-^-\x = -, ^-^r- ■ \aa pp J aa pp Taking (1) {/3' — a) — (2) (/3 — a), so as to eliminate y, we get m{^-oi){/3'-a'){^'-a)={{n/3-ma){l3'-0L)-{ml3'-na')[^-a)]x - (n^ - ma) (^' - a') a + (?n/3' - na] [^ - a) ^8', or m {(/3 - a]{l3"' - a 13') + [^' - a:)[oi' - ayS)} = [n- m)(/3/3' - aa> -■n{(;S'-a>^+(;S-a)a'/3')l+m{(^'-a>'^+(/3-a)/3"0], .-. {n-m){ (/3' - a') a/3 + (/3 - a) a^'] = (h - w) (/3/3' - aa') a-, and dividing by aa'/S/3', /J l_\ ^ g + g' _ /3-F /3' "^Vgg' m'J da /3/3' ' 4. (g). Shew that the integral parts of (3* + 1 )'""'' and (3* + 1)'^'" + 1 are respectively divisible by 2'"'^^ where m is any integer whatever. The integral part of (S^+ir^^ is (3* + If'""^ - (3*- 1)"""S since it is a whole umnber, and (3* — 1)^'"+^ is less than 1. Now generally ^.«., _ y-.^. = [x-y) [[x^'+fl + try (a;-*- ^4/"'-0 + a^/(a.— * + yn + ... + x'f (A), and if £c = 3* + 1, y = 3* - 1, x-y = 2, xy = 2, .-. (3* + 1)"'""'' - (3* - 1)'^'"*' = 2 [{(34 + l)'-"" + (3* - 1)'""} + 2 {(3* + l)"^"-^ + (3i - I)"-""-'''} + . . . + 2"']. 76 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. This part of the question, then, reduces itself to shewing that (3i+ 1)'-"" + (3*- l)'-"" is divisible by 2"'. Again, the integral part of (34 + l)'''"' + l is {Si+1)''"'+{S^-1)'% since this is a whole number and (3*— 1)^'" is less than 1 ; hence the second part of the question reduces itself to shewing that (34+1)'"' + (3i- 1)'^"* is divisible by 2"'^\ and therefore mcludes the first part. Now (3* + 1)*" + (3* - 1)"" = (4 + 2.34)'^" + (4 - 2.34f ", = 2''"'{(2 + 34)'-'"+ (2-34n, which is evidently divisible by 2^"^\ Also generally _4«+2 I 4'i+a / 2 1 2\ f / 4» , 4»\ 2 2/ 4«-4 . 4n-4\ + a.yK- +/"-), -...+ (-)VY"}; .-. (34 + 1)*"+''' + (34 - 1)*'^'^ = {(34 + l)'-' + (34 - 1)'} [{(34 + 1)*" + (34- 1)*"} - 2'-^{(34 + l)*«-^+ (34-1)^"-^} +...+ {-Y2'"l which, since (34 + 1)'" + (34 - 1)*" is divisible by 2'"^', is divisible by 2'"+", and therefore a fortiori hy 2'"'"+' or 2^''''*'^*\. Hence, whether ?n be of the fonn 2n or 2n + 1, (34+1^"+ (34 -If" is divisible by 2'"^*,. and both parts of the proposition are true. (/5). Prove that for a given integral value of a, there are (1), a integral values of b which will make the integral part of (a + b'-Y'"^' divisible by 2'"+^ ; (2), a integral values of b which will make the integral part. of {a + b'~Y"'^' + 1 divisible by 2'"^^ ; (3), 2a integral values of b which will make the integral part of {a + UY" + 1 divisible by 2'"+\ (1). The integral part of (^4+ af'*' = (&4+ af"^' - [U-af'^-'j provided b^ — a < I and > 0, i.e. if b lie between a^ and [a + 1)"^, which gives only 2« values of b. 1849.] ALGEBRA. 77 Now by equation (A), + [h - «^) [{b'- + ay-'-' + {hi - aY"'-'} +...+ (/._ aTl Now since b is only to have (a) values, we may make b — (i^ even, and then the problem is reduced to shewing that {bi + ay"+{bi-af" is divisible by 2", which will h fortiori be tme if {b'- + ar + {bi-aY% {the integral part of {b^' + af' + 1} be divisible by 2"^', and therefore (1) reduces itself to (3). (2). The integral part of {a+b^-f"'^'+l={a+b^-Y"'^'+{a-biy"'^\ provided a — b^ < 1 and > ; therefore b must lie between n^ and [a— 1)'\ and it will appear by a precisely similar process to the above that (2) reduces itself to (3). (3). The integral part of (a + b^Y'" + 1 = {a+b'-f"' + {a - 7>i)*", provided a — b^< 1 and > — 1, i.e. if b lie between [a— 1)'^ and {a+ ly, giving only (4«) admissible values of b. And if we flirther make b — d^ or d^ — h even, the number of values of b will be reduced to (2a) . Then, these conditions being satisfied, (3) can be proved as in (a), only writing 5* for 3* and a for 1. Hence the propositions enimciated are tnie. 1849. 1. A quantity of com is to be divided amongst n persons, and is calculated to last a certain time if each of them receive a peck every week ; during the distribution, it is found that one person dies every week and then the coni lasts twice as long as was expected : find the quantity of com and the time that it lasts. Let X = the number of pecks of corn, y = weeks it is expected to last, then - = the whole number of persons = n ; y or X = ny (!)• 78 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1H49. Also n = number of pecks distributed in tlie 1st week, V — 1 = 2nd n - r + I = J-"" and it lasts 2i/ weeks ; therefore n + {n — I) + {n — 2) + ... + {n — 2?/ + 1) = whole quantity of corn, or {{2{n-i/) + l}?/ = x (2); .-. by (1) 2{n-y) + I = n, n + I _ 7i[n + \) ^ - 2 ' and 2y = n ■\- 1, which determines the quantity of corn, and the time it lasts. 2. If ,^, C,. be the number of combinations of m things taken ?• together, and jy be less than m and ??, shew that We have (1 + xY{\ + .-r)" = (1 + a-)'"^". Now the coefficient of £c" in (1 + .t)" is vOfx'. also the coef- ficient of x^ in (1 + a;)"'"^" must be the sum of the jiroducts of the coefficients of a;' in (1 + .x)'" multiplied into the coefficient of af"" in (1 + x)" taken for all values of r from to j>. Hence 3. If ^Gr be the number of combinations of n things taken r together, prove that if a be an integer greater than 1, then will „,(7,„ be greater than („C,.)". 1849.] ALOEBILV. 79 The total number of combinations which can be made out of ?w things is „„C,.„. But if we divide the na things into a sets of n each, and restrict ourselves to those combinations containing ra things which can be made by takuig r out of each set, we shall get, since ,, (7, combinations of r things may be made out of each set of w things, [„C,.)" combinations. Hence, since (7.„ includes every possible mode of fonnationa, and („6'^)" only one particular one, it is clear that 4. If ^ be greater than unity, 1 1 1.2 1.2.3 and if it be less, 1 1 1.2 L^__ « 1-j, 1 +i? (1 +p) (1 + 2p)^ ^ (1 +p) (1 + 22j) (1 + 32)) ■ We have in general, when^ is greater than unity, 1 \__ _ M + 1 1 p — \ ^; + w 2^ -\r n' 2^ — X"* 1 1 ?? + 1 1 J) — ^ J9 + 7? ^? + 7i '/> — 1 ' therefore, putting successively n — 1, 2, 3 , 112 1 + jj — 1 2.) ■\- \ ^ + 1'^> — l' 1 +^f^+ ^j + 1 ^^ + 1 V7? + 2 y> + 2 ^ - 1 1 1.2 1.2.3 + T— ^T7 n^N + ■^ + 1 "^ (^+l)(^, + 2) ^ (^>+l)(7.-+2) •;>-!' 1 1.2 1.2.3 + 7 — r-^, — r^x + 7 — , ,x/ , ■ nN/.. ■ ox +•• 2>^\ (i?+l)(i^ + 2) " (^j+ l)(iJ + 2)(^^ + 3) 80 SOLUTIONS OF SENATE-HOUSE PKOBLKMS. [IHot*. Kj) be less than unity, _^ I _ ^ {n+_l)p _i_ . 1 —J) 1 + np 1 + lip ' 1 — /' ' therefore, putting ?« = 1, 2, 3,... successively, I _ 1 2p 1 1 -i> ~ 1+p l+p ' \-p ' - ^ 2p / 1 3/> 1 \ ~ 1 +/?''" 1 +p li + 2^ "'' r+ 2p • 1 -pj ' ~ l+p'^ [l+p)[l^2p)^' "^ (l+^)(l + 22^)(l + 3/^)^' -f 5. A bag contains three bank-notes, and it Is known that each of them is either a £5, £10, or £20 note ; at three suc- cessive dips into the bag (replacing the note after each dip) a £5 note was drawn : what is the probable value of the contents of the bag ? There are six possible states of the bag, viz. (1). 3 £5 notes in which case the value would be £15. (2). 2 £5 and 1 £10 £20. (3). 2 £5 and 1 £20 £30. (4). 1 £5 and 2 £10 £25. (5). 1 £5 and 2 £20. £45. (6). 1 £5, 1 £10, and 1 £20 £35, and these are all a priori equally possible. Now the chance of the observed event in case (1) is 1, (2) or (3) (ror|, (4) (5) or (6) (i^or^, therefore the probable value of the contents is 1 X £15 + A (£20 + £30) + sV(£25 + £15 + £35) 1 + 2 X I, + 3 X i — * ~ 4fi 1 27 _ 4?nio — ^mi = £19 15.?. V^^d. ALGEBRA. 81 1850. 1. Prove that the sum of the fractions which are intermediate in magnitude to any two nmnbers vi and n, and liave 3 for a denominator, is n' — ni^. The fractions, together with the intermediate whole numbers, will be 3w + 1 ^m + 2 3n - 1 3^"' 3 ' 3 . /3m + 1 3« - 1\ (3^2-1) - 3»« whose sum is I — 1 — — 1 , _ [m-\-n) [^[n-m) - 1} _ _ ^ and the sum of the intermediate whole numbers is (m+1) + (?» + 2) +...+ («- 1), » — 1 — m = [[m + l) + [n-l)^ , (■w + r?) {ii — m— 1) ^ 2 ' therefore the sum of the fractions is (w + n) (3 [n — m) — 1} {m -f n) [n — m — 1) 2 2 ' = n' - m\ 2. There are a number of comiters in a bag, of which one is marked 1, two marked 2, up to r marked r ; a person draws a coimter at random, for which he is to receive as many shillings as the nmuber marked on it : find the value of his expectation. Since the person is as likely to draw one counter as another, the value of his expectation total value of contents of bag number of counters in bag ' 1'^ + 2^ + ...+ ,.« = l + 2+...+ >- '^'^^'''^'' 82 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. Now I' + 2' + ... + r' = „ , o'{r +l)(2r+l) 6 r(r + l) 1 +2 +... + r= \^ ' \ r(r + l)(2r + l) r(r + l) hence his expectation = -=^ '-^ 1 r — - = — - — shillings. 3. If a, J, c be in harmonic progression, shew that 111 1 ^ - + - + 7 + 7 = 0. a c a — b c — o T * 7> 1 1 1 >S' /3 + a' /3-a' then - 4- - = 2/3, a c and a - b = -p: -^ , _ a -"/3(/3 + a)' 7. 1 1 _ a "/3(/3-a)' ^._ _J_ _^ _J_ ^ /3(^-a) _ /3(/3 + a ) ' ' a — b c — b a a ' = - 2/3, 1111.. .-. - + -+ 7 + 7=0. a c a — b c — b 4. If there be z counters of which z are marked m\ z .n,... with or without other marks; 2!„,^, w, w, with or without other marks; 2;^,,„ ,,,y... marked w, w, ^j, ^...; the number mimarked is s — 22 „+ 22;,„,„— 22;„,,„,p+..., 2 involvmg all combinations. 1850.] ALGEBRA. 83 Let Z>,,, denote the operation of selecting from the counters those marked with ??«, D^ those marked with w, &c. Then it is manifestly the same thing whether we first select from the heap those marked ;/?, and then from these, those also marked n ; or whether we first select those marked w, then from these, those marked m ; or at once select those marked 7n, n. Tliis may he symbolically expressed thus : D D = D D = D ; similarly, we have in general D D D ... = D D D ... = ... = D ...: (1). Also 1 — Z),,, will denote selecting those unmarked with m^ 1 — D^^ those immarked with ??, &c. Hence the whole number unmarked _ ( i _ 2) j (i _ i)J . . ,z^ = (1-22) +2i) -2i>,, „+...)z', since the spnbols D^^^^ A.v have been shewn (equation (1)) to be commutative ■^ m ' ^ m,n ^ m,n,p ^•••1 the required number, 5. Prove that a;' + / + (a; + y)« = 2 (.x" + xy^- ?/)* + SxY [x + yf (ar* + xy + f] ; and if x^ + xy -{• y^ = a, xy {x + \j) = i, and n be any positive integer, shew that ^^»+y^-+(^+^)^»^2a" + n(n-2)a-6- + "("-'^^;;;'^^"-'^ a-^- /^(»-r-l)(7^-r-2)...(ri-3r + l) ,_„.^,. 3.4. ..2r (a). Let 2 be a quantity, such that X + y + z = 0', then x"" + f + [x^-yT = x'" + /" + z'\ g2 84 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. and x'' + xy + y^ = {x-\- yj — xy^ = -z{x + y) -xy, = - {yz + zx + xy)j xy{x + 7/) = -xyz- therefore, taking the notation of the latter part of the question, yz + zx + xy = — a, xyz = — by therefore x, ?/, s, are the roots of the equation .•. {^ — x){^ — y){^ — z) = ^^ — af + hj identically j -i)(-D(-i)--M' l-|.(«-jy, .-. log(l-|)+log(l-|)+log(l-|)=logjl-^,(a-| x + t/ + z , 1 a;' + / + ^' , , 1 x" + /" + z'" , •• f +2 f ''"••■"'"2m f" "'"• 1 / 6\ 1 / i\^' 1 / i\" = rr-|; + 2rr-|j +-+s|-^«-|j+-; therefore equating coefficients of ^j , i_ (aj--"- + y2» + 2'-*«) = 1 a" + -^ {n-l){n-2) ^„^,^, 2n n n—1 1.2 J_ (n-2)(n-3)(n-4)(7^-5) , ■^71-2 1.2.3.4 "" ^ +••• 1 [n-r){n-r-l)...{7i-?,r+\) »i — r 1.2. ..2r n 2 2.3.4 (9i-r-l)...(n-3r+l) ^ 2.3. ..2« "* ^ +• 1850.] ALGEBRA. 85 .-. x'" + /" + [x + T/y = 2a" + n{n- 2) a"-'b' {n-S){n-A){n-5) ^,.^^, _^^ n + - 3.4 n{n-r-i)...{7i-Sr+l) ^ ^ 3.4.. .2r "" ^ ■^•••• Hence putting n = 4, we get a;' + y + (a; + yY = 2 [x" + xij + ff + S{x'' + X7/ + f) xY {x + yf. /aV'^'' 6. (a). If a be less than h, prove that ij-j is increased by adding the same quantity to a and b. (/3). And if w be greater than 1, shew that by means of this fonnula prove that ^ (a^ + a, +. . .+ a J" > n\a^. . .a^. (a). Since a is less than &, we may put a = h — c, where c is a positive quantity less than b ; then a and 5 will be increased by the same quantity if we increase Z>, c remaining constant j ^y "V~^j = a: suppose; •. loga;= {2b-c)\og(^l -^ , / c\ /, 1 c 1 c' 2 l\c /2 l\c' /2 1 ti-i which is manifestly negative, since the coefficients being of the form ( - j are positive ; and the absolute magnitude of the series is diminished by increasing b if c remain unaltered ; hence ( V 0+6 r) , is increased by adding the game quantity to n and h. 86 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. (/3).* First, let J be greater than unity = 1 + - suppose; we have then to shew that nj L n—1 J V " — 1 , these quantities when expanded by the Binomial Theorem become , + ^ + __^' + —i *+•■■' , n — \ ., V n—\j\ M— 1/ 3 . and 1 + a; + — 7-T x- + -—- x +... ; all the terms of both series are positive, and each term of the first series greater than the corresponding term of the second, or 1 + 7J "(^ + ^1) '^^^ U) >toj ' when a is greater than h. Cb . X Next, suppose j less than unity = 1 — - suppose, we have to shew that l-i^^ > J % or 1 =" w/ I w — 1 J V n — 1 (. -ii-i 1 j > (1 — x)''": these quantities when expanded by the binomial theorem become 71 — \ „ V 71— l) \ 71— \, 1 + ^^+ 1.2 ■'" + 1:2:3 ^ +■■•' 7^ .y \ 71 \ 71/ ., and 1 + X + ^^ a.^ + ^-^^^ ^^ + . . . ; * 'This part of the solution is given by Mr. Thacker in a recent number of the Vamhridfie and Dublin Mathematical Journal, No. xxv. p. 8L 1850.] ALGEBRA. 87 all the terms of both series are positive, and each term of the first series greater than the con'esponding term of the second, and therefore the proposition is true in this case also. Now let the n quantities «j, a^v^n? ^® ^^ ascending order of magnitude ; then V n / ' V na^ ) ' ^"^ 1'+ [n-l)a, I ' by the above, ^rt. -I- «,+...+ a n-l > a similarly ( -^ 5__^ « j > a_^ f _! * " w-l ' ' n-i f ^ I ^ , I ^ \ n-2 ? > n-l ' H , > a ,a : hence by multiplication or [a^-\-a^^-..,+ aJ' > n{a^a^..,aj. 7. If - be the r'^ fraction converging to — , and n' be the r*^ remainder in the process for finding the successive quotients, prove that m p _ p n q pq ' Let ^ , ^ be the {r - 2)'^ and (r - 1)* converging frac- 9'r-2 2'r-l tions respectively; m the r'^ quotient, n' the (r— !)'•» remainder; then i^ = »'^>r-, +i^r-25 88 SOLUTIUNS OF SENATE-HOUSE PKUBLEMS. [1850. And the fx"actioii — may be derived from - by wTitiug rii-\- —r, for 111 In the above expressions for p and q ; in '«' + :^' ) i^r-i -^Vr^ Now — is in its lowest tenns, and n is prime to n", .-. m = [m'n" ■^n)p^^^ + n"p^_^^ n — [m'n" + n) q^._^ + n'q^_,^^ , m p mq ^ np anci '^ -^ . n q nq Jig- * 8. Find the probability of drawing a black and a white ball the same number of times from a bag which contains an equal number of each ; the balls being dra^\ai one by one and replaced after each draAving, and the nmnber of drawings being the same as the number of balls in the bag, but this number is unknown, any number from 2 to 2n being equally probable. Suppose there are 2x balls in the bag; the number of drawings will then be 2a;, and the number of possible ways in which the balls may come out = 2'''''". Of these the number of favourable cases equals the number of permutations of 2x things taken all together, of which x are of one kind and x of another, _ 1.2. ..2a; ~(l.2...a;y^' _ (a; +1) (a; + 2)... 2a; ~ 1.2. ..a; ' therefore chance of proposed event on this supposition _ J_ (a- +1) (a; + 2)... 2a; ~ ^r 1.2. ...r I851.J ALGEBRA. 89 Now the chance of there being 2x balls in the bag = - what- n ever number > 0, > «, x may be. Hence the chance of tlie proposed event ^ ^^ !^l!i . .1 in+l)in + 2)...2n n (2M 2M.2 2'"' 1.2 1851. 1. If y , y7 be fractions in their least terms, the denominators of which do not exceed a given nmnber ??, the fonner fraction being given, and the latter detennined from it by taking for a and b' the greatest values of x and y (?/ not greater than w) which satisfy the equation hx — ay = 1, then of all the fi'action in their least tenns, the denominators of which do not exceed w, the fraction j-, exceeds y by the smallest quantity. ,,,. , a a ha — ah' 1 Wehave _ _ _ = _^^ = _ , since a', h' are values of x and y in the equation hx — ay = 1. Let ^ be any other fraction in its least tenns whose de- nominator does not exceed /i, then a a ha — a/3 ^~h^ ~ir^ ~ h^ m being some integer greater than 1 ; then a and ^ are values of X and y in the equation hx — ay = m. Now this equation is satisfied by X = ma ± qa^ y = mh' ± qh^ where q is any integer. Again, the successive values of x and y which satisfy the ccpiation ;,,y _ ^,y ^ 1^ = Ta say, 90 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. differ by a and b respectively ; hence, as b' is the greatest value of y, less than n, in this equation, b' + b> n, therefore h fortiori mb' + b> n. Hence /S cannot be of the form mb' + qb^ since it is less than n. Neither can it be of the fonn mb' ; for then a would = ?««', and p would not be in its least tcnns. Hence yS must be of the form mb' — qb : .•. /36 = [mb' — qb)b < mbb'j a a m 1 13 b ^ mhb' ^ bb' ' a a or of all the fractions in then* least terms, whose denominators t do not exceed ??, y? exceeds r by the smallest quantity. 2. The sum of the series -^ + -^^ + -^^^ + ... to infinity, where S is positive, is greater than (2^ — 2) "^ [^^ ~ ^)i ^^^ 1®^^ than 2* -=-(2^- 1). Let 8 be the sum of the given series. Then 1 1 /_2_ ^ _8_ > p+3 + 2 V2^* "^ 41+a "*" 8^3 "*■ ^ P "^ 2 V2^ ^ 4^ "^ 8"^ "^ >lH-i ' 2 2^-1 2«- 1 1851.] ALGEBRA. 91 Again, 1 /J_ J_\ /J_ _1_ _1_ Jl_\ 1 1 1 ^ jd ^ 2'» 4* 1 < 2«-l 2^ — 1 2^ Hence S lies between -i — ~- and 2* - 1 2^ - 1 * 3. Solve the equation 1 1 — X I I - a ^ X + a 1 = ; 1 — X X 1 — a a and thence infer the resolution of the first side of the equation into factors. We see at once that the given equation is satisfied by a; = a. Again, for a write ; , then 1 _ 1 _ _ 1 -a 1 - a ~ 1 ~ ~~ar * I — a 1 — a , = • = - a. a 1 1 -a Hence the given equation becomes 1 1 - X , 1 \ - a ^ X + a - 7 + — — = 0, I — X X 1 — a a of which a or is a root. 92 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Similarly, ; — is a root of the given equation. But a_-l a 1 a - \ 1 - a ' a therefore the roots of the given equation are a, , — ; and the first side may therefore be put into the form x[l —x) 4. From the equation ah[c + d-e -/) + cd{e +/- a-h) + ef[a + h-c-d) = 0, detennine, in terms of &, c, (7, e,y, the ratios a — c : a — d, and a — e : a—f\ and shew that the relation between the six letters may be expressed in the form P = Q^ where P and Q are each of them the product of three differences of pairs of letters, or in the form R= 8^ where R and S are each of them the product of four differences of pairs of letters. (a) From the given equation we get _ cd[h - e -/) + ef{c + d-b) ^ ^~ b{c + d-e-f) - cd + ef ^ _ c[b-d) (e+f) - he' + ef{d-b) + 6'd •'• "" """ b[c+'d-e-f) -cd + ef ^ "^1 b{c + d-e-f)-cd^€f =.-(b-d] ( '-')i'-f) (1) ^ ' b{c + d-e-f) -cd+ef ^^' Snmlarly, a-d=-{h-e) ^^ f^ ^^^_^_^ J ^^ ^y 5 — c _b — d c — e c — f - d b — c' d— e' d — f (2). 1851.] ALGEBRA. 53 In the same manner it may be shewn that a ~ e _h - f e — c e — d ,, ^irj-h^e'T^c'J^d ^^^' (yS) From (1) we see, by interchangiug c with e, d with /, which does not alter the original equation, that ^ ^ ' l>[e-\-f-c-d) -i- ef- cd^ a — c _ h — d {c — e){c —f) ^ ' ' a — e ^ ~ f (e — c) (e — c?) ' ••• («-«) (*-/) {e-d) = [h-d) {c-f) {a-e), which expresses the relation between the six letters in the fonn P=Q. (7) Again, by (2) we get («-c) [b-c) {d-e) (d-f) = {a-d) [h-d] (c-e) {c-f), which is in the form B = S. 5. If d^ + 1 be exactly divisible by j), and — be converted into a continued fraction, until two consecutive reduced fractions, — , — , , are found, such that p^ > n < n. then n ^ n ' ^ ' j^ = [na — mjpf + w". It is manifest that {na-mp) =np [-- -) ., ., (m mV by the property of Continued Fractions. -r, m m 1 But — = + — -, ; n a nn .'. (na — mpY < ^, ; and n''^ > p ; 94 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. therefore, h fortiori^ [na — mpf < p^ and n^ < p ; .*. {na — inj})'^ + n^ < 2_i). But {na — mj)f + n^ = n\d^ + 1) — 27nna.p + m^p\ which must be divisible by p, since a^ +1 is so : and {na — mpf + ri' has been shewn to be < 2^, and it must be positive ; therefore we must have {na — mpY + n^ — p. 6. If {a+yS+7+...}^ denote the expansion of (a+/3+7+...)", retaining tho^e terms iVa"/3V^''-" only in which & + c + ^+ ... l!(>p— 1, c + d+ ... ■;^ p — 2^ &c. &c., then *This theorem may be put into a rather more convenient fonn by wi'iting x — a. for a- ; we have then to shew that {x - a.y = x''-n {ay {x + /S)"'^ + ^ll^zil (« + ^Y {x + ^ + 7)""'' or, writing a^ for a, a,^ for /3, &c., (a. - «,)• = a;" - « (a.)' (* + «J- + '-^^^ (a, + «,f (.r + a, + ocj""' - " '""lilT"'' («.+«^+='.l' (-+«.+«.+=•.)"-' + (1). ' The proof of this depends on the expansion of the quantity {a, +«,+ ... + «,}". * For tlie solution of this problem we are indebted to Mr. Cayley. 1851.] ALGEBRA. 96 Expanding by the Binomial Theorem, (a^ + a,+ ...+a,)'' To pass to {a, + a^ + . . . + «/,}''. The sum of the indices of Og, ttg, ... a^,, are not to exceed^ — 1 ..., and generally the sum of the indices of a,., a;.^^, ... a^,, are not to exceed p — r + 1. Hence in (a.^ + ag + ... + a^)*", the required conditions will be satisfied, if only the sums of the indices of a^_,.+2, ap_,.+3j ■ • • «p do not exceed *• — 1, and the sum of the indices of a^.^^,, a^,^^^, ... a^ do not exceed r — 2, and so on. And this will be the case if, considering a^ + ag + . . . + a^_^+j «^ one quantity^ we replace (a^ + a3+ ...+a^)'by K^a + «3 + • • • + Vm) + V-+2 + • • • + a;,}^ Hence {a, + a,+ ... +a,|^ = a,^ + |a/"M(«2 + "3+ ••• V^+i^f)!' + f ^^ «/"' {(a. + "3 + • • • + a,_,.) + a^_^Y + ... + pa^{a, + a^ + ... 4 a,}""' (2), the last term (a., + a^+ ...+ OLp)" being of course rejected alto- gether, since the sum of the indices exceeds^ — 1. Now the coefficient of - — - — ... (- a,)*", on the 12 r left-hand side of equation (1), is cc"~'. On the right-hand side, it is, expanding each of the quan- tities in the brackets ( } by (2), (aj + a, + ... + a,J"-'--^((a,+ ... + a,J]'(a^ + «, + ... + a,,,)"-'-^ n-r-1 r+2/ 96 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [I8iil. or, if wc write y fur x + a.^+ ... + a,^,, to for a.^ + ... + a,,^,, the coefficient on one side is (^ — w)""'', on the other «-r " ' f 11/ , \M-'-l » — r » — y — 1 , ,,, , Ml-,-!! + —J 2 ^"^ "^ "''■^•'^ ^^ "^ ^'^' ^ °''*='^ ' + Therefore we have to shew that n — r n — 7' — 1 , lo / , , \»i-)--> + _ _ {a, + a ,,r {y + a .,, + a ,3)" "-... the theorem itself, wiiting n — r for n. The coefficients of a^**, on each side of the equation (1), are obviously equal. If then the theorem hold for the indices 1, 2...(m — 1), it is proved to hold for the index n. But it obviously holds for the index 1 ; therefore it holds miiversally. 7. If = a + — ^-— , and — —, — = a + , , : mp + X ]} + It mj) + X p + 7^ ' then supposing ic and u to vanish when x vanishes, 1 1 . ,s -7 — = m (a. ~ a. . u ti Since u and u' vanish when x vanishes, we have 1 B 1 . )8 — = a+— , — , = a + — , mp J) tnp p Hence, by the first of the given equations, p + u 1 Tnp + X m X .'. p + u = au {mp + a?) + ^ H oc = w ( 1 - mB + oLx) + » — ; 1851.] ALGEBRA. 97 .'. u (m8 — ax) = — , m m'/3 1 .'. ma = - . X u Similarly, by the second of the given equations, m^^ , 1 ma = —: X u the required result. 11, -, = m a — a u u H ( ^'H ) PLANE TRIGONOMETRY. 1848. A cannon-ball is moving in a direction making an acute angle 6 with a line drawn from the ball to an observer; if V be the velocity of somid, and nV that of the ball, prove that the whizzing of the ball at the different points of its com'se will be heard in the order in which it is produced, or in the reverse order, according as /? < > sec 6. The whizzing will be heard in the ordei' in which it is pro- duced, or in the reverse order, according as the sound or the ball moves more quickly towards the observer. Now the velo- city with which the ball moves towards the observer = wFcos d ; hence the whizzing will be heard in the natural or reversed order, according as V >< n V cos 0, or no sec 0, the required condition. „ j^ sin (a — jS) _ sin (a + ^) sin yS sin ^ ' shew that cot /3 — cot = cot {ol + 0) + cot (a — /3). Since sin (a — /3) sin (a + 0) sin /3 sin 1 1 sin /3 sin (a + 0) sin sin (a — yS) ' sin {a + 0-^) _ sin (g - /3 + 6') ^ sin yS sin (a + 0) sin sin (a — /3) ' .'. cot yS - cot {a + 0) = cot + cot (a - /3), or cot 13 — cot = cot {a + 0) + cot (a — /S), the required result. PLANE TRIGONOMETRY. 99 1849. 1. If COS a = cos ^ cos <^ = cos yS' cos 0', and sin a = 2 sin ^(/> . sin ^0', shew that tan \a = tan \j3 tan ^^8'. Since sin a = 2 sin \^ sin |^<^', .-. sin' a = (2 sin' :|<^) (2 sin' ^ <^') = (1 — cos 0) (1 — cos 0') ; , / cosa\ / cos a ,1 — cos a = 1 :x 1 - cos/Q/ V cos/37 ' .'. cos a = sec /3 + sec /S' — cos a sec /8 sec /8', , sec B + sec yS' cos B + cos /3' and cos a = :, ^^ 7^ = 7^ p=; ; 1 + sec p sec p I + cos p cos ya ' 2 <* _ 1 ~ cos a _ 1 — cos y8 — cos /8' + cos /3 cos y8' 2 1 + cos a 1 + cos /8 + cos /8' + cos /8 cos yS' ^(l-cos^)(l-cos^;) g ff (l + cosy8) (l+cosyS) 2 2 ' a /8 yS' and tan - = tan — tan — . 2 2 2 2. Find •=• from the equations [a + h) sin d + [a - h) cos 6 = {ci' + Jr)\ a sin' e + b cos' = {Zah)K Squaring the first of the given equations, we get {(i'+V') [mi'd^- cos'^) - 2ah [cos'S-sm'S) + 2(a'- J') sin^cos^=a'+ b'; .-. (a' - &') sin 2^ - 2rtZ> cos 2$ = 0. Let - = tan , then this equation gives sin 2 (^ - 4>) = 0, whence 2 (^ — ^) = or tt ; .". 6 = (f) or (f) + ^TT. 11 -J 100 SOLUTIONS OF SENATE-HOUSE PROBLEMS, [1849. Taking ^ = , the second of the given equations gives a siii^ <^ + i cos' = (3aJ)*; 'fTI + ^7-^ = ('«^^^-' a J \ a • • a = (3«&)*, and 72-3y+l=0; o o « 3 1 , , , ••■5 = 2±2(-')' (')• Again, taking ^ = ^+^7r, the second of the given equations gives a cos^ (f> — b sin'* )*, a — b ,„ ,, , 1 = {^^^)H {a'-^bj .-. d'- W= {a' + bf {3abf, a* - Id'V + b^ = Sab{d' + V'), (a^ + yy _ ^.d'b'' - dab (a' + J'^) = ; ... (a^ + J2 _ 4^J) ^^2 _^ J2 ^ ^j^ ^ . therefore, fii-st, a''* + J^ — 4a5 = 0, ^'Z Kb. and ^ = 2±(3)i (2): 1850.] PLANE TRIGONOMETRY. or, secondly, d' + h' -V ah = 0, and ••• ?=-i±i(-3)* 101 (3). The six values given by (1), (2), and (3), are all the values which Y admits of. o 1850. 1. If through the angles of a square four straight lines be drawn externally, making the same angle a with the successive sides, so as to form another square, find its area. Let a be the side of the interior square : then the area of the exterior square = interior square + four triangles each equal to ^d^ sin a cos a, or = d^ + 4 x ^a^ sin 2a = d\l + sin 2a). 2. Shew that 2tan-Mtann45°-a) tan 1/3} = cos"' /J^^1^L±£^^ ^^ ^ / 2A-J Vl + tanacosyS; ' if a be < 45°. In general cos (2 tan"' j:;) = 2 cos^(tan~'cc) — 1 2 1 Let X = tan* (45° — a) tan^/3, then ^ ^ 1 + tan(45 -a)tan'^/3 1 — tana 1 — cosyS _ 1 + tang 1 + cos^ 1 — tana 1 - cos/3 1 + tana I + cos/3 _ tana + cos/3 ~ 1 + tana c<>s/3 ' 102 SOLl'TK.lNS OF SHNATE-HuUlSE rKOliLK.MS. [I«o0. .-. 2 tan-' {tan*(45" - a) tan^/3} = cos- ( tan« + cos^\ If a were > 45°, the given expressions would become imaginary. 3. Draw AB and AC (fig. 58) at right angles to one another, and make AB equal to twice AC] produce CA to D until CD is equal to CB: prove that BB will be the side of a regular pentagon inscribed in a cii'cle, of which AB is the radius. Also, if with centre B and radius BA we describe a circle AEF^ of which ABF is a diameter, and make AE equal to AB^ then FE will be the side of a regular pentagon circumscribing a circle, of which A C is the radius. (a) Let AC = a, then AB = 2a ; .-. CB=CB= [AB' + A Cy = bhi ; .-. AD= (54-1) a, BD = {AD^ + ABy = (6 - 2.5* + 4)*a = (10 -2.5*)* a , (10-2.5*)* = 4a ^^ — 4 = 2 sini7r.2a = 2 sin lir.AD ; therefore i?i) is the side of a regular pentagon inscribed in a circle, of which AD is the radius. {^) Again, FE'' = AF'-AE' = {2.ADY - AB' = 4(6-2.54)rt"'' - 4a' = (20 - 8.5*) a' = 2^(5-2.5*) [aY = {2tsinl7ryAC'; therefore FE is the side of a regular pentagon circumscribed about a circle, of which AC is the radius. 1850,] PLANE TRIGONOMETRY. 103 4. If (?, hj c, be the sides of the triangle ABC, 2^1 ?> ^ lines bisecting the angles drawn to the opposite sides, and p\ g-', r these lines produced to meet the circle which circumscribes the triangle; shew that COsi^ COsii? COsiC 111 — -I- = — -k^ = — = — I 1 — J) ^" •""^ r a c p cos^-4 4- q cob\B + r cos^O = a •\- h + c. (a) Let AD (fig. 59) be the line bisecting the angle BAG, i>, D' the points in which it meets BG and the circumscribing circle respectively. Draw DG, DH, perpendicular respectively to AB, AC] then DG = DH = p m\^A, and D G.AB + DKA G=2 (area of triangle) = AB.AC sin A', .'. p sin^-4 {h + c) = be sin J. = be 2 sin|^^ cos^^ ; 2 cos4^ 1 1 p be .,. ., , 2cosi^ 1 1 Similarly, ^— = - + - , •^ ' q c a 2cosiC 1 1 i. = I • coshA cos^B cosA(7 111 p) q r a b-^c (/3) Again, join BD', CD': these lines will be equal to one another, since they subtend equal angles at the circumference. But BD" = c' + p' - 2pc cos^^, GD'^ = V' + p" - 2pb cos\A ; .*. c" — 2p'e cos^^ = ¥ — 2p'b cos^^ ; .•. 2p' cos^-4 = J + c. Similarly, 2q cos ^5 = c + «, 2r' cos^(7 = rt + i; .'. p cos^^ + q cos ^5 4 )•' cos| C = a -\- b -{■ c. 104 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. 1851. 1. If ABC be a triangle right-angled at (7, E the point in which the inscribed circle touches BC^ and F the point in wliich the circle drawni to touch AB and the sides CL4, CB produced meets CA : shew that if EF be joined, the triangle FEC is half the triangle ABC. Let r, / be the radii of the inscribed and escribed circles respectively ; then, since is a right angle, CE=r and CF=r\ and triangle FEC = \rr' [s — a) [s — h) (s — c)]* {s[s — a) (s — Z>)]* = i s adopting the usual notation, = i(6"-a) {s-h) 2 = ^ (Z> + c — a) [a-^c — b) = 1 [a^ + F -{a- h)'] -.' c' = a' + b'\ = ^ah = I the triangle ABC. 2. Shew that sin/3 sin 7 sin [y — /3) + sin 7 sin a sin (a — 7) + sin a slnyS sin {^ — a.) + sin (7 — /3) sin (a - 7) sin (/3 — a) = 0. We have, in general, sin^ sin 5 slnC = ^smA{cos{B-C)-cos[B+C)} = i{sm{B+C-A) + sm{C+A-B) + sm{A + B-C) -mi{A + B+C)}. Hence, if ^ = /3, B=y, C=y-B, sluyS sin 7 sin (7-/3) = i (sin 2[y — /3) + sln2y8 -'feln27}. Similarly, sin7 sina sin (a — 7) = 4 {sin 2 (a — 7) + sin27 — sin2a}, sin a sin^ sin (/3 — a) = ^ {sin 2(/3 — a) + sin 2a — sin 2/3}. 1851.] PLANE TRIGONOMETRY. 105 Again, if ^ = 7 - ^, i? = a - 7, C = /S - a, sin (7 — /3) sin (a — 7) sin (yS — a) = i{sin2 (iS - 7) + sin2 (7 - a) + 8in2 (a - yS)} ; therefore, adding these equations, we get sinyS sin7 sin(7 — yS) + sin7 sina sin (a — 7) + sina siii/9 sin (/3 — a) + sin(7 — ;S) sin (a — 7) sin(/3 — a) = 0. 3. The equation sina; = has not any imaginaiy roots. We have — ^ 2 smx = e""" — e '^. Now, every imaginary quantity may be expressed under the form a H — */3. Substituting, then, this quantity for a*, we get -Ij: — ix --, -o — i, a £ — £ =£"£/* — £ ' S.P = cos a (e"'' - £^) + -* sina {e~i^ + z^) ; therefore, if sina; = 0, we must have cosa(£'/^-£/') = 0, sina(£-/^ + £'^) = 0. These require, either that cosa = and £~^ + e'' = 0, which cannot be satisfied by any real vakie of /3 ; or that sina = 0, and £~^ - e'' = 0, which can only be satisfied by yS = 0, shewing that a; = a, a real quantity : whence the equation sin a; = has not any imaginary roots. 4. If the cosines of the angles A^ B^ C\ of a plane triangle be in arithmetical progression, shew that s — a, s — h^ ^ ~ ^1 "^^ be in hannonic progression, s being the semi-siun of the sides. We have cosvl = 1-2 sin'^^, cos5 =1-2 sin'^ B, cos C = 1 - 2 sin'^ (7; therefore, if cos^, cos^, cos 6*, are in arithmetical progression, sin'*^^, sin'' ^5, sin'^'^C, are so; lUG .SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. ^ (^s-a){s-c) _ {s-a){ s-b) {s-c ){s-b)^ ac ab CO 2h c a + ' s — h s — c s — a or 2[^--(5-Z>)] ^ s-[s-c) ^ s-{s-a) ^ s — b s — c s — a '' 2s s s + s-b s — c s — a s — b s 1 1 + c s — a whence , 7 , , are in arithmetical progression ; .-. s — a , s - b , s — c, are in hannonieal progression. 107 SPHERICAL TRIGONOMETRY. 1848. 1. In a right-angled spherical triangle, shew that sin a tan^^ — sinb tan^J5 = sin (a — b) : shew also that if ^ be the spherical excess, . 1 T-, sinAasini& , „ cosia cos^b cos^c cos^c (a). We have in general , , 1 — C08-4 tan*^ = ; — 5 — , ^ sin^ ' , . . sin(7 . sina and sin>4 = — — sma = -; — , suic sine since C is a right angle ; .•. sina tan^^ = sine (1 — cos^), = 8inc(l — tan6 cote) by Napier's rules, = sine — tanZ> cose : similarly sin 5 tan ^5 = sine — tana cose; .'. sina tan^^ — sinJ tan ^5 = cose (tana — tanZ*), cose . , ,, = 7 sm a — 6 . cosa cos 6 But by Napier's rules, cose = cosa cos^, .•. sina tan^^ — sin J tan^^ = sin (a — b). (^). Again, sin'^^a sin^^i + cos^'^a cos'^^J = :^{(1 — C08a)(l — cosi) + (1 + cosa)(l + cosJ)}, = ^(1 +co8a cos J), = ^(1 +co8c) by Napier's rules, = cos"''^o; 108 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. fsin^asin^bV /cos^a cos|/a' • 217;. . 2 if /,\ I - ^_ \ _|. / f — — ±_\ = sin^i!/ + cos^ii...(l) : V cos^c / V cos^c ami in any spherical triangle w^ r,\ cos^ia — b) .^ ^^ ' cos^(a4-J) also .1 + 5 = 180° + ^ - C, hence cotMC-^) = ^i^|cot|C; ^^ ^ cos(^a + ^>) ^ ' therefore, since C is a right angle, 1 + tan^JS* 1 + tan^a tan ^5 1 — tan^^" 1 — tan^a tan^J ' . tanij5; = tan^^a tan^J, sin^^ sin^a 'm\\b cos|rt cos^Z* cos^^ cos^^c ' cos^c ' ), sin^^ = sin^a sin^J cos^c ' cos4^ = cos^a cos^J therefore, by (1 cos^c the required formnlse. 2. If three small circles be inscribed in a spherical triangle, having each of its angles 120°, so that each touches the other two as well as two sides of the triangle, prove that the radius of each of the small circles = 30°, and that the centres of the three circles coincide with the angular points of the polar triangle. Let ABC (fig. 60) be the triangle, draw the great circle AD to D the bisection of ^C; let be the centre of one of the circles, through draw the great circle EOF perpendicular to BG and intersecting AD in E^ FE will be a quadrant ; ajso join ^0 by a great circle BOH^ 50^ will bisect the angle ABG\ draw OG the great circle perpendicular to AD. OG and OF Avill each be equal to r the radius of the small circles : let BG=2a, BF=x, FD = y. Then, in the right-angled triangle ABD, coa A BD = tani5Z> cot ^5; 1848.] SPHERICAL TRIGONOMETRY, 109 or, since L ABD = 120°, 1 1 — tan'' a — h = tana — — , ^ 2 tana ' 1 — tan''a 2 ' .-. tan'^a = 2, and cos a = —^ . Again, in the right-angled triangle OBF^ smBF= cot OBF.tan OF, 1 or smcc = — ^ tanr, and in the right-angled triangle FO G, OF = 90° — r, and / OFG = FD = 2j, and sin06^ = sinO^sinO^^?, or sin?' = cos?- sin^, .•. amy = tanr ; and cosa = cos(x + j/), = (1 — sin'''a;)*(l — sin^i/)* — sin.r siny, or = (1 - ^tan'''>-)4(l - tan'r)* - ^ tanV ; 2 .*. cos'' a -I- rj cosa tan^'r + ^tan*r = 1 — ^tan^r + ^ tan*?-, or i + I tanV = 1 - | tan^ r, .*. tan'"'?- = ^, and r = 30°. Again, let BOH = h, BO = z- then, in the right-angled triangle BHCy cos BC = cosBH.cosHC, or cos 2a = cos J cosa, .*. 2 cos"''a — 1 = cosa cosi, ^^ - 3 = 51 C086, •.• cosa = -. , and cos5 = — wi) 3* ' • -"^^^-34 1 110 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. also in the right-angled triangle OBF^ sin OF = sin OBF.m\ OB, 3* or \ = — sin z : 2 2 ' .'. sin2; = — r = — cosA, 3* ' .-. J = 90° + 2;, and 0/f is a quadrant. Again, joining OC, we have cosOC = cosO^ C0SJ5C + slnO^ sin 00 cos OB C, = cosz cos2« + sinz sin 2a cos 60°, _ _ 2^ 1 1 2.2* 1 ~~3*'3'^P~3~2' = 0, and 00 is a quadrant, so is also 0-ff; therefore is the pole of -40: similarly, if 0', 0" be the other centres, 0' is the pole of CB, and 0" of BA ; therefore 0' 0" is the polar triangle of ABC. 1849. 1 . If P be the perimeter of a spherical triangle, of which the angles are A, B, 0, and the spherical excess F, prove that ^.^^^^ [sin^E smjA - ^E) sin {B - ^E) sin ( 0- ^E)]^ ^ 2sin^^ sin^iJsin^O By the expression for the sine of a side of a spherical triangle in temis of the angles, {sini^sin(^-i^)sin(J5-i^)sin(0-i^)}i=isinasin5sinO, = ^smh sinO sin^ = |sinc sin^ sinJ5, 1 2 = ^(sina sini sinc)^(sin^ sin5 sinO)^ (1). Again, siniPslnfiP-a))i . ( sin(^P-Z>) sin (^P-c) ]* I ^ [ sin 5 sine J ' COsi^ = ^± »un^2. ( smo smc with similar expressions for the cosines and sines of ^B and ^ ; cos"!^ cos^'iPcos'-'iO sin^'iP sini--4 siniP sinAO sin a sin?; sine ' I Hoi.] SPHERICAL TRIOONOMETRY, 111 sinasinisinc ~ sin'^^ sin''^i?sin''^C ' _ sin''^ sin'^i? sin'^6' . 1 r» 1 / • • 7 • \3 (sin^ sinjB sin (7) .". sin^i^= i(sma sino smc . , . — . , „ . — rr^, ^ *^ ^ sm^A sm^B sm^C^ _ {smEsmjA - ^E) sm{B - ^E) smjC - ^E)]i ~ 2sln^^ sin^Z? sin^C ' bj (1); the required formula. 1851. 1. If ABC be a spherical triangle, right-angled at C, and cosvl = (cosrt)^, shew that b + c = ^tt or |7r, according as b and c are both less or both greater than ^tt. By Napier's rules cos^ = cosrt sin 5, but by the conditions of the problem cos^ = cos'''a, .•. cos a = sin 5. Again, by Napier's rules cose = cos a cos J, = sini? cos J from above, sinJ5 . , , = . , smo coso, smo ' = — — smb cos J, since is a right angle ; .•. sine cose = sin?> cosJ, or sin2e = sin 26; and b is not equal to c, as then B would be a right angle, and A would equal «, which is contrary to the equation cos^ = cos^/; ^euce 2& + 2c = TT or Stt. Now b and e are both greater or both less than ^tt, since cos^ or cos'^a = tan J cote ; therefore 2J + 2e = tt or 37r, according as b and e are both less or both greater than ^tt. 112 THEORY OF EQUATIONS. 1849. 1. Given y = .rs", prove that - = 1 + 717 + ^ + ••• +^T^'+ ••• X [2 [3 \n One root of the equation ?/ - a-c^ = is the coefficient of - m the expansion of — log [ 1 ) . (See Murphy's Theory of Equations^ p. 77, Art. 62, and p. 80, Ex. 3.) Now -log 1 = — +-^ + ...+ -+ ... V y I y ^ y ^ y Expanding the exponentials, we see that the coefficient of - is ^x' i'x^ n"-'x 2 1.2.3 1.2 .„ n ' which is therefore a root of the given equation. \«-i -u- y ^ 2x i^xY inx) Hence ^ = i + — + ^-—L + ... + v — L_ + ... X [2 [3 \n 2. If a^j, x^...x^ be the roots of an algebraical equation, and no two of them be equal, then 1 1 1 1 jn-1 7n being a positive integer less than n. 1849.] THEORY OF EQUATIONS. 113 (a). Here f{x) = {x-x;){x-x^),..{x-xj, we may therefore assume 1 A, A A f{x) X - X^ X - x^ X — X 1 -4j, A^y..A^^ being independent of a;; .-. l=A^{x-x;){x-x^)..,{x-xJ+A^{x-x;){x-x^)...{x-x^){x-'X^)+... + A^^ {x — x^)...{x — a;,^_j) identically ^ Hence, putting x = x^^ 1 = A^{x^-x^){x^-x;)...{x^-x„), = Af{^.) (1); similarly 1 = AJ'{x^) (2), i = A/'K) W- Hence 1 ^ 1 1_ _ 1_ _ / (^) ~ (•» - a^i) />i) (^ - a'J f\^,) '"'^ ix- xj f{^J identically. Therefore putting a; = 0, which makes /(a?) =^„, J_____l 1__ _ 1 \_ 1 1_^ 1 _ 1 / ^ 2CJ and similarly for the other fractions. Hence 1 _ 1 / ^ ^ \ _1 H-? + ^+-.l+z>,VJl+f + |{ + ...) 1 /, rr cr' \ + ... +-rrr-s 1 + --f -^ + ... (1). 1 114 SOLl'TIONS OF SENATE-HOUSE PROBLEMS. [1851. B^t J_ ^ 1 ^ 1 f{x) x" +2hx-' + ... +2)„ ^n /j ^l\ _^ _^ A V X '" x^ a; V •>c Hence, if m be a positive integer, less than «, the coefficient of ^ in 7^— r = ; and the coefficient in the right-hand member X J-[x) of (1) is 1 — J — ? i- -I — « — • ^ ffl-l ^ m-i ^ m-i hence J. . + ^^tt — r 4- • . . + ■^, — ^ = 0. 1851. T,^ 1 1 — i^ « 1. If x+ = - 3;?, 1 — X X -^ then (x + y-^ - 0)'-' i^) = - 27 (^; - a>) (^ - a>y, where (o is an imaginary cube root of unity. 1 1 — X Call the quantities a;, and ^ , ^/j, j/^? ^^d 3/3 5 then y^+y, + yz = - ¥ (i), a^ 1 and y,y, + yjj,^y^, = Yzr^-x~ ^^"^^ = 1 1 + a; - 1 1 — a; X = y, + y,■^y^-^ = -3(i^ + l) (2), also y,y^y^ = - 1 (3). Again, x^ 1 (1 - a;)" y^y^ + y^yz + 2/3V1 = j-^r^ - ^^fz^ + ^ _ 3x{\-x) _ ~ x{l-x) ~ ^*^' 1H51.] THEOllY OF EQUATIONS, 11/ and y'^y^ + y^y^ + y'^y^ + y'^y^ + y~y^ + y; y^^ = -y^ {3 (p + 1) + ^,3/3} + -^7(2), = 9p (^; + 1) + 3, by (1) and (3) ; ••• y"y, + y"y. + y'y, = 9i? (i> + 1) + e ... by (4) and (5). And [y^ + mj.^ + a)'V3)' = j/^ + y/ + ^3"^ + 3^?/,y, (?/, + 6)yJ 4 similar tenns + ^y^y^^ {3/1 + (*" + ^) 3/2} + similar terms = (-3i^r- 9(&)-l) + 3( + &)'•* = — 1, = - 27 (/> - ft)) (p - ft)7'' 12 ( no ) GEOMETllY OF TWO DIMENSIONS. 1848. 1. With two conjugate diameters of an ellipse as asymptotes a pair of conjugate hyperbolas is constructed ; prove that if one hyperbola touch the ellipse the other will do so likewise ; prove also that the diameters drawn thi'ough the points of contact are conjugate to each other. Let the equation to the ellipse, referred to the conjugate diameters, be 1^ + ?-=' W- And to the hyperbolas ^1/ = ^" (2), ^I/ = - c' (3). (a) In order that (2) may touch (1), we must have ^ _^ , ^ a perfect square, in which case we shall have also X" xy y^ a c a perfect square, and (3) ^v^ll also touch (1). (/3) . If the above expressions be perfect squares, we see that 4 _ J_ - ah ••• ' =Y' and - = ± - ; X a 1848.] GEOMETRY OF TWO DIMENSIONS. 117 and \i x'y be the coordinates of the pohit where (1) meets (2), ,, he' W ^ =2- Similarly, if x"i/" be the coordinates of the point, where (1) meets (3), „„ b ,,,, a y'=2^ -^ =2- And if r\ r", be the lengths of the corresponding semi- diameters, CO the angle between the axes, r"' = x'^ + 2^'^ — 2x1/' cos w = ^[d' + b^ — 2ab cos to), r'"' = x"^ + y'"'' — 2x"y" cos g) = \[c^ + i^ + 2ah cos oj), ... r"' ^r"' = ce + h'; therefore ?•', r" are conjugate to each other. 2. Shew that the curve which trisects the arcs of all seg- ments of a circle described upon a given base is an hyperbola whose eccentricity = 2. Let AB (fig. 61) be the base, a its length, AC = CD = DB = r, CAB = 0, we then have r {1 + 2 cos 6) = a, or, referring the curve to A as origin and AB as axis of a;, x' -\- y' = [a-2xY'^ .-. 3j?' - / - 4aa7 -|- a' = 0, the equation to an hyperbola, the squares of whose axes are to one another in the ratio 3:1, and whose eccentricity therefore = (3 + 1)^ = 2. 3. Let Z) be a point in the axis-minor of an ellipse whose eccentricity is e, S the focus, the centre of curvature at the D 9 extremity of the axis-minor ; with centre D and radius = — — 118 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. describe a circle ; shew tliat this circle will touch the ellipse or fall entirely without it, according as D is nearer to or further from the centre than the point 0. Let C be the centre of the ellipse, and let CD = h^ also let a, J, be the semi-axes of the ellipse, its equation will be a o Also DS' = d'^ + li^^ so that the equation to the circle will be 72 a;"' + [y + ]if = a^ + 4 (taking D below C), 1 - e' or .-r^ 4- y + 2/^ J/ = «' + K' — ^— . Where this meets the ellipse, we have y W aV «■' o' e' "' or ^-^/ - 2% + A'' -^ = 0; .-.3/ = -^- h. If this value of y give a real value for a-, the circle will touch the ellipse, if not, it will fall entirely without it, since its radius f ct' + -5 ) is greater than a, and therefore, a fortiori^ than DB\ which is less than h. In order that the value of x may be real it is necessary that y be not greater than Z>, therefore \-e' , e' or he' <-^-h, ? CO^ i.e. as D is nearer to or further from the centre than 0. 4. PSp is any focal chord of an ellipse, A the extremity of the axis-major ; AP^ Ap meet the directrix in two points Q^ q : shew that I QSq is a right angle. We may prove this property for any point in the ellipse by a process exactly similar to that of Part I. Conies^ 1848, 3 ; except that we have the equations sin PBS = e sin PSR. sin PRN^ sin QR8 = e sin QRS sin PRN, instead of those there given. This theorem may also be proved by the method of Reci- procal Polars. (See Salmon's Conic Sections^ chap. XIV.) Take the polar reciprocal of the whole system with regard to the focus S. To the ellipse w^ill correspond a circle, to the point P, 2J^ two parallel tangents Rt^ rt\ (fig. 62) variable in position. To A (or any point in the curve) will coiTespond a fixed tangent tt\ and to the directrix the centre S'. Hence to AP^ Aj) will correspond the points t^ t' respectively, and to Q^ q the lines St^ St'. But it is easy to see that the lines Sf, St' are at right angles to one another ; therefore the line joining the points Qj j, subtends a right angle at the focus S. 5. In the given right lines AP^ AQ, (fig. 63) are taken variable points ^j, q, such that Aj) : pP :: Qq : qA • prove that the locus of the point of intersection of Pq, Qp is an ellipse, which touches the given right lines in the points P, Q. Let AP = o^ AQ = 5, Ap = a, Aq = yS; then the conditions of the problem give a:a-oi::h-^:/3, or - + f = 1 (1). an 120 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. Take AF^ AQ, as axes; then the equations to Pq,pQ respec- tively, are M = ' (^). M- (^)- o a whence, eliminating a, yS from (1), we get X y a x^ xy if ^ X y a ab 1/ a h the equation to the locus of the intersection of Pq^ p Q^ which, since the square of half the coefficient of xi/ is less than the pro- duct of the coefficients of x^ and 3/^, is an ellipse. When £c = 0, we have J, J ^ + 1 - 0, •'• 3/ = ^ shewing that the ellipse touches AP in P. From considerations of symmetry it is evident that it also touches AQ m Q. 6. A parallelogram is constructed by di'awlng tangents at the extremities of two conjugate diameters of an ellipse ; prove that the diagonals of the parallelogram form a second system of conjugate diameters, and that the relation between the two systems is reciprocal. 1848.] GEOMETRY OF TWO DIMENSIONS. 121 Let the equation to the ellipse referred to the conjugate diameters be <+C = i (.). b The equations to the tangents, drawn at the extremities of these diameters, are a; = a, x = — a^ y = h, y = -h] therefore the equations to the diagonals of the parallelogram thus formed, are ^ ., M (^)> I=-f («)■ At the points where (2) meets (1), we have a b therefore the equations to the tangents at these points are M=±^' W' therefore these tangents are parallel to (3). Hence the diagonals fonn a system of conjugate diameters. Again, the equations to the tangents at the extremities of (3) M = ±^' («)' and at the intersection of (4) and (5), we have either cc = 0, or 3/ = 0, shewing that the diagonals of the parallelogram, formed by the lines (4) and (5), are the first system of conjugate diameters; hence the relation between the systems is reciprocal. 7. PSp is any focal chord of a parabola whose vertex is A, prove geometrically that AP^ A_p will meet the latus-rectum in two points Q, q, whose distances from the focus are equal to the ordinates of the points j7 and P respectively. 122 auLUTIUNS OF SENATE-HOUSE PROBLEMS. [1848. Draw P3/, prn^ (fig. 64) ortllnates to the points P, p re- spectively. Then since SQ is parallel to Pil/, .-. SQ:AS:: MP : AM, .-. SQ:4.AS'':: MP: ^AS.AM, :: MP: MP'] .-. SQ.MP=^AS\ Now SP=2A8-\- SM, = 2A8-\- SPcoaPSM', .-. SP{l-cosP8M) = 2^^, .-. Pil/(1 - cosPSM) = 2A8s{nP8M: similarly pm[l -\- cos 2)8 m) = 2 A8 sinjy 8m j .'. PM.pm = 4:A8'\ = 8Q.PM from above ; .-. ^wi = 8Q, similarly PM = 8q. Or the distances of Q, q, from the focus are equal to the ordinates of the points ^j, P respectively. 8. From a given point in a conic section, draw geometrically two chords at right angles to each other which shall be in a given ratio. The construction which we shall give depends on the pro- perty that all chords of a conic section which subtend a right angle at a given point P of the curve, intersect the normal at P in a fixed point. Draw PK (fig. 65) the normal at P, and draw PU^ PV any two chords at right angles to one another. Join UV, cutting the normal in K. Then by the property above enunciated, if PQ, PR are the required chords, QR will pass through K. Again, PR t3in PQR = -^^ a given ratio, hence the angle PQR is known ; on PK we describe a segment of a circle containing an angle equal to PQR ; let it cut the ellipse in Q. Join PQ, and di'aw PR at right angles to it, PQ, PR will be the required chords. 1848.] GEOMETRY OF TWO DIMENSIONS. 123 9. DeteiTnine the equation to the conic section wlilch passes through five points whose coordinates are given ; and thence shew that the equation to the conic section which passes through the five points whose coordinates are 1,-1; 2, 1; -2, 3; 3, 2; -1,-3, is 6iy - llxy - 65a;^ + 36^/ + 174a; - 151 = 0. Let ajj, y^ ; x^, y^ ; a-,, y^ ; a„ y^ ; a,, y.^, be the coordinates of the five given points, which we shall call yl,, A^^ A^j A^j A^ re- spectively. Then the conic passing thi-ough A^y A^^ A^j A^^ circumscribes the quadrilateral, whose sides are A^A^j -^3^4j A^A^^ A.A^. The equations to these sides are ^-^2 y-y^ =0 ^2 - ^3 X - ^3 ^3 - ^4 X - X, ^4 - ^5 X — ^5 y, - y. y-yz y, - y. = 0, y-y. = 0, y,-y. y -y. = 0. ^5 -^2 y,- y^ Now the equation to a conic, circumscribing a quadrilateral, the equations to whose sides are ii^ = 0, u^ = 0, u^ = 0, u^ = respectively, is ^ ,, ^ -^^.^ 2 4 3 0/ \ being an indetenninate parameter. Hence the equation to a conic passing through A^, A^^ A^j A^^ '^ 'X- x^ y - y\ (X - x^ y-Jh} ■^2 - ^3 yt- yJ ' ^^4 - a^6 y,- yJ fx-x^ y - yy ^fx-x^ y- y.^ v^3 - ^4 3/3 - y,^ ' \^S - ^2 3/5 - 3/2. = \ The quantity X is detennined by the condition of the conic passing through the point A^{x^y^) ; this gives ' ^1 - -^4 y, - y , ^4 - ^5 y*- Vi '^1- 3_ 3/1- ^ '^2- 3/2- ■yJ r^ -•^3 _^ -y. y^ - yj v^6 - ^2 .'/.-, - .'/■> 124 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. Eliminating X between tliese last two equations we get, clearing the quantities within the brackets of fractions, {[x-x^){y-y,)-[y-y,][x-x,)\\{x-x:^{if-y^)-{tf-y,)[x-x^)] {(^,-«'2)(3/2-^3)-(3^-2/J(^2-^8)}{(-'^-'^4)(3/-y5)-(yi-3^4)(^-a^6)l { (-^ --^a) [y-yy (3/ -3/3) (•^3--^4) 1 { (-^ --^J iy-y-y (i/-y.) i^^-^^)] ' the equation to the required conic. The reduction of this to the symmetrical form would be very tedious, and we shall therefore leave it in the above shape. In the numerical example ^, = I7 3/1 = - 1 ; ^2 = 2, 3/, = 1 ; a-, = - 2, ^3 = 3 ; ^4 = 3, 2/4 = 2; x^ = -l, y^ = -3. Hence x^ — x^ = — \ 'y a^j — cCg = 3 ; x^ — x^ = — 2 • x^ — x^ = 2j 3/,-3/2 = -2; y,-y, = -4^; y,-y, = -^] 3/^-3/5 = 2, a;^ — X3 = 4 ; x^ — x^ = — b-, x^ — x^ = A ', x^ — x^ = - S^ ?/.2 - 3/3 = - 2 ; 3/3-3/4 = 1; 3/4-3/5 = 5; 3/5 - 3/2 = - 4 ; therefore the above equation becomes {_ 2(a;-2) - 4(3/-!)} {5{x-S) - ^y-2)] {(-2)(-l)-4(-2)H5(-2)-4(-3)} |l(^ + 2) - (-5)(y-3)} {-i{x+l) - {-S){y + S)] . {l(3)-(-5)(-4)}{-4(2)-(-3)2} (8 - 2a; - 4j/)(- 7 + 5x- 4.y) _ {- \Z + x -\- 6ij){b - Ix + 3j/) •'• 20 ~ 34 ' .-. 17(5a;-4?/-7)(a; + 2j/-4) - 5(a; + 5?/- 13)(4a;-33/- 5) ; .-. 65a;' + llxy - 61/ - 174a; - 36y + 151 = 0, or 61/ - llxy - 65a;' + 36^/ + 174a; - 151 = 0, is the equation to the conic passing through the five given points. 10. Two chords AB^ AC are drawn from a given point A in a cm*ve of the second order so as to contain a given angle, shew that BC will always touch a curve of the second order. 1H48.] GEOMETRY OF TWO DIMENSIONS. 125 Let the equation to the given conic section, refen'ed to A as origin, be Ax^ + 2Bxi/ + Cf + 2l)x + 2Ey = (1), and let ax + ^y = I (2) be the equation to -BC, a, /3 being variable parameters. At the points of intersection of (1) and (2), we have Ax^ + 2Bxy + Cy' + 2 {Dx + Ey){ax + ^y) = 0, or ( C + 2E^)y' + 2 (5 + ^a + D^] xy^ + (^ + 2Z>a) x' = 0...(3). This may be considered as a quadratic in "- , and if ^j, ^2 ^® ^^^ roots, ij, t,^ will be the tangents of the inclinations to the axis of X of AB^ A C respectively. But AB^ A C include a constant angle, iarC^m suppose; hence we must have -^ = m. 1 + tA ' or ^ \,^' r^-^ = irc. (I + M2) Now from (3), by the the theory of equations B+Ea^D^ _ A + 2l)a C+ 2EI3 ' '' C + 2^/3 ' .-. 4i{{B+Ea + I)^Y-{A + 2l)a){C+2E^)] = m' { {A + 2Da) + C + 2E^]\ .'. 4.[E' - AC-\-2{BE-CD) a + 2{BD-AE) yS + E^a' + D'0' + 2{B+DE)a^], = m'{A + C + 2{I)a + E^)]% which may be written under the foi-m aa' + 2hafi + c^' + 2(f7a + e/9) + 1 = (4), a, 5, c, <7, e being certain determinate functions of A^ B^ (7, Z>, J?, and in. Now consider the conic section whose equation is ^V + 2B'xy + Cy + 2[D'x + E'y) + 1 = (.5). 126 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. Where (2) meets this, we have ^V + Wxij + Cy + 2{D'x + E'7/){aa: + ^i/) + [ax + ^i/Y = 0; .-. {A + 2D' a + a') x' + 2 (5' + E'a + D'^ + ayS) xy + {C'-{-2E'^ + ^')f = 0. In order that (2) may touch (5) the roots of this equation, considered as a quadi'atic in - , must be equal ; we must therefore have [B' + E'a + Z>'/S + a^Y = [A' + 2D' a + a') ( C + 2E'^ + /3'-') ; ... B" - A'C + 2 {B'E' - CD') a + 2 [B'D' - A'E) yS + {E"'-C') oi' + 2{B'- D'E) a/3 + [D" - A') ^' = 0, which agrees with (4) if E"' - C'_ _ B' - D'E' _ D" - A' _ B^ -A'C'~ ' B" -A'G'~ ' B'' - A C B'E' - CD' _ B'D' - A'E' _ B"'-A'C ~ ' B'-'-A'C ~^' which five conditions can be satisfied by means of the five disposable quantities A'^ B', C, D', E . Hence BC always touches the conic whose equation is (5). This theorem may also be proved by the method of reciprocal polars. For taking the polar reciprocal of the whole system with regard to ^ ; to the conic will coiTespond a parabola, and to -S, C (two points the line joining which subtends a constant angle at the origin) will correspond two tangents containing a constant angle. The reciprocal theorem then is : If two tangents be dra\\Ti to a parabola, including a constant angle, the locus of their point of intersection is a curve of the second order.* This may be proved as follows : * Taking the polar reciprocal of this systena mth regard to the focus of the parabola, the theorem to be proved is the following : If a chord of a circle subtend a constant angle at a given point of the curve, it always touches a circle, which is knoAvn to be true. 1848.] GEOMETRY OF TWO DIMENSIONS. 127 Let y = tx + - be the equation to any tangent to a parabola. This may be written ,, „ t--y.t + -^0 (1). XX This equation, considered as a quadratic in t^ gives the tangents of the inclinations to the axis of the two tangents di'awn to a parabola through a point {xy). In order that these may include a given angle tan~S?i, we must have, if ij, ^^ be the roots of (1), t — t -^ = m ; 1 + f,t, therefore, by the theory of equations. 1 + ^ X or y^ — 4acc = nf[a + x)'\ the equation to the locus of xy^ which is therefore a curve of the second order. 11. Pf J) are the extremities of two semi-conjugate diameters of an ellipse E^ whose semi-axes are «, b ; upon FD describe an equilateral triangle PDR^ so that the point R may fall with- out the ellipse ; the locus of R will be an ellipse E^ : assuming the above result, shew that if E^ be similarly treated, as also all the successive ellipses, the axes A^^ B^ of the a;'^ ellipse E^ so described will be comprised in the fonnula [a + 5)(coti7r)-' ± (a - J)(cot^7r)*". In the figm-e (66), let CP^ CD represent the equal semi- conjugate diameters of the ellipse E^ and let D' be the other extremity of the diameter through D. Join PD^ PD'\ on them describe the equilateral triangles PDR^ PD'R ; then i?, R' will be the extremities of the axes of E^. Join CR^ CR'. Then CR = ^A^, CR = \B^: 128 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. also ^, = t&nPCR = -, and FV'.CV = ^ab; CV a' ^ ' .-. OF' = i«, and PF' = ij. And CR' = CV + V'B', = _ + 2iZ>cos-. O TT Similarly Ci? = ^ + 2*a cos - , 2* 6 l(^, + JJ = (« + &)(2*cos^ + i), AT ^X 7^ 1 3^+1 2* Now 2*cos-+-, = -^^ = 3P33 3^+l\| /2 + 3hi /l + cosi7r\| 3^-1/ V2-3V Vl-cos^TT/' = lcot^)*. Similarly it may be shewn that 2^cos|-i = (tan^, , .-. i(5, + A) = (« + ^)(cot^), i(5,-^J = (a-J)(tan^y, a formula connecting the axes of any two successive ellipses ; .-. ^^ = (« + 5)(cot^) 4- («_J)(^tan^) , A = (« + ^) (cot ^) - (a - &) (tan ^) . 1849.] GEOMETRY OF TWO DIMENSIONS. 129 Now assume B^ = {a + b) (cot ^)*' - -^ (« - b) (tan ^)'", A=(« + ^)(cot^) +-'(«-^)(tan^) ; tlieu, by the formula already given, ^^^ = -—T-" (cot - ) + ^.r-^ (tan-) , 2 V 127 2 V 127 = (« + J)(cot-j +_-(a-J)(tan^j ; •Ml A A + B / 7r\*- A - B f ir\^ similarly J,,, = ^-^ (^cot - j + ---^^ (^tan -j , = (« + J)(cot^) +_-(«- J) (tan ^j . If then the assumed fonn hold for E^ it is proved to hold for ^^j. But it has been shewn to hold for E^^ therefore it holds universally. 1849. A is the origm (fig. 67), 5 a point in the axis of?/, BQ a line parallel to the axis of ar ; in AQ (produced if necessary) P is taken such that its ordmate is equal to ^^: shew that the locus of P is a parabola. Let AB = a, AP = r, BAP = ^ir - 6^ then the ordinate of P= rsin^, also BQ = «cot^; .'. r sin^ = acot^, or r^ siv^ 6 = ar cos 6 ; therefore, putting r sin = y^ r cos^ = a-, y' = c^, shewing that the locus of P is a parabola, a® b^ 2. K from points of the curve — ^ + -a = (a* — b^j tangents be drawn to the ellipse -7 + 75 = 1, the chords of contact will '^ a b be normal to the ellipse. K 130 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. If at the 1, 7] be the point (.r^), current coordinates of the normal its equation will be to the ellipse X a' v-y " y ' 01 r - a^ 1, T2 2 ' • a" — h^ X 1/ — (^ y If ar^, y^ be the coordinates of the pole of this line with respect to the given ellipse, we have _ g" 1 - ^' 1 y^ - y^ -a^ y- Eliminating jc, y between these two equations and the equation to the ellipse, we get ? + p = («■'-*')■'' the equation to the locus of the pole of the normals to the ellipse. Hence if from points of this curve, tangents be di'awn to the given ellipse, the chords of contact will be normal to the ellipse. 3. An oblique cone stands on a circular base; prove that one of the axes of the section made by a plane passing through the centre of the base and pei-pendicular to the axis, is a mean proportional between the other axis and i)seca, where D is the diameter of the base, and a the angle between the axis and a normal to the base. Let MPM' (fig. 68) be the section through the centre of the base and perpendicular to CO the axis of the cone ; let the section intersect the circular section RPR' in the line NP. Let a and h be the axes of the section, then F _ NP' _ RN.NR ' a' ~ MN.NM' ~ MN.NM' ' 1849.] GEOMETRY OF TWO DIMENSIONS. 131 Let lCAO = ^, CBO = y; also the z MO A is given equal to a. Hence BN _ sin(/3 + a) NR' _ sin(/3'-a) , MN~ sin/3 ' :?^~ sm 13' ' Z)"'' _ sin (/3 + a) sin(y8' — a) /-^ '' ^~ sin/3 • siu^Q^ ^ ^' and a = JL¥' = .1/0+ OJ/'=^l-.-^S^ + _.J^l ..(2). 2(sm(/3 + a) sm(/3'-a)] Again, since AO = OB, sm AC sinBCO sinCAO ~ smOBC cos (8 + a) cos f/S' — a) or ^; — pj-^ = — \ — 7^7 — ^ = p suppose ; snip smp -^ ^^ ^ ;:> + sina ,^, ^ - sina .•. cGt/3=^^— — -, cota =~ . cos a cos a Substituting in equations (1) and (2), we have 12 -jj = (cosa + cotyS sin a) (cos a — cot/3' sina), (1 +2^ sina)(l —]) sin'a). cos a _ 1 — ^^sin"''a^ cos'^ a ' , D I cosa cosa and a = — -. y t , . 2 VI +p sin a 1 — ^?sina/ I) cosa 1 — p sm a .-. = jr, i>seca, or h'' = aD sec a, and 5 is a mean proportional between a and Z^seca. 4. Let P,, P^, Pj,, (),, (2.,? ^3 be six points lying in a conic section; let the areas of the triangles P^Cl^Q^, ^iQaQ^i ^iQiQ^i be denoted by ^,, P,, O,, and the areas of the triangles formed K2 132 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. by putting /!,, P^ successively in the place of P, be denoted by ^'11 ^27 ^2) -^3) A) ^3 I'espectlvcly ; then will A,\B,C, B,CJ ■ A^\B,C\ BfiJ ' A,\B,G, B,C„ Let Mj, Mj, Wg denote the distances of any point from the lines ^•2^35 ^3^15 QiQ;^ respectively, then ?/, = 0, tt,^ = 0, u^ = will be the equations to these lines themselves. And the equation to any conic section passing through Q^^ Q^^ Q^^ may be written under the form ^ + ^^ + -^ = (1), W, M, «3 Aj, X,^? \ being constants whose values will be determined by the condition of the conic passing through P^, P^, Pg. Wg', ^3' be the values of u^^ w.^, u^ respectively at Pj, Let u u. u Then A = 4^.^3-<, A = ^^3^x-<, ^, = i^x^.-<, with similar expressions for ^.^, P^) ^2 5 -^35 ^35 ^3 • J <'? < ^j 1 5<"5<" ^- 1 \ 1 \BA - i^.cj + 1 A U<^2 1 8 1' f 1 1 \ " ^2^3-a^, 'Q,qA<\^. ^3 ^2 "<'i 1 / 1 % ^ 1 7J^.," Ct 1 >3" «2«*2/i 3 ^ And since P.P^Pg all lie in (1), we have = 0, A/, A.„ A„ _ -V + -T, + -7, =0, %, M, M3 A.. A-„ A<- ^ -7^ + -^, + -^ = ; i*. W., t*„ 1849.] GEOMETRY OF TWO DIMENSIONS. 133 whence eliminating "^^W by cross-multiplication, we get 1/1 1 \ 1 / 1 1 + 1/1 1 , + — , -7^ - -^r-. =0; whence dividing by ^Q.,Q^.Q^Q,.Q,Q^, 1/1 1\ 1/1 1\ 1/1 1, 5. The equations of three straight lines are u (= X sin 6 — 1/ cos ^ + c) = 0, «, = 0, m^^ = ; prove that the equations of the four circles, to each of which these lines are tangents, are . . 0-$^ . . 0- 0^ w* sm -^— — i + ?/j* sm — - — ^ + u^ sm -i— — = 0, a a i w* sm -^-- — - + w * cos ? + ?/ * cos — = 0, 2 2 2 * ^. - ^1 I ' Q-Q. * ^, - ^ w* cos -^^—r — i + w * sm — - — ^ + ^C cos -^ = 0, 2 2 2 * ^2 - ^1 . i ^ - ^. * • ^, - ^ . M* cos -^-— — J + u^ COS — - — ^ H- ?<2* sm -^ = 0. ^ ^ ^ The equation to any conic section, touched by these three lines, may be wi-Itten, \u^ + \u^ + \u} = (1). K we reduce this equation to the form of an equation of the second degree, the coefficient of the terms of two dimensions will not contain c, c^, c.^; hence, to find the condition of this repre- sentmg a circle, suppose c, c^, c^ to be Indefinitely diminished, the ratios c : c^ : c^ remaining unaltered. The three lines u = 0, u^ = 0, ?/,^ = 0, will then all pass through the origin, and the circle touching them will degenerate into the origin ; Its equation will therefore be x^ + if = 0, or 1/ = ± — * .r. 134 s()h:ti()ns of senate-house problems. [1849. Equation (1) will therefore become, dividlug out by a;, \ (sin d + -i cos (9)4 + \ (sin 6^ + -* cos ^J* + \ (sin e^ + -4 cos ^Ji = (2)* Hence, by Demoivre's theorem, equating real and imaginaiy parts separately to zero, ff ff X sin - + \ sin -^ + \ sin ^ = 0, 2 2 2 ( ^^y e e \ cos - + X, COS -~ + \, cos -^ = 0. 2 2 -^ 2 Eliminating X, X^, X,^ by cross-multiplication from (1), (3), Ave get 1 • ^., ~ ^, i. • G ~ ^., i • ^1 ~ ^ /x w* sm -^'— : — - + w,^ sm — - — - + uj sm -^— — = 0. 2 ' 2 ^ 2 Now if in this equation we write tt + ^ for ^, tt + ^^ for 6^, and rr + 6,^ for 0^, successively, by which substitution equation (2) is not altered, we get the equations to the remainmg circles : these are X • 0.^ — ^, 1 — 0„ 1 0,-0 ^ tt* sm -~- — ^ + M * cos — - — ^ + uj cos ^ — 0, 2 2 2 u^ cos 0-0^ ^ . 6' - ^, , ^ - ^ 2 -^ 4- M 4 sin — - — - + u} cos -^ — = 0, "^ 2 2 , — 1 — i ' — w* cos -^— — - + Wj* cos ^ + Mjj* sm -^-— — = 0, ia i^ ^ the required equations to the circles. 1850. 1. If at a given point two circles intersect and their centres lie upon two lines at right angles to each other through the * This method of investigating the condition that equation (1) may repre- sent a circle, is due to Mr. Leslie Ellis. It may be shewn in precisely the same manner, that if ip (m, u^, u^) = be any equation of the second degree, the condition that this may represent a circle is (^"^e, £-i9i, t~i6j^—o. 1850.] GEOMETRY OF TWO DIMENSIONS. 135 point; prove that, whatever be the magnitude of the circles, their common tangents will always meet in one of two straight lines which pass through the given point. Take the given point as origin, and the lines on which the centres lie as axes. Let a, )8, be the radii of the circles. Then the intersection of the common tangents must always lie on the line joining the centres of the circles, whose equation is ? + ^=l. From considerations of symmetry it is easy to see, that if this intersection always lie on one of two fixed lines passing through the origin, the equations to these lines must be x + y = 0, X — y = 0. Hence, if such be the case, the equation to one of the common tangents must be X y X + y - 4- I + ^ -1=0, a ^ 7 where 7 is a constant to be determined. In order that this line may touch the circle whose radius is a, it is necessary and sufficient that its distance from the centre of the circle, whose coordinates are a, 0, be a. We must there- fore have 1 i\' /I 1 Y" ' a 7/ Vy9 7/ .}__(]_ ly (I 1 ' ■ 7' ~ Va 7/ V/3 7 shewing that 7 is a symmetrical function of a and yS, and there- fore that if this straight line touch one of the circles, it must also touch the other. Hence the intersection of the common tangents always lies on the line a; + y = 0, if a and yS have the same sign, i.e. if both centres lie on the positive, or both on the nega- tive side of the origin. If one centre lie on the positive, the other on the negative side, similar reasoning will shew that the intersection of the common tangents lies on the line x — ?/ = 0. 136 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. 2. A number n of equal confocal parabolas are ranged all round the focus at equal angular intervals ; shew that the product of the distances of all the points of intersection from the focus is — 5— , I beinff the latus-rectum. Taking the common focus as pole, the equations to the para- bolas will be ' = -b ("' 4 sm - 2 -—77B~^ <^'' V2 ^ n) * ™ (2 + it) 4 sm - H TT V2 n If ^j, 0^ be the two values of 6 at the intersections of (1) and {m + 1) it is manifest that since these intersections lie at the extremities of the same chord passing through the pole, Also we easily see that u, = , .'. C/ = TT . ^ n Hence, if r^, r^ be the corresponding values of ?•, / I ' . , mTT ' ■■^ , ., mTT ' 4 sm -— 4 cos — - 271 2n • •. rr = — . ^ ^ . „ mTT 4 sm — n 1850.] GEOMETRY OF TWO DIMENSIONS. 137 Similar expressions holding for the intersection of (1) with each of the other parabolas, we have product of all the distances of intersections of (1) with the other parabolas ^ ' Sin — sni — sm tt w n n -r, , . TT . 27r . n — \ n rJut sm - Sin — ... sin tt = -^, ; n n n 2 ^2(«-l) therefore the above product = - — 5— . n To get the product of all the distances of intersections, we n have merely to raise this quantity to the power - : (not n ; since each intersection would then be counted twice over) ; therefore product of all the distances of intersections = — s- • 3. The locus of the points from which a circle is projected into a circle, upon a plane inclined at a finite angle to that of the given circle, is an equilateral hyperbola. Let (fig. 69) be the centre of the given circle, AB that diameter of it in which it is cut by a plane through 0, pei-pen- dicular to the line of intersection of the plane of the circle, and the plane of projection. Let CD be the corresponding diameter of the circle in which it is projected. Join CA^ DB^ and produce them to meet in E^ E will be the point from which the given circle is projected. Draw EG parallel to CD and equal to AB^ terminated by EC^ ED : let EG, AB intersect in P, then must AP = GP^ BP = FP. Through draw two lines OX, F, parallel to those respectively bisecting the angles APE, APG, and take them as axes. Let «, h be the coordinates of P; — «, — 5 of ^ ; X, y those of E. Then x will be the abscissa of P, and it is hence casv to sec that n -\- 2.r, — h will be the coordinates of G. 138 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. Hence, f , 77 being cuiTent coordinates, the equation of ED will be f — a 7} — h a + 2x — a — h — b^ ^ — a 7} — b or -= — — (1) X b V / And tlie equation to EP is ^ = x (2) At E^ the intersection of these two lines, we have f} = V-, and from equations (1), (2), x~b' .'. xy = aJ, shewing that the locus of E is an equilateral hyperbola, of which OX^ Y are the asymptotes. 4. Prove that y = nx -\ ad iniinitum is the ^ X + X + ... '' equation of a hyperbola. Find the position and magnitude of the axes, and write down the equation of the conjugate hyper- bola under the same form. Since y = 7ix + y — nx = X -{- X -\- 1 1 X + X +... 1 X -Y y — nx^ - 1 y — [n—l] x^ ••• {^J-nx){y-{7i-l)x] = 1 (1), the equation to an hyperbola, whose asymptotes are represented by the equations , , ^ "^ ^ y = nx^ y = (7? — 1) x. The axes bisect the angles between the asymptotes, therefore l.SaO.] GEUMETliY UF TWO DIMENSIONS. 139 their equations are y — nx _ y — [n—\)x , {\ + ny " {1 + {n-\y\^ ^^^' y — nx y — i^n—\)x (Tm?? " " {i + (/i-ir}i ^^^ respectively. Hence the positions of the axes are known. To find the magnitudes of the axes, we have, combining (1) (2), y — nx = , {l + (n-iyf y - [n-l)x = ^ 5^ ^ ; (i+«'r . {i+(^-iri^-(i+^-o^ • • »^ — ' • _ n{l + (n-l)']^- (n-l)(l + ny {l-flw-irni+n")' = 2 [(1 + 71^)4 [1 + {n- lY]i - [n' - n + 1)], which gives the square of the magnitude of one of the semi-axes. Similarly the square of the other semi-axis may be shewn to be = 2 [(1 + riy {1 + (n - 1)^}4 + {n' -n + 1)]. The equation to the conjugate hyperbola is {y-nx)[y-{n-\)x] =-1, which may be written 1 11 — nx — , ^ X ■\- y — nx^ _ 1 1 y X — X — 1 1 .r — X — .. is the equation to the conjugate hyperbola. 140 SOLUTIONS OP SENATE-HOUSE PROBLEMS. [1850. 5. From a point (fig. 70) are drawn two lines to touch a parabola in the points P and Q\ another line touches the parabola in R and intersects OP, OQ in 8^ T; if V be the intersection of the lines joining PT^ QS crosswise, 0, i?, V are in the same straight line. Let OP = a, OQ = h, then the equation of the parabola referred to these lines as axes, is !)' + (!/- (')• Let 08= oi, 0T= /3, then the equation to 8T will be M- c^)- To find the condition that (2) may touch (1), we proceed as follows : Square each member of (1) and multiply it crosswise by (2), we thus get X y _ f/^"\- (y a "^ ^ " |W "^ V^ This may be considered as a quadratic in \ ] •, and in order that (2) may touch (1), it is necessary that its roots be equal ; hence its first member must be a perfect square. The equations to PT, Q8 respectively, are - + ^ = 1 a b where these meet, we have the equation to a line passing through the origin, and through the intersection of PP, Q8^ that is to OV. 185U.] GEOMETRY OF TWO DIMENSIONS. 141 Again, to find the equation to OR^ we have since the first member of (3) has been made a perfect square, therefore squaring, ^ (" " -) = 2^ (jg - -^j ' '1 1\ /I r ^^Kl--^)+K^-^)='' the equation to OR^ which agrees with that ah-eady formed for V] hence 0, i?, V are in the same straight line. 6. A series of circles pass through a given point 0, have their centres in a line OA^ and meet another line AB. From ilf, iV, the points in which one of the circles meets the lines OA^ AB^ are drawn parallels to AB^ OA, intersecting in P. Shew that the locus of P is a hyperbola, which becomes a parabola w^hen the two lines are at right angles. Take as origin, OA as axis of x, let the equation to any one of the circles be x^ +f = 2rx (1), and that to AB xcosol + ?/siua = a (2) ; hence the coordinates of M are 2/-, 0, and the equation to the line through J/ parallel to AB^ is xcosa + ?/sina = 2rcosa (3). Now the circle in general cuts AB in two points, either of which may be denoted by N. The equation to the line through either of these points parallel to OA (the axis of ic), will be ob- tained by putting the ordinate of that point = 0, and therefore the equation to the pair of parallels will be obtained by elimi- nating X between (1) and (2). This gives (« -y sina)'"* + y' cos' a = 2r cos a (a — 3/ sin a) ; .-. y'^ + 2 (rcosa — a) ^sina + d^ — 2arcosa = 0...(4). To find the equation to the locus of P, the intersection of (3) with either of these lines, we must eliminate r between (3) and 142 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. (4), whence we get y^ + (iccosa + ^sina — 2a) ?/sma + d^ — a (a?cosa + y slna) = 0, or ^■'(1 +8in'*a) + xi/ cosa sin a — a [x cos a, + 3^ sin a) + d^ = 0. This is the equation to the locus of P which is evidently in general an hyperbola. If however the lines are at nght angles a = 0, and the above equation becomes y'^ — ax + d^ = 0, representing a parabola. 7. If from the focus of a parabola, lines be drawn to meet the tangents at a constant angle, the locus of the points of inter- section will be that tangent to the parabola whose inclination to the axis is equal to the given angle. Prove this in any manner, and shew that if m be eliminated between y = mx -\ — , and V = [x — a), the result contains a factor which answers to the locus. Also explain briefly the origin and signification of the other factors. (a). Let 4a be the latus-rectum of the parabola, a the iji- clination to the axis of any one of the series of tangents, yS the constant angle at which the lines through the focus meet the tangents. Then taking the focus as pole, and the axis as initial line, the equation to the tangent is r = acoseca cosec(^+ a). That to the line through the focus is 6/ = 7r-(a + ;8). Eliminating a, we get as the locus of the intersection of these "^^^ r = acosecyS cosec(^ + yS), representing the tangent whose inclination to the axis = /3. /3. From the equation m + t . y = [x — a) (1), we get m [x — a + ty) = — t{x — a) + y ', 1850.] GEOMETRY OF TWO DIMENSIONS. 143 therefore combiuing this with y = mjc + ^^ (2), y — tix — a] fy + X — a y = ^ !^ 1 X + — -. -, a, ty + X — a y — t[x — a) y— tix—a) , , ill— tix—a) ty + x — a - '^ ^ {x-a)+a]^ ^ ^ + ^ ty + x-a^ [ty + x — a y — t[x—a))^ {{x-a)^ + f}{l+f) . ty + x — a) [y — t[x — a)} ' X — a)'' ■+ y' ty — fx — a ^f+[x-ar^^ {[x-aY + f]{l+f) . ty + x — a i^ty + x — a)[y — t[x — a)]^ = 0, X — a + ty ' y — t{x — a) This is satisfied by y = tx + - ^ representing the locus found above. It is also satisfied by [x-aY + y' = % which requires that cc = «, ?/ = 0, representing the focus. This would be obtained by making m = (—1)* in equation (2). Its signification therefore is, that if tangents be drawn to the imaginary branch of the parabola, got by making x negative, and lines be drawn through the focus of the real branch cutting these tangents at a constant angle, the point of intersection of these lines will only be real when the tangent to the imaginaiy branch of the parabola passes through the focus of the real branch. 8. Within the evolute of an ellipse is inscribed a similar ellipse ; within its evolute another similar ellipse, and so on ad infinitum ; shew that the sura of all the areas TT id'+V') 4 ah Let mrt, mh be the semi-axis of the first inscribed ellipse, then m will be a homogeneous function of a and h of no dimen- sions, and therefore the same function of ma and mh ; hence 144 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [18oU. wt^/, ni^b will be those of the second, and the sum of all the areas TTuh 1 - m' ' we have therefore only to find the value of m. Now the equation to the evolute of the ellipse is {axf+{hyf = {d^-hi (1). In order that this may touch the ellipse \ina) \mh) they must have a common tangent at a common point. Now the equation to the tangent to (1) at [xy] is In order that this may touch (2) at {xy)^ we must have 'a^i 1 X (S^^.-(3.. = K-0-'. 2 7 27,a J X ina ' • 2 2 2 z3j 7 ma a — y^ _ mW' therefore, by (2), m = ^r-^ 5 therefore if S equals the sum of all the areas, Trab S = a'- b'\' a' + ¥ 4 ah 1850.] GEOMETRY OF TWO DIMENSIONS. 145 9. Find the points ^,, A^^ A^...A^^^_^, A^^^ in a parabola, such that the tangents at these points are parallel to the focal dis- tances SA^^^^ SA^, SA^y..SA^_^^ ^^„t-\-) respectively. TiQl A^A^KV [^g. 71) represent the parabola, A''/S'X its axis, A^T^^ ... A^T^^ the tangents at ^„...^,,, respectively, and let lA^SX = a^, then lA^T^X = ^a^. Hence, by the conditions of the problem, we must have W=^m (1), K = «. (2), K = a,-, (''), ^m-l = am-2 (w-1). The last equation will be TT + ia,,, = «„,_, (m), since this will satisfy the condition of A^^^T^^ being parallel to SA^^^_^^ and it is manifestly inconsistent with the preceding equations to have ^a^^^ = a^^^_^. Hence, multiplying generally equation (?') by 2'""^, and add- ing all equations thus formed, the quantities a^, a^, ... «,,_ , dis- appear, and we get ^ + i««, = 2"'-'a,„ ; - 27r •■• «m - 2'" _ 1 • whence ., = ,-J^ , 2'-+«_. and generally, a,. = ^„. _ ^ , and the positions of the points A^^ A^...A^ are determined. 10. From the focus of an ellipse lines are di-awn to any four points in the curve, and the reciprocal of each line Is multiplied by the sines of half the angles between any two of the remaining lines ; prove that the sum of the first and third of these products taken in order is equal to the sum of the second and fourth. L 146 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. Let L = the latus-rcctum, e the eccentricity of the ellipse ; then Its equation, refeiTcd to the focus as origin and the axis- major as prime radius, will be 1 _ 1 — e cos^ and let ^,, 6^^ 6^^ 6^ be the angles between the axis and the successive lines, r^, ;-^, rg, r^ the lengths of the lines. Then — sm -2-— — -* sm ^— — ? sm -^ ^ y-j 2 2 2 2 ^0-0, 0-6. (^-f> = ^ (1 — e cos ^j) sin -^-— — * sm -^-— — ? sm Z ^ >^ 2 2 2 = ^ (1 - e cos ^J (sin [6^ - e,) + sin (^, - ^J + sin (^3 - ^,)}. O- ., , 1 . ^, - ^. • ^.> - ^4 • ^4-^1 bimilarly — sm ^-- — sm -^— - — sm -^-— — - •' r^ 2 2 2 = ^ (1 - e cos^3) {sin (^, - e^) + sin (^, - ^J + sin(^^ - ^J]; therefore if ^8' denote the sum of these two quantities, and If the symmetrical fmiction sin {6^ — 6^ + sin {6^ — 6^ + sin (^3 — 6^ + sin (^.^ - 6^) -}- sin [d^ - ^J + sin [6^ - 6^ be denoted by <^, and if we put c^ for cos^^, s^ for sin ^j, &c., + ^3 k^l - ^1^.2 + ^4^2 - ^^4 + ^'^1^4 - ^4^1)}] = ^ [0 - &A[^-h)+ ^-f^i^-^l) + C3^4(«l- «2)+^4^lK-*3)}]- Similarly, if S' be the sum of the second and fourth products, ^S" will be obtained from S by writing 6,^ for 6^^ 0^ for ^.^, 6^ for ^3, ^j for 6^ ; hence (/), which Is a spmuetrical function, will remain unchanged, and whence It appears that S = /S", or the sum of the first and third of these products is equal to the sum of the second and fourth. 1851.] GEOMETRY OF TWO 1)1 .M KXSloNS. 147 11. If lines be drawn tlirough any two of the points A^ i?, Cy.. and other lines through any two of the points f/, i, Cy.. all in one plane, prove that the intersections of AB with «7>, of AG with «c..., will all lie in one straight line, pro- vided that the lines through the intersections of any two of the first series of Hues and the corresponding intersections of the second series all pass through the same point. Conceive the points -4, i?, C, ... to lie in one plane, and ((,b,Cy.. in another; then, since the lines joining the intersec- tions of any two of the first series of lines and the corresponding- intersections of the second scries all pass through one point, Aa^ Bbj Cc... all pass through one point 0. Now consider any quadrilateral, as ABah^ whose angular points are any two points of the first series, and the correspond- ing two of the second series. Since the lines Aa^ Bb intersect, they are in the same plane, therefore also AB^ ab^ are in the same plane, and must therefore intersect,* and their intersection must manifestly lie in the line of intersection of the planes ABC..., abc... Similarly the intersection of any other pair of lines, as A C, ac, lies in that line. Hence, if we suppose the planes ABC..., abc..., to be indefinitely nearly coincident, the proposition enunciated follows at once. 1851. 1. Having given a focus and two tangents of a conic section, shew by means of reciprocal polars, or otherwise, that the chord of contact always passes through a fixed point. Let a circle be described passing through two fixed points, A, B, and let F be the intersection of the tangents at A, B. The locus of P will be a fixed straight line, pci-pendicular to and bisecting AB. Now take the polar reciprocal of this system with respect to any fixed point S. The reciprocal of the circle will be a conic section whose focus is S, and which has two fixed tangents (the reciprocals of A, B). Hence the reciprocal of P, which is * If these lines happen to be parallel we may still consider them as inter- secting in a point infinitely distant. L2 148 SOUTTIOXS OK SENATE-HOUSE PROBLEMS. [1851. tli(^ chord of contact of these tangents, will always pass through a fixed point, the reciprocal of the locus of P. 2. Shew that there will be two pairs of equilateral hyper- bolae which pass through two given points -4, B^ and touch two given straight lines, and that the chords of contact of each pair meet in AB^ and are equally inclined to AB.^ Take the middle point of AB as origin, AB as axis of a*, let h^ — h^ be the abscissse of A^ B, respectively, and let the equations to the two given tangents be 5 + f_l = 0, ^+1-1=0, and that to their chord of contact - + ^—1=0, a p ' where a, /9, are indeterminate parameters. Then the equation to a conic section touching the two given lines may be written mider the form M->)(M-')=^(M-' \ being an indeterminate parameter. Two equations for the determination of the three arbitrary quantities X, a, yS, are given by the conditions of its passing through u4, B. We thus get ^o(^')=Ks-y «- a J \a J \a. The third equation is given by the condition of the curve being an equilateral hyperbola. In order that this may be the case, it is necessary that the sum of the coefficients of x^ and y-* = 0. This gives — - ^. + 777 - 3i = 3 . aa a. bo p * A shorter solution of this problem, due to Mr. Gaskin, will be found in the Appendix. 1851.] GEUMETUY OF TWO DlMENS-lONS. 149 Combining (1) and (2), we get ^h — aV (h — a) (h — a) Kh + aj (A -f a) {h + a) ' This is a quadratic for the determination of a. Let a,, a„ be its roots. Subtracting (1) from (2) and eliminating \ by (3), we get 1 1\ n i\ 2 /I 1 ■ ,? + w) \a a'J a \aa' bh\ Hence, for each vahie of a there will be two of yS, equal and of opposite signs. Let them be denoted by ^8^, — ^,, ^^^ — /S^ respectively, then we get two pairs of equilateral hyperbolae, whose chords of contact respectively are - + !- = 1, --i = 1 5. It is easy to see that the lines represented by (4) intersect in AB (the axis of x) and are equally inclined to AB. The same will be the case with the pair of lines denoted by (5). Hence the chords of contact of each pair meet in AB^ and are equally inclined to AB. 3. If from a point of an ellipse a line be drawn to the ex- tremity of each axis, and a parallel to the same axis be drawn through the point in which such Hue meets the other axis, the locus of the intersection of these parallels is an equilateral hy- perbola. Trace the coiTesponding positions of the point on the hy- perbola, and the point on the elhpse. Let a, J, be the semi-axes of the ellipse, take the axes of the cui've as coordinate axes, and let a cos a, h sin a, (a being a variable parameter) be the coordinates of the point tlu'ough which the lines are drawn. The equation to the line drawn through this point to the extremity of the axis-major, is X — a cosa y — ^ sin a a cosa —a h sina ' 150 S moves round through BA'B'A. 1851.] GEOMETRY OF TWO DIMENSIONS. 151 4. Dctenniue the values of a', m\ n\ such that the relation {{x-af + /}* = m (x^ + y'y- + n may be equivalent to the relation [{x - a)" + y'Y- = m {x' + y'^Y + n. The transfonnation fails (1) in the case where the curve represented by the given equation is a conic section, (2) has a double point. (a). The equation {{x-a'Y + f]^ = in {x' + f)^ + n\ when transformed to polar coordinates, and rationalized, becomes r^ — 2a.'r cos^ + a.'^ = (?«';• + ?«y, or (1 -m")r' - 2a'r cos^ - 2m'n'r + a' - ?i" = 0. The equation {(a; -a)' + /}* = m {x'+iff- + n, similarly transformed, becomes (1 - 7n') r^ - 2a.r cos 6 — 2mnr + ai' — n" = 0. In order that these equations may be identical, the coefficients of r^, r cos ^, r, must bear the same ratio to one another as the constant temis, 1 — ni'' OL m'n a^ — n^ 1 - m^ a mil a'"* - -n' lation a ^ a" - n"' Q ■> a a^ - 71^ 1-^ we get a' a = -? , »" '-^' 1 - 1 - in' I - n' ^ 1 - m' ' ^' 152 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Also wc have n a mil 7)1 a 'e — ^ , or — — ^ i a vin VI n 71 ' a • m' ^ a ' and n' = m, a a iiice a 1 - nf ' 1 a' - n' • a! ^^ a 1 — m^ 711 a"'* — n^ n ^ a 1 — nf '" (1), (2), (3). The required values of a', m', w', are determined from (1), (2), and (3). (y8). The given equation, when rationalized, will in general be of the fourth degree in x and y. In order therefore that it may represent a conic section, it is necessary that the terms of a degree higher than the second should disappear of them- selves from the rationalized equation, which requires that 7)1 = ± 1. If this condition be satisfied, the given equation DGCOmGS {{x-ay + f]i± {x'+7/'y = 7i, shewing that one of the foci of the conic section is the origin, that the coordinates of tlie other are a, 0, and that the axis- major = 71. The transformed equation must therefore, since it represents a conic section, take the form {{x-a'Y + 7/}i±{x'+f)i = 7l\ which gives a', as the coordinates of the second focus, 7i as the axis-major. Hence we must have a' = a, w' = ;«, and the transformation fails. 1851.] GEOMETRY OF TWO DIMENSIONS. 153 The transfonnation also fails if n = a, for we then get m' = 1 , a = 0, n = 0, and the transformed equation becomes an identity. In this case, the given equation, transformed into polar coor- dinates, becomes (1 - vf) r - 2 (a cos ^ + ma) = ; whence we see that if cos^ = — m^ we have r = ; but the equation cos = — m is satisfied in general by two values of ^, whose sum = 27r ; hence the origin is a double point. If therefore the curve have a double point, the transfonnation fails agam. 5. If be the centre of a reflecting circle, Q a radiant point, and the line from Q to produced to meet the circle be con- sidered as the axis, then, if a be the radius, u the distance QO^ the inclination to tlie axis of the radius through the point of incidence of any ray, and the inclination to the axis of the reflected ray, pcos(f) = acosd + wcos2^, psmcj) = asm0 + usm20j where p = {a' + u^ + 2au cos 0Y is the length of the incident ray. Let P (fig. 73) be the point of incidence of any ray, M the point in which the reflected ray cuts the axis. Let and draw PN perpendicular to the axis. Then QN= /3 cos (<^ - 21/r) = QO + ON = u + «cos^, PN = p sin (0 — 2\|r) = a sin 0. Again, yjr = (ji — 0, .'. p coa{'20 — (j>) = u + acos0 (1), p8in(20-<^) = asin^ (2): (1) cos 2^ + (2) sin 2^ gives p cos<^ = a cos + u cos 2^, 154 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. (1) sill 2^ — (2) cos2^ gives psincf) = aamO + usm20j the required equations. 6. Using the notation of the last question, and assuming the tiTith of the theorem stated therein, shew that if fi-om the point of incidence of each ray there be drawm, in a direction opposite to that of the reflected ray, a line equal in length to the incident ray, the locus of the extremities of these lines is a curve cutting the lines at right angles, and the equation of which, referred to the radiant point as origin and the axis QO SiS axis of x^ is x^ + y^ — 2ux = 2a{x^ -i- y^)^. Shew that the origin is a double point, and trace the curve : shew also that the equation may be expressed in the form {{x — af + /]* = m [x'-\-y^)^ + n. ' (a). Produce MP to i?, making PR = QP^ then we have to find the locus of R. Let x^ y be its coordinates, then we readily see that x= QN -\- pcos . X ^ + w.^ x'' 2 , -2 — *<^* "T ^:''" / 2 , 2^ X +y G-^ +,y . 0.'^+/' 1851.] GEOMETRY OF TWO DIMENSIONS. 155 .'. x^ + }f -2ux = 2rt(x'+y')*, the required equation to the curve. (y8). This equation, transformed to polar coordinates, becomes r — 2u 0,09,6 = 2a, .-. r = 211CO&0 + 2a (1). Hence the radius vector of the curve exceeds by 2a the radius vector of a circle which passes through the pole, one of whose diameters is prime radius, and whose radius = u. If therefore we draw such a circle, and produce 0-4*, the radius vector of any point A in it to i?, making AB = 2a, the locus of R will be the required cm've. The cm-ve will pass through the origin when cos^ = , which condition, if a < u, is satisfied by two values of 0, one less, the other greater than tt. Hence if a < w, i.e. if Q be outside the circle, two branches of the curve pass through the origin, which is therefore a double point. When ^ = 0, /■ = 2 {u + «), and when 6 = tt, r = — 2 [u — a) ; hence the curve will have the form represented in fig. (74), where Qq = 2{zi + a), Qq =2{u-a).-f Again, l QRM= L 0PM (since QR is parallel to OP) yfr = (f>-0, sm cos^ — cos^ sin^ tSinQRM = COS0 cos^ + sin<^ sin^ u sin a + Mcosi by the result of question 5. * Tlie line OA must ahvaj's be produced in the positive direction of the radius vector, therefore when > .Vtt, OA must be produced backwards. t This is the form of the figure when Q is outside the circle : if it be witliin it, the curve docs not pass through the origin, and the loop Qq' docs not appear. The origin will then be a conjugate point. 15G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. And if L be the angle between the radius vector and tangent, 1 flu u sin cott = — T7i = — L dO a + u cos 6 ' = tan()i?J/. Hence the curve cuts the lines PR at right angles. (7). The equation [{x - a)' + y'Y- = m {x' + y^ + n is equivalent to (1 — m^)r^ — 2mnr — 2ar cos^ + a^ — w^ = 0, and this coincides with (1), if -2 = «? -. -2 = ^1 a. - n =0] 1 _ ,n^ - "' 1 _ ni' a u — a m = - . 71 = a = ,/<- — >*— , u u therefore (1) is equivalent to ,2 2^ 2 N J w — a\ ,J^ u , ., .,,, It — a '■^\* ^ — :— +3/ =-i^"+rf + ) M / j a u which is in the required fonm. 7. Given the centres of three circles, each of them touching the other two externally, determine the radii. How many systems of circles are there when the centres are given, but the circles touch externally or internally at pleasm'c ? Let a, &, c, be the distances between the given centres ; then '•2 + ''3 = ^) ^3 + '\ = h r^ + r.^ = c, b + c — a similarly r^ = 2 ' c + a — b 2 ' a + b — c which determine the radii. 1851.] GEOMETRY OF TWO DIMENSIONS. 157 If the circles touch internally or externally at pleasure, there will be four systems. For the circle described with any one point as centre may include the other two, which must touch each other externally, thus giving three systems. Or they may all touch externally, giving four systems in all. 8. The locus of the point from which two given circles subtend equal angles is a circle. Let -4, A' (fig. 75) be the centres of the two given circles, P a point from which the circles subtend equal angles. Draw the tangents P T, Ft to the circle whose centre is A ; PT\ Ft' to that whose centre is A'. Join P/1, PA\ AT^ At, A'T\ A't'. Then LTFt = AT Ft'. And AF, A'F respectively bisect the angles PPf, TFt', .-. z AFT = L AFT: also the right angle ATF = \k\e, right angle A'T'F', therefore the triangles TAF, TA'F are similar, therefore AF'.A'F:: AT: AT (1). Divide AA in 0, so that AO \ A :: AT : AT take as origin, A OA as axis of x. Let AO = a, A = a', and let X, y be the coordinates of P. Then by (1) ... d'%x^a)^^f]=d^[{x-d)'^f], or (a — a')[x^ -\-y'^) — 'iaax = ; shewing that the locus of P is a circle passing through 0. 9. The lines joining the corresponding points of two similar and similarly situated figures in the same plane intersect in a point. All sections of a conical surface of any degree by parallel planes are similar and similarly situated figures, and every gene- rating line passes through coiTCsponding points. Hence, con- versely, the lines joining coiTCsponding points of two similar and similarly situated figures in parallel planes, pass through one point (the vertex of the conical sm'face of which they are 158 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. sections). Let the planes l)e now made indefinitely nearly coincident, and the proposition enunciated follows at once. 10. Given any three of the four lines Ox^ Oy^ 0])^ Oq^ (fij^. 76), the foiu'th may be determined, such that if «, h be the points in which a line tln-ouf2,h a point Q in Oq intersect Ox^ Oy^ and n\ h' the points in which another line through the same point Q intersects Ox^ Oy^ the point of intersection of the lines ah' and ah lies on the line Oj). Let II = 0, V = 0, be the equations to any two lines passing through the point 0, and let u — \v^ u = Xv, u = X^r, w = \v, be the equations to Ox, Oy, Oq^j Oq, respectively. Also let It) = be the equation to Qa, and n — \ii — fiic = that to Qa. Then the equation a{u — \v) — [u — \gV — [xw] = 0, where a is a disposable parameter, represents a line passing through a. In order that this may pass through h, the above equation must be identical with l3{u-\v)-w = 0, y8 being also a disposable quantity. In order that these equa- tions may be identical, we must have a — 1 a\,. — \a and (Xy — \j){u — \v) — {\y — \.) (« — \v — fiw) = 0, or {\ - \){u - Xv) + (\ -X^)fiw = 0, is the equation to ah. Similarly (\ - \) (^ - K^) + {\ - \) /^^^' = 0, is that to ah'. Where these intersect, we have (X^ - \,) [u - Xv) + (X - \) (u - Xv) = 0, or {X^ + Xy-2\)u + X,{X^ + Xy)v = (1). In order that this may lie in the line 0/j, whose equation is ti - X^v = (2) 1851.] GEOMETRY OF TWO DIMENSIONS. 159 (1) and (2) must be identical ; hence \ i\ + \- 2\) + \ {\ + \) = 0, or {\ + X,]{\ + \) - 2\X, = 0, an equation from which, when any three of the quantities X^, X^, \ , X^, are given, the fourth may be determined, so that when any three of the lines Ojc, Oy^ Op^ Oq are given, the fourth may be determined so as to satisfy the required conditions. 11. The radii vectores from the focus of a conic section to two points of the curve make equal angles with the line drawn from the focus to the point of intersection of the tangents at the two points. Let a, (B be the angles which the radii vectores respectively make with the axis-major, then the polar equations to the tangents, referred to the focus as pole and the axis-major as prime radius, will be 1 2 - = - {^cos^ + cos(^ — a)}, 1 2 - = y {e cos^ + cos [6 - y8)|. r i Where these meet, we must have which will be the equation to the line through the focus and the intersection of the tangents, which evidently bisects the angle between the radii vectores. Hence the proposition is tnie. For a demonstration of this theorem by the method of re- ciprocal polars, see Salmon's Conic Sections^ chap. xiv. 12. If two triangles be circumscribed about a conic section, their angular points lie in another conic section. Let u = 0, V = 0, t« = 0, be the equations to the sides of one triangle, and let the sides of the other triangle respectively opposite to these be represented by ?/ + &,t' + Cj?c=0.,.(l), a^« + t' + r,?r=0...(2), a^u + \v+ir=0...{?,). 160 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Then, since these two triangles are circumscribed about a conic section, it will follow that if (ly^'J denote the line joining the intersection of v = and (3) with that of i« = and (2), with similar notation for the other corresponding lines, {v^io^), {^o^v^), (?//',) all pass through one point. Now the equation to (^'g^t'J is V 10 w + - + — = that to 2v,u„ IS — -\- V + — = U V to t/.-y,, — I h w = (A). The elimination of w, r, lo between these equations would give the necessary condition that the three lines denoted by them should pass through one point, or that the two triangles should be cii'cmnscribed about a conic section. Now the equation to any conic circumscribing the triangle (123) can be put into the form a [a^u -\-v-\- c^w) {a^u + h^v + w) + ^ [a^u -\- b^v + w) [u -{■ h^v + c^w) + 'y[u-\- l\v + c^w]{a^u -{- V -\- c,^ic) = (4). Here the coefficient of li^ is proportional to « + -+-, ofv^to ^ + /S + f , n -2. . ^ /3 of 10 to h — + 7. c c ^1 2 Hence, if we give to a, yS, 7 respectively the values which w, V, w have at the intersection of the lines (A), each of these coefficients will vanish, and equation (4) will be reduced to one involving vw^ wu, uv only ; It will therefore also represent a conic circmnscribing the triangle whose sides are u = 0, V = Oj 10 = Oj and consequently, if two triangles be circum- 1851.] GEOMETRY OF TWO DIMENSIONS. 161 scribed about a conic section, their angular points lie in another conic section.* From the above proof it is not difficult to see that the converse (which is also the reciprocal theorem) is true. 13. If the angles ^, cj)' are connected by the equation cos)u, = cos^ cos<^' — sin^ sin^'(l —c^ sin"''/i)-, and sin0, sin^' arc the abscissae of points on an ellipse, the semiaxes of which arc 1, (I— c'^)^, then the tangents at these points meet in a point, the locus of which is an ellipse confocal with the given ellipse. Let f , r) be the coordinates of the intersection of the tangents, then the equation to its polar is 1 — c* Let sin^, (1 — o'')* cos^, be the coordinates of the points where this line meets the given ellipse, then ^ . _ 77 cos ^ the roots of this equation in 6 are <^, ^'. It may be written in the forai (|sin^-ir = ^^(l-sin^^); 1 -' .•. sm)S the equation to the locus of (^, t;,) which is therefore an ellipse, the squares of wliosc semiaxes are respectively l-(l-c''sin» ^ l-(l-c''sinV)^ , 1 + cos /x ' ^ ' cos /i — (1 — c"* sin'V)- ' 1 — cos/A — (1 — cos/i) (1 — c*^ sin'^yu-)* ^ or ; g , sm /i 1 — c'' sin'^/i — cos/u. — (1 — cos /a) (1 — c"* sin"''/u,)* ^ siu'^/i ' therefore, if c' be the distance from its centre to its focus, ,a _ 1 — cos yu, — (1 — cos /a) (1 — (? sin^/i)* C — ;: — o sm fx 1 — cos/A - c'"^ sin'''/u, — (1 — cosyu,) (1 — c^sin^/i)* sin'"' /A ••• C' = «5 whence the ellipses are confocal. 14. If cc, y^ z, r«, are linear functions of the coordinates of any point, such that no three of the lines represented by the equations « = 0, 2/ = 0, z = 0, ?i; = 0, meet in a point, the equation w + [yzf + [zx]^ + [xy)^ = is that of a cui've of the fourth order having three double tangents, x = 0, y = 0, z = Oj and three double points, y = z = w, z — x = w^x=y = w. Shew also that the six points of contact of the double tangents lie in a conic section. Where the line a? = meets the cun'^e, w + [yzY- + {zx)^ + {xy)^ = (1); we have also w + [yz]^ = 0, or w'' = yz , (2). 1851. J GEOMETRY OF TWO DIMENSIONS. 163 It hence appears tliat the line x = meets (1) only in the points where it meets the conic (2), that is, in two points only. But (1) when rationalized takes the form («SPtake SQ = \L (the semi-latus-rectum); draw QR^ ST perpendicular to ;SP, meeting the tangent at F in R and T respectively; also draw SY perpendicular to the tangent meeting it in 1^; and let PZ7, QZ drawn parallel to the transverse axis meet 8Y in U and Z respectively : it is required to prove one of the following properties : (1) P is a point in the latus-rectum. (2) QR passes through the point U. (3) PU=e.PS. (4) SR = e.ST. (5) SY.SZ={^L)\ (1). Let the inclination of SP to the axis-major be a, then the polar equation to the tangent at P will be 1 2 - = -^{(cosi9 + cos(^-a)}; and that to QR, r = \L sec (^ — a). At P, the point of intersection of these lines, we must have ^ = |7r; therefore R is a point in the latus-rectum. (2). The equation to /ST Is /, sin a tan = , e + cosa 16G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. or ill rectangular coordinates, jc ama = y {e + cos a). Now the ordinate of P is i^L sin a 1 + e cos a ' therefore the equation to PU is ^L sin a y = 1 + e cosa At U the point of intersection of these, we have _ ^L (e 4- cosa) 1 + e cosa ' .'. X cosa + y sin a = ^L. Now this is the rectangular equation to QB ; hence QR passes through the point U. , V m, 1 • n r^- \L cosa (3). The abscissa of P is . ^ ' 1 + e cosa That of CThas been shewn to be ^L (e-fcosa) 1 + e cosa ' and P, ?7, have the same ordinate ; hence PU==e.PS. (4). Since i2 is a point in the latus-rectum, SR is perpen- dicular to PZ7, and ST is perpendicular to 6T, Pii to SY^ whence it readily follows that the triangles STR, SPU are similar ; .-. PU'.PS:: SR: ST', but PU=e.PS, .: SR = €.ST. (5). The polar equation to QZ is r sin ^ = ^L sin a, 1851.] GEOMETRY OF TWO DIMENSIONS. 167 and that to ^F is zj sin a e + cosa . ^ sina .*. sma = ;H-2e cosa + e')*' therefore at Z^ the intersection of these lines, r = 6'Z = ^X (1 + 2c cosa + e')*. Also SY=^\L ^ {(e + cosa)'^ + sin* a}* 1 ^ (1+26 cosa 4- e')*' .-. 8Y.8Z=[\L)\ ==^L 17. If -4j, A^...A^/y rtj, «2---'^i5 ^^ ^^^^ angular points of two polygons of n sides each, which circumscribe a given circle, and Pj, P^...F^ the points of intersection of their first, second... h"' sides respectively ; shew that Shew also, by means of projective properties or otherwise, that the same equation is true when any conic section is substituted for a circle. From 0, the centre of the circle, draw perpendiculars Oi?,, OB^...OB,^, 0\, 0\...0h^,, on the sides A^A^, A^A^...A^A^, a,a^^ ^,«2*'-^n-i^«? respectively. Through draw any line OX, and let generally B,OX = a„ h^OX = /3, (fig. 78). Then A^OX = i (a,_, + aj, .-. AOB, = A,OB,_^ = ^ (a, - a,J. Similarly, a, OB, = M^r - A-J, and P,C>Z=i(a, + ^,); .-. P,0j5, = P^OX - B^OX =h{0r- aJ- Also PA, = a (tan ^. OP. + tan P, OA" ) , a being the radius of the circle, = a{tan^(a^ - a,_J + tan^(|S, - a,.)| _^^ sin^(/3, - a,_J cos-^ (a^ - a^_,) cos ^ (/3^ - aj * 168 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Also F.l = a (tan yl,.,, OB,. - tan P, OB) = a {tan^ (a,.,, - a,) - tan^ (/3, - a,,)} = a sin^(a,^, -^, ) The expressions for P^a,. and P^a^+j, will be got from these by simply interchanging a and y9 ; . Pr, -n sin^(a. -/3,_J • • ^^''^ - "" eosi (^, - /3,_J cosi (a, - ^J ' Hence cos^(/?,.^j - ySJ cosi(a, - ^,. P.^...P.a. PA..^,.Pa ^ sin I (^, - a,_,) sin ^ (a, - ^,._, ) cos | («,^^ - g,.) c os ^ (^,.^, - ^ J , sin^(^,,^j-a,.)sin^(a,.^j-/3,.)cos-^(a,-a,_,)cos^(^,.-^,_J From the form of this expression it is easy to see, that if we give r every value from 1 to n inclusive and multiply the re- sulting fractions together (observing that instead of a,.^j, /3,.^,, we write otj/SJ, every factor will appear both in the numerator and denominator. Hence P,^,.P,a,.P,^,.P,a,...PA-i^A _ 1 PA,P.a,^PAs'^.%'"KA-PA ' or P^A^.P^a^.P^A,^.P^a^. . .P^.^^A jf f T> i 1 .-. . OP,..OA,..sm P,.OA, It tor P,.A, we substitute ^^^ -, and make similar substitution for each of the other lines, each member of this equation will, since OB^ = a, be divisible by OA,. Oa,. 0P\ . . 0A„. Oa. op: and there will remain merely a relation between the sines of angles subtended at 0. The property just proved must there- 1851.] GEOMETRY OF TWO DIMENSIONS. 169 fore be tnie for any figure into which the circle can be projected, that is for any conic section. (See the article on the Method of Projections, in Salmon's Conic Sections^ Chap. XIV.) 18. Prove one of the two follo%viiig properties. (1). When one of the foci of a conic section and two tan- gents are given, the locus of the other focus is a straight line. (2). When the centre of the conic section and two tangents are given, the locus of the focus is an equilateral hyperbola. The proof of these theorems depends on the property, that the product of the pei'pendiculars from the foci on the tangent at any point of a conic section is constant and equal to the square of the semiaxis minor. (1). Let J.P, AQ (fig. 79) be the two given tangents, S the given focus, H that whose locus is to be foimd. Draw /S'F, HZ perpendicular to AP\ SY\ HZ' io AQ'^ then SY.HZ= ST. HZ', .-. HZ'.HZ:: ST : SY, a constant ratio ; therefore the locus of His a straight line. (2). Take the centre as origin, and let the equations to the given tangents be cc cosa + y s'moL — a = (1), a^cosa' + i/ sina' — «' = (2). Let I, 77, be the coordinates of one focus, then — ^, — v^ will be those of the other. Now the length of the perpendicular from I, 7], to (1) is ^ cosa + T) sina — a; similarly, that from — ^, — 77, is — I cosa — 77 sina — a. Hence we get (I cosa + 77 sina)'"' - n' = ^\ /3 being the semiaxis minor. Similarly it may be shewn that (^cosa' -f- 77 sina'V^ — «'^ = ^'-^ 170 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. .•. (^ cosa + 7} sina)"'' — a" = (| cosa' + rj siiia')"' — «'^ or (I cosa + r) sina)'^ — (^ cosa' + r/ sina')"'' = «■* — « '^ the equation to the locus of |, t;, which is therefore a rectangular hyperbola, the equations to whose asymptotes are I (cos a + cos a') + t; (sin a + sin a') = 0, f (cosa — cosa') + ■j? (sina — sina') = 0. 171 DIFFERENTIAL CALCULUS. 1848. Find the equation to that involute of a cycloid which passes through the cusp, and shew that in the immediate neighl)om'hood of the cusp it becomes the curve 2a{4:yY = {3x)*y a being the radius of the generating circle. Let X, y be the coordinates of any point in the cycloid referred to the cusp as origm, and base as axis of ic, s its dis- tance measured along the arc from the cusp ; ^, rj those of the corresponding point in the involute. The equation to the cycloid will be X = a(^ — sin^), y = a[\ — cos^), and we have, since the tangent at {pcy) passes through (f?;) at a distance s from [xy]^ ^ dx dy dx Now 32 = rt(l — cos^) = 2asiii''^^, do -^= asin^ = 2a8ini^ cos^; da dd ^ ' and s = 4a(l - cos|^) ; .-. f = a(^-8in6') - 4a(l-co8^^) sin^^, = a(^ + sin^-4sm^^), 7) = a{l — cos^) — 4a (I — cos^^) cos^^, = «(3 + cos^ - Icos^^), the equations to the involute. 172 SOLUTIONS OF SENATE-HOUSK PROBLEMS. [1849. In the iiiinicdiatc neighbourhood of the origin where 6 is small, these become and 77 = a j 4 — - 6' . + 24-^ f-.f; 6" = %.4- Hence, eliminating 6^ a'-- =(f)' or the required curve. ■ 2«(477)^ = m% 1849. 1. K P be a point in a cycloid, and the con-esponding position of the centre of the generating circle, shew that PO touches another cycloid of half the dimensions. Let a(^ — sin^), a(l— cos^) be the coordinates of P, as in the last problem ; then a6 and a will be those of 0. The equation to PO is X — aO y — a sm V cos V or X cos Q -^ y ^VQ.Q — a{6 cos Q + sin 6). Differentiating this equation with respect to Q as variable parameter, — a; sin ^ + ?/ cos ^ = a (2 cos Q — ^ sin &) ; .*. a; = a(^ — sin^ cos^), = ia(26'-sin2^) (1), and ?/ — r; ( 1 + cos'* &) = \a\\ +cos^^) (2). 1849.] DIFFERENTIAL CALCULUS. 173 Equations (1) and (2) shew that the line OP always touches a cycloid whose cusp coincides with the cusp of tlie original cycloid, and generated by a circle of half the size of its gene- rating circle. 2. Find the locus of the ultimate intersections of the lines defined by the equation a;cos3^ + ?/sin3^ = a(cos2^)* (1), where 6 is the variable parameter. Differentiating (1) with respect to ^, a;sin3^ - ?/cos3^ = asin2^ (cos^)* (2). Squaring (1) and (2), and adding, x' J^ f = d'cos2d (3). Again, (2) -i- (1) gives ccsin3^ — ?/cos3^ , ^^ — z — ^ . ,^ = tan 2^, a;cos3c7 + ysm'do or — ^— — — — = tan 2^; 1 + tan3^^ X .-. ^= tan^, X 1 1 /«\ -1 -1 2 1 — tan*'^ and by (3) x' -\- i/ = a' - = a + tan' x'^-f d' + f or {x^ + yr-^^\^'-f)^ the equation to Bcmouilli's Lemniscate. 3. If e be the eccentricity of a conic section, r the distance of any point from the focus, p the radius of curvature at that point, and ds an element of the arc of the curve, then , Jr' , fd'r ' =d?-^PW' 174 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. Let the equation to the coiiie section referred to its focus as origin and axis-major as axis of a?, be x' + f = [ex + cY ; .*. r = rx + c, dr , dr _d'C _ e ds ds J ('hf\ dx \ \ dxj d'^r d dr I ds' dx ds ' ds ' dx e^ it dx'' dx 1 -||)T'I' dy^ e dx "^-(IJ drV , fdS'Y , ^ + \dx) dlX ds) ^P Uv ~'' 7±V' "^ [dxJ = e\ 4. If u be a function of the independent variables x, ;/, z^ given by the equations ^*=/(^, (1), s = F{Ix + w??/ + nz + ht) = [mz — ny) + % [nx^ ~Iz) + yjr [ly — mx)^ and if V + m' + n' = F ; shew that , du du du , du aa; dy dz dt where -j- is obtained from (1) by considering s constant. 1850.] DIFFERENTIAL CALCULUS. 176 We have also =z n')(^ — wi/r' ; dt ??^' — 7>li|r' I similarly | = |/(.) +-0)| (j-^^' - «f ) - |/((), Multiplying these equations by ?, wi, w respectively, and adding, remembering that 1^ + m" + w" = ^'^j ,^M c7m f?M , du . ^ J- + ?w -J- + w -J- + « -^ = 0, GKC dy dz dt ' du „,, , smce -J means/ [t). 1850. 1. A paraboloid of revolution with its axis vertical contains a quantity of water, into which is sunk a heavy sphere, and the water is just sufficient to cover the sphere ; find the form of the paraboloid that the quantity of water with which this can bo done may be the least possible. Let a be the radius of the sphere, I the latus-rectum of the paraboloid; h the height to which the water rises when the sphere is siuik : then if C be the content of the paraboloid of height ^, V the volume of the sphere, Q the quantity of water, g = o-r; and we have to make Q a minimum by the variation of I and h. 176 SOLUTIONS OF SENATE-HOUSE PK0BLEM8. [1850. Now C=l'rrlh', therefore IK^ must be a minimum. Now from the vertex the equation to the section of the paraboloid is / = ^x ; that to the section of the sphere is [x- {h-a)Y + y' = d\ In order that these may touch one another, we must have [x — {k — a)Y + Ix — a\ a perfect square, which requires that 4:{h'-2al) = {l-2{h-a)]% or r - U[h-a) + 4a' = (l). Hence we must make JK'' a minimum subject to the con- dition (1), which may be written £+_2ar. 4Z ' therefore we have to make (Z+2aV = mmmium I 4 1 or ^ T = ; .-. ? = fa; which determines the form of the paraboloid. 2. If a circle be described touching a curve at any point (r, 6) and passing through the pole, shew that the equation to the circle will be do r The general equation to a circle passing through the pole and the point (r, 6) is r — r sec(^— a) cos(^' — a) (1), 1850.] DIFFERENTIAL CALCULUS. 177 a being the angular coordinate of the diameter through the pole. This equation may be put in the form cos{6'-a + (^'-^)| j« ^= )• '■ !^ ~ cos [d — a) = r {cos {& -6) - tan {6 - a.) sin {& -6)]. Now from (1), 'ia> = ~ ''' ^^^ (^ ~ '^) '^'^ (^ ~ ^) = -j^ , when 6' = 6^ do since the circle touches the curve at the point (?•, 6)\ •'■ *■ r tan (^ - «) = ^ ) and r = r {cos [6' - ^) + "4 sin [6' - 6)] , d sm[e'-e) ~ '* dd r ' , d sm{0'-0) ^ or r' + r' ^ ^ ' = 0. dU r 3. If a parabola roll upon a line, the focus will trace out a catenary. The following more general problem admits of very easy solution: A given curve rolls upon a straight line, to find the locus of any point to which the curve is referred as pole. Let AB (fig. 79) be the given straight line, A any fixed point in it. Let CF be the rolling cm've, C the point which has been in contact with A, S the pole, P the point of contact in the position represented in the figure. Join ^SP and di'aw /SF per- pendicular to AB. Let A Y = x, YS = y, SF = r, s the arc of the cun^e described by S. Then the tangent being manifestly pei*pendicular to SF, we have '^ = cohFSY=^, ds r * For tills solution, wc are indebted to Mr. Goodwin N 178 H(.>LUTl(>NS OF SENATE-HOUSE PROBLEMS. [1850. Let the equation to the rolling curve be r'' =f{p)i then the equation of the required locus is ^■'{•H- (I)} =/(.). In the case of the parabola, we have '■=?^ dx y c ds r yl the differential equation to the catenary. 4. Find the forms of the curve whose equation is ,ii I ^i- xy = m [x + y — « ], according as nt^ is > = or < -^ a^. Arranging the equation as a quadratic in cc, we have and a^ = ^ + A {/ + 4/n' («'-/)}* (1). Hence we may use x = -^.^ as a guiding cm've ; its fonn is shewn by the dotted curve. To consider the equation or f - 4w* . f + ^m\t' = 0. This equation, considered as a cubic in y\ will have three 3 3J real roots or one, according as oti' is > or < — — a'"*.* If it has three real roots, one of them is negative, and the corre- 3 3- * If m = -j— «^, it vdH have three real roots, tAvo of them being equal. 1850.J DIFFERENTIAL CALCULUS. 179 spending values of y imaginaiy. If it has two equal, and there- fore two real roots, the equal roots are positive and the other negative, giving only one positive value to y'. If it has only one real root, it is negative and y imaginary. Again, differentiating the original equation, we get dy 2m'x — y^ dx ^fx - 2m'y ' therefore 1, when y = 0, and therefore x = a^ dx Also, taking the lower sign in equation (1), X = 2m' 2 m \ y^ -1 = — m 2 2 '"V &c. = if ?/ = GO : hence the axis of y is an asymptote to the cui-ve, which we thus see has the fonn represented in figs. 80, 81, 82, according 3.3- as m' is > = or < ~ — a^. 4 5. Trace the curve whose equation is 2a [r — cf = cd {2a — cO) when c = — , TT and prove that as c increases indefinitely the cui've approximates to a circle. When 6 is positive, c6 must be less than 2a or 6 loss than TT, and 6 can receive no negative value. Also, any value 6^ of gives the same value for ?• as tt — 6^. Solving the equation we have r = c± [cd[2a-ce)]^ 2a = — + 2a K(-^)r <■)• N 2 180 SOLUTIONS OF SENATE-HOUSE PHOBLEMS. [1850. The quantity affected with the ambiguity lias its greatest a value when - = ^, when r receives the value 2a r = — ± a IT the latter of which is negative. Again, putting ^ = 0, we have r = c] and differentiating, therefore when r = c, and therefore ^ = 0, dr 33 = °"- Hence the cui've is of the form shewn in fig. 83. Also, as c and therefore a is indefinitely enlarged, equation (1) becomes >• = — , representing a circle. TT 6. Find the locus of the consecutive intersections of the curve whose equation is x^ + y'^ = 2ax' + 2by' (1) ; a and h having any values which satisfy the equations 2„ (.!-,) = (."-/) I -2., (2), 2i(4-,)=.'-y' + 2.,| (3), X and y being the coordmates of any given curve. The problem is best solved by introducing polar coordinates. Let X = r cos 6^ y = r smO] .-. dx = — r &mdd6 -+ cosOdr^ dy = r co^6dd + shiddt", .'. xdy — ydx = r^dd^ [x^ — y^) dy — 2xydx = r^ cos2ddy — r^ sm20dx = r^ cos Odd — r^ s,m 6 dr^ {x^ —y'^) dx + 2xydy = r^ sin Odd + r^ cos 6 dr. 1851.] DIFFERENTIAL CALCULUS. 181 Hence equations (2) and (3) become 2a = r cos 6 — sin 6 -f^ , da 2b = ?• sin 6 + cos -j^ : au da . . ^'r .-. 2^=->-sin^-sm^^, 2 -77: = r cos c? — cos U ^7^ . Also equation (1) transformed into polar coordinates becomes r' = '2a cos 0' + 2b sin 0'. Differentiating this equation with respect to 0, considering r' and 0' constant, we have = 2 -77. cos^ +2-77= sm^ dtf dv = r&m{0'-0)-cos{0'-0)'^,- tan (^'-6') = dff' From this equation, when r has been substituted in terms of 0^ from the known equation to the curve, we can find in tei-ms of 0'', and thence r and -^ will be known in tcnns dtf of 0'j and the required equation to the curve will be r = 2a cos^' + 2b sin^' dr = rcos{0'-0) + Hm{0'-0)%. d0 1851. If (f) (c) be a rational and integral function of c, the coeffi- cients of which are functions of any number of variables ^, ?/,,.. then if 8 denote differentiation with respect to the variables, and the quantity c be eliminated from the equations [c] = 0, 182 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. 3(/)(t')=(), the result may be represented by tiSPSQ... = where P, Qy.. are the roots of the equation {c) = 0, and tc = is the result of the elimination of c from the equations (p (c) = 0, '{c)=0. Since 0(c) is a rational and integral function of c, and P, Qy.. are the roots of the equation 0(c) = 0, we have (c) = [c — P){c— Q) identically. dP dP Let P^ denote ^- dx. P. -7- diu... then ^ c?^ ' "' dy '^' g0(c) = 0(c) Slog 0(c) =-*w(^^^^-^^--) =-{(^.+^.+-)(o-^)(c-^)-+(^.+^.+..-)(c-^)(«-^)+"l- Hence the result of the elimination of c between 0(c) = and S0(c) = 0, is = product of the expressions (P^. + P^+...) (P- Q) (P- /?)..., ((?.+ (?,+•••) («-P)(^-P)...,&c. = (P,+P^+...)(^,+ ^,+...)...(P-(?)(P-P)...(^-P)((2-P)... = v suppose. Again, 0'(c) = (c-^) (c-P)...+ (c-P) (c- P). ..+... ; .-. ,,= (P-^)(P-P)...(^-P)(^-P)..., and SP=P, + P, +..., 8^= ^,.+ ^^4-...; .-. v = uBPBQ.... Hence the result of the elimination may be represented by u8P8Q... =0. ( l«-i ) INTEGRAL CALCULUS. 1848. The comer of a sheet of paper is tunied down so that the sum of the edges turned down is constant ; find the equation to the curve traced out by the vertex of the angle ; find also the area of the curve. Let r, ^, be the polar coordinates of the vertex, refeiTed to the origmal position of the vertex as pole, then the lengths of the respective edges are ^r sec 0, ^ r cosec ^, respectively ; therefore the equation to the curve, is l^?- (sec ^ + cosec ^) = constant = a suppose, or in rectangular coordinates, {x + 7/) {x' + f) = 2axy. To find the area, turn the axes through an angle ^tt, then we get _acos26'_ *" ~ 2icos^ ' therefore if A be the area of the loop traced out by the vertex, A=-l rW 1 r\^ i-^ cos' 2^ ,^ 27j ff" COS a 4cos''^-4+-^] iW cos a/ 2cos2^-2 + 6ec'<9) J ^ sec (f) a = — T I (a'^ sin''' <^ + Z>^ cos"' <^) c?^ Again, let ^' be the area of the ellipse which has the same greatest and least diameters, then if these diameters be 2rj, 2?\^, A = Trr/^- cos'' sin"'' 6 4 "I" 74 JNow r = c a' = -5T5 («' cos'''0 + &''' sin'*^) («■'' sin''*^ + h" cos'^0) cos 20 d'b' (V 2 / V 2 The maximum and minimum values of r will be got by putting cos20 = and 1 successively; •*■ "*> ~ a* 2 ' ''•' ~ "■' ., _ TTC^ d' + h' *'• ~ab ~~2~ ' the same value as that previously got for A. 18 1 lies between -r^ — -tt^i ■> f = I — ) and -, w 73-^ , approaehiner much nearer to the V j^^yx'j (a-l)(« + ir" ^^ ^ former limit when n is large. (1). In general 1 1 + ^,« + °L(^) ,n-Y +. [m—pY [m + 2^) > Now + 1 +...+ {n + l-rj))" [n + 1 - {r - 1)2)Y {n + iy 1 +... + + + l + {r-l]2)Y {n+l+rjjy + + l—rpY {n+l + ypY + l{n+l- (r-l)|>] = + 1 {71 + 1 + {r- 1)2)Y_ 1 +...+ {n + iy > -, r- + 7 rr- +•••+ -, 7T- 1 from above, [n + \Y [n+lY {n+\Y^ ' 2/- + 1 ^ [n+lY' P + P " {n+l-rpY [n+l-[r-l)2iY Now let f'p = \ - Pi then this becomes P , P , , i^ +...> 212? + pj {n + lY r« + +...+ {n + i +2)Y [n + ^ + 22)Y {n + i -pT {» + 1)'" ' therefore, a fortiori^ P + ...+ T P (" + i+7?)^ ■■■ (n + i)* [n + lY 1848.] INTEGRAL CALCULUS. 187 Let j9 = dx^ then this becomes r"^<& 1 / > n.h^'^ (/^ + 1)^' similarly j^^^^->^^^^^, > We shall thus obtain an infinite series of inequalities similar to the above. Adding them all together, we get "^ dx 1 1 1 + 7— T^r, +••■ r dx_ „,_ (a-l)(« + ir {n+iy (n + 2) Again, if ^j — ^iV being any integer, + ^"~ri Tirvr +••• (/ te™is) (/t + 1 +i^)" (» + 1 + 2^7)^ < -;^ — - — c- + , ^r - +... (;»' terms), («+l)" («+l)' ^^ ^' 1 < (n + 1)^ For /> write dx^ then this becomes ^''+''' ^ 1 / whence, as above, we get '" ^ _ 1 1 ^ ~ (a-l)(« + l)''' ^ (« + 1)^ ^ (7i + 2) [ — = + 7Z-n7V.+-"5 We have / — - = -, rr ^7 — -Tvsrr - 7 — rsv^l ? 1 r 1 .V 2,1+2) V^^2«+2 2 f a - 1 (a-l)(n+l)'-* (2n4-2 ( a-l)(a-2)(a-3) 1 ■^ 6 (2n + 2)''^' 1 («-2)(«-3) . (/?+ 1)' 24(« + i; 188 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. r^ dx 1__ ^ (a-2)(«-3) R . r ^ - _L_ \ \ L_i •'„. ^""(«-i) t("+ir (n+2rf' 1 -T TT^U (a-1) (n+ir i ^1 , 1 n+ 1 a-1 (a-l)(a-2) 1 a-1 (w+ip i(n + l) 2 (n+1 1 a-2 1 (w + 1)' 2 (w+1)^"^ ""^•■^^a; a-2 +. \ r rx+- (n + ir J„,, a- 2(,i + i; It hence appears that when n is large, -. p- approaches r"+l ^:p r""^^ dx much more nearly to the limit I — than to I — , whence the latter part of the proposed theorem readily follows. 4. If f[x) be positive and finite from x = a to x = a + h^ shew how to find the limit of for n = 00 , and prove that the limit in question is less than T I f{^) ^^^) assuming that the geometric mean of a finite number of positive quantities which are not all equal is less than the arithmetic. Hence prove that s-'^o'"'* < /„ £'"'*, unless u be constant from £C = to £C = 1. Let log/(ir) = -Fix), then log {/(«)/(« + ', a) .../(« + '^ hj^\ = ?/ suppose. 1849.] INTEGRAL CALCULUS. 189 When n is infinite, let - = dxi then ' n 7/ = ^ {F{a) + F{a + dx) +...+ F{a + h)}, 1 /•' = J I F{a + x) dx ; ...{/W/(., + i ;,).../(„ + ^A)f approaches to the limit e^-^" iog/t««)''' ^^ ^^f„ iog/(«)rf«^ -^^^ smce the geometric mean of a finite number of positive quantities which are not all equal, is less than the arithmetic, {/(«)/(« + i a). .,/(« + !i^^ a)}* as long as n is finite, and this will hold up to the limit when n is indefinitely increased : but in that case therefore the required limit < y I f[x)dx. Hence if /(o^) = e", « = 0, and ^ = 1, s-^i""' < /^s"(7a', unless ?i be constant from x = to ic=l, in which case they are equal. 1849. 1. Investigate the series &' tt' ^ cos 20 cos 30 „ for values of between — ir and tt. Let COS0 — ^cos20 + |co830 — ... = n, sin0 - ^sin20 + ^cos30 - ...= v, lyO SOLUTIONS OF SKNATE-IJOUSE PROBLEMS. [1849. then if lie between — tt and tt, U +-iv = £-*^ - i£-*'^' + ^3 *, = l0g(l+£-^) = l0g(£ ^U £-') + l0g£-*S = log(2cosi0) + -H^; therefore equating imaginary parts, V = sin0 - ^sin20 + |cos30 -... = ^0: integrating with respect to 0, 11 R^ - COS0 -f -5 COS20 - -2 COS30 + ... = -+ (7. To determine the constant, put 0=0; I 1 ^ Now 1 - |. + ^, -■••= 1 + I + y +••■- 2 (j! + p +••• II , /. 1 1 . =l+2^. + 3.+-.-Hl + 5. + 3.+. ..osi„e=«|l-(|)]{,-(l)}..., and sind = 6 — — — ^ +... : equating coefficients of ^^, 1 /. 1 1 fl'2 TT^ 1 1 and — =T7; — cos0 + ^ cos 20 — -r, cos20 +•... 4 12 2'' S'' 2. If a line be drawn through the centre of an ellipse, cutting the major axis at an angle 0, and the curve at an angle 1849.] INTEGRAL CALCULUS. 191 = - f ^ + tan'^^ ; , _ _V cos"' B + a sin' •■• ^^"*P~ "(«•''-//■') sine COS© ^ (a-'+^>'-')(cos''g+sin-'0)-(ff''-^/)(cos'''e-siu-'g) 2(«'''-Z>-) sin cos _ fl--' + ^-^ - (^-^ _ ^^) C03 2g («■■'- 6^) sin 20 ' .-. (a' - 6') (sin20 tan<^ + cos20) = d' + //, and («' - />') cos {29 - (f)) = («' + Z»"') cos<^. (2). Again, since _ b"" cos' + a" iiW 9 ^"^9-- {d'-h-^)^[n9cos9 ' 192 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. = / X?n7r - tan -r-r^ — ,2. . ^ ^h at/ (2) 1 C"' = - I onvdO by adding (1) and (2) where ?« is a constant integer to be determined. For this purpose we observe, first, that its vahie is inde- pendent of any relation between a and h • and secondly, that if a = b the ellipse becomes a circle and (jy always = ^tt. Hence in this case (Pde= ^dB = — and m = - 1. Hence also, in all cases, m = — 1 and f J ^dB = -- 3. Through a given point B (fig. 84) of the axis of a; a line is drawn parallel to the axis of y : to any point Q of this line another straight line is drawn from the origin and produced to P until PQ = BQ. Find the equation to the locus of P, trace the cui've, and find the whole area included between the cm've and the asymptote. Extend the geometrical description so as to include the whole of the curve given by the equation. Let AB = a, AP = r, PAB = 0. Then QP = r — a sec B, QB = a tan ; .*. r = a(sec0 + tan0), the polar equation to the curve. 1849.] INTEGRAL CALCULUS. 193 DifFcrentiating, -jji = asccO (t&nO + secO) da = r sec 0, .". r^ -J- = rco&O = a (1 + sinO). dr When = - and — , r = oo and r^ -^ = 2a and : hence 2 2' dr ' the axis of ?/, and the line DCD parallel to it such that AC =2a^ are asymptotes to the curve : f\ r. I dr ^ ^ = ^' '• = ^' rTe = ^^ > 0, < 2 "■? ^ i^ positive ; „ TT 1 + sin . t/ > — < TT, r = — a J. — IS negative ; 2 ' COS0 ^ ' rt Stt 1 - sin . t/>7r<— -, r = — a jr— is nesrativc : 2 ' COS0 ^ ' B = ^Stt, r = cc , „ Stt ^ 1 - sin0 . > — < 27r, r = a ^ - is positive : 2 ' • cost^ ^ the negative values of give no new branch of the curve. Hence the cun^e is of the form represented in (fig. 85). To find the area [A) included between the curve and the as}Tnptote D CD'. Produce AF to meet the asymptote in R ; then the element of the area A BA = ^{AE' - AF') 8A = ^pasecOy - a' (sec0 + tan0f} SO = ^a" (3 sec' - 2 sec tan - tan" 6) BO = ^d' (2 sec'd - 2 sec0 tan0 + 1) SB ; .-. A = ^d' f2tan0- -\ + b] + C ' \ cosy / ^ [ \l + smBj j 194 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. from 6 = to = ^TT gives ^A, .-. ^ = (^7r-f 2) d\ the required area. If P' be the point where AP cuts the branch BP' of the curve, it is evident from the tracing of the curve that QP = QB : hence the curve may be described as the locus of the point P on the Hne AR whose distance from Q equals QB. ( 195 ) GEOMETRY OF THREE DIMENSIONS. 1848. 1. If three chords be drawn mutually at right angles through a fixed pomt within a surface of the second order whose equation is u = 0, shew that 2 ^- will be constant, where R and r are the two portions into which any one of the chords drawn through the fixed point is divided by that point. Prove also that the same will be true, if instead of the fixed point there be substituted any point in the sm'face whose equa- tion is w = c. We shall prove the second part of this only, since it mani- festly includes the first. Let the equation to the surface a = 0, referred to its centre and axes, be Ax" + Bf + C£' = 1 (1). Let a, /8, 7 be the point through which the lines are drawn ; then, since it always lies on the surface u = c, we have Ad' + ^/3'^ + Ct^ = 1 + c (2). Let l^m^n^^ ^i^h^t h^^^a'hi ^^ ^^^ direction-cosines of the lines, then their equations are -,— = ^ -= - = Px say 3), h ~ -\ ~ >\ ~^' ^^' xj-a y - ^ z-y ^'^ =^r=''' <"' 02 106 SOLUTIONS OF SENATE-HorSI-: PROBLEMS. [1848. ('.»'.»,)7 (^.;'».2*'.2)^ ih"',",) lacing, since (3), (4), (5) are at right angles to one another, subject to the conditions K' + f: + K =1 (6), < + < + <=! (7), < + < + < = 1 (8). Where (3) meets (1) we have, substituting for xyz in tenns ^ (a + /.p,)'^ + ^ (/3 + m,p^Y + ^ (7 + n.p.Y = L The roots of this, considered as an equation in p^, are R^r ; hence Rr ~ Aa' + ^y8^ + Cy' - 1 ^ Ai;' + ^m;-^ + c^;^ c Similar expressions resulting from (4) and (5), we get by (6), (7), (8), ^ 1 A+B+C ^ Rr~ c ' which is constant. 2. Find the locus of the foot of the pei-pendicular let fall from the origin on the tangent plane to the surface xyz = a^ ; point out the general form of the required sm'face, and find the whole included volmue. The equation to the tangent plane to the given surface at a point {xyz), is ^, V, ^, ^ + ^ + - = 3. xyz The equations to the pei'pendicular on this plane from the origin, are At the intersection of these we have ^^1 = yy^ = ^^1 = - — 3^ ; therefore, since xyz = a^, we get as the equation to the locus required, (^^-^ + y^^ + ^-^ = 21a'x^y^z^. 1848.] GEOMETUY OF THREE DIMENSIONS. 197 The form of this surface will be that of four similar sym- metrical pear-shaped portions, meeting in a point at the origin, and lying in the octants + + +, -\ , - + — , h- The equation to the surface, transfonned to polar coordinates, becomes r" = 27rt^ cosO slu^0 cos^ sin^. And if V be the volume of one of the portions V = ^JfJr' smSdrdddi/, between proper limits, 9 . /-*"/■-" = - aM I COS0 sin'0 cos^ sm 2 J -' 9 ; = - a^ j cos (p sin (p dcj) = ^ a' 16 therefore If V be the whole volume of the surface, V= 4F' = -a^ 4 3. A plane moves so as always to enclose between Itself and a given surface S a constant volume ; prove that the envelope of the system of such planes is the same as the locus of the centres of gravity of the portions of the planes comprised within S. Conceive the plane to receive a small twist about any straight line passing through the centre of gravity of the portu)n com- prised within S; then, whatever portion is cut off from the enclosed volume on one side of this line, an equal portion will be added to it on the other,* so that, by the conditions of the problem, the plane will pass from any one position to the con- secutive one by turning about a line passing through the centre of gravity of the portion comprised within S. Therefore the en- velope of the planes will be the same as the locus of the centres of gravity of the portions of the planes comprised within S. * See Cambridge and Dublin Mathematical Journal, vol. iii. p. 181. 198 SOLUTIONS OF SENATK-IIOUSE PROBLEMS. [1848- 4. OA^ OB^ OC, are three straight Ihies mutually at right angles, and a lumuious point is placed at C; shew that when the quantity of light received upon the triangle A OB is con- stant, the cui"ve which is always touched by AB will be an hyperbola whose equation referred to the axes OA^ OB^ is [y—mx) (x — my) = m&^ where OC = c^ and m is a constant quantity. With C as centre, and CO as radius, describe a spherical surface, then the quantity of light received on the triangle A OB Is the same as that received by the spherical triangle CA'B' intercepted between the planes COA^ COB^ CBA^ and will therefore be proportional to the area of that surface. But if S be this area, S = ^irr' {A OB' + OAB' + OB' A - it) = 27rr' ( OAB' -f OB' A - ^tt), since A OB is a right angle. Therefore if the quantity of light received by the triangle be constant, OAB' + OB' A must be so, = 2a suppose. Let the angle OB' A = a + 0^ then OAB' will = a-d^ and the equation to the plane ABC will be cos(a+^) x-\- cos[a — 6)y+ {1 -cos'(a + ^) -cos''(a- ^)}*^=7>; p will be detennined from the consideration that where this meets the axis of s, we have s = c ; .-. {l-co3'''(a+^)-cos'(a-^)li = c, therefore the equation to AB is cos {c(.+ 6) X + cos[a— 6) y = (1 — cos'"' {a+ 6] — cos^ (a - 6)]^ c = (-cos2acos2^)4 c (1). The quantity (— cos2acos^)* is real, since 2a is greater than a right angle and less than two right angles, and therefore cos 2 a negative. Putting tana = w, tan^ = t, (1) becomes (1 - 7it) x+ [l + nt] y = {{n' - 1) (1 - f]}^ c, and we have to find the locus of ultimate intersections of this line, subject to the variation of f. 1848.] GEOMETRY OF THREE DIMENSIONS. 199 Clearing the equation of radicals and arranging according to powers of f, it becomes f [re [x-yf + in' - 1) c'} + 2tn [if-x') + (x + yf - {n' - 1) c' = 0. Eliminating t between this equation and its derivative, we get {[x + yY + {n^ - 1) c'] [n' [x - yY + {n^ - 1 ) c'^} = n^ (f - xj, which may be reduced to [x + yr-re[x-yf={n^-\)c\ or {(1 -n)x + (1+ n) y} {(1 + n) ic -I- (1 - n) y\ = («' - 1) c' ; 1 f . n — \ therefore puttmg = »i, [y — mx) [x — my) = hik? is the equation to the curve always touched by AB. In a manner similar to this may be solved the following problem, set in 1851. Let a spherical surface whose centre is the origin of coor- dinates meet two of the coordinate planes in the great circles Zx^ Zy\ also let the points P, Q be taken m Zx^ Zy respec- tively, so as to make the surface of the spherical triangle PZQ constant: shew that the curve which is always touched by the great circle PQ has for its equations x^ -\- y^ -{■ ^ z= d^^ and xy = ^d^ sin^, where E is the spherical excess of the triangle PZQ. The geometrical conditions of this problem are the same as those of the foregoing. Writing z for c, we have as the equation to the surface always touched by the plane through the centre, 2 7r + 1 ^ ^ ' + x' + / + z' cos2a. 200 SOLUTIONS or Si:NATl->Ht>i:SE PROBLEMS. [1848. But -rr + E= A' OB' + OAB' + OB' A In the previous notation = -^TT + 2a ; .-. 2a = ^TT + E, and our equation becomes xy = \ (a;"'* + ^^ + 2:''') sinE". But since the curve is traced on the sphere, w^e have x^ -\- y^ + z^ = a^, and we get as the equations to the curve, X + y '\- z =^ a , xy = ^a" sinjE", the required equations. 5. If be a given point in a surface of the second order, and OA^ OB, 00, any three chords passing through mutually at right angles, shew that the plane ABC will always pass through a fixed point. Take as origin, and the three lines OA, OB, OC, xn. any position as axes ; let the equation to the surface be Ax'->rBy'+Cz'-\-2Ayz+2B'zx-\-2C'xy+'iA"x+W'y+2C"z={). Then the length of OA will be the value of x, when y = 0, 2 = 0; hence 9 A" 0A = -^-: A 2B" 2C" similarly OB = ^- , OC = y^ , and the equation to ABC will be Ax By Cz ^ ^ ,,, y+^ + ^, + 2 = (1). And the equations to the normal at are _ 3/ _ 2 A" ~ 2B" ~ 2C"~ 2 [A"" + B"^ + C"''f where r is the distance from the origin of the point [xyz). (2), 1849.] GEOMETRY UF THREE DIMENSIONS. 201 Where (I) and (2) intersect, we have, dividing eaeli term of (1) by the corresponding member of (2), r Now we may establish one relation among the nine coeffi- cients of the equation to the surface. Let then [A"^ + B'"^ + C"'^) be constant, then the above equation shews that r varies inversely as ^ + 5 + a But it is known that if the equation be transformed into the form Px' + Qy' + Bz' + 2P"x + 2Q"y + 2B"z = 0, the quantities P, Q^ B, are the roots of the equation (-S'- A) [S- B) {8- C) - A" [S-A)- B" [S-B]- C" {S- C) - 2A'B'C' = 0. Hence, by the theory of equations, A+B+C=P+Q + B, a constant. Hence ?*, the distance from of the point in which the plane ABC intersects -the normal at 0, is constant, therefore the plane ABC always passes through a fixed point. 1849. 1. If planes be drawn through any two generating lines of an hyperboloid which intersect, shew that they will cut the surface in another pair of generating lines. Let the equation to the hyperboloid be '2 2 2 -+^--=1 (1) Now a plane, drawn through two intersecting generating lines of an hyperboloid, touches the hyperboloid at their point of intersection. Let then x\ y\ z\ be the coordinates of this point, then the equation to the plane will be XX yy zz +"/-^: = i c^). 202 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. a;', y\ z\ being subject to the condition or h' c ^+^-:;ii = i (3), (1) + 2 (2) + (3) gives m-m'-m- (^)- a condition which must be satisfied by the coordinates of any point where (2) intersects (1). Now (4) may be put into the form ic + aj'Y'* . _ (z-\-z'^ (y-^y a which may be written shewing that where (2) meets (1) we have either fi±£:+2=i.fyif: + y±l] and ^^-2 = \ {'-±i - ^f] , O/ \C U J Ct rC \ C J representing one generating line, or ^E±ii + 2 = i' (-£±1 _ 2^) and i^^' - 2 = '., f ^ + ?t±l^) a \ c J a fc \ c b ) representing another. Hence if planes be drawn through any two generating lines of an hyperboloid which intersect, they will cut the surface in another pair of generating lines. 2. If u =f[x^ y, z) be a rational function of cc, y, 0, and if w = be the equation to a surface, for a point («, J, c) of which all the partial diiferential coefficients of u as far as those of the (w — 1)* order vanish, shew that the conical sm-face whose equation is [(x-o) ^ + (3/-^) I + (^-^) j} /(«, h c) = 0, will touch the proposed surface at the point (a, J, c). 1849.] GEOMETUY OF THREE DIMENSIONS. 203 Tx X — ay — hz — c ,. Let -^j—=^ = — — 1 / 111 /rt ^ ' be the equations to any line passing through (a, Z>, c). Denoting each member of ( I ) by r, we shall obtain the other points of intersection of (1) with u = by writing a -f- Ir for cc, h -\- mr for y, c + nr for z in the equation u = 0. This gives, developing by Taylor's Theorem, 7 HI 7 HI 7 n« which, since -^ = -^ = -j-y^^ = for all values of m less than n becomes, dividinsr out by , ' ^ ^ 1.2. ..w' If the line (1) touch the surface ?< = at the point («, J, c) equation (2) must be satisfied by making r indefinitely small; (2) will then become a condition to be satisfied by the direction-cosines of (1) in order that it may touch m = at the point (a, 5, c). To obtain the locus of all such lines, we must eliminate /, w, n from the above equation by means of (1). This gives j(^-a) I- + (2,-i) I + (.- c) gV{a, J, c) = as the equation to the conical surface which touches u = at the point (^/, />, c). 204 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. 3. A rod AB is fixed to a universal joint to A^ and another rod BP is conneeted to it by a universal joint at B: all di- rections of the rod being equally probable, find the chance of P lying between two spherical surfaces of given radii, whose connuon centre is A ; and shew that the chance of P lying within a given elementary portion of space containing the point P, varies inversely as AP^. Let AB =a,BP= h. The chance that P will lie between two spherical surfaces of given radii, is the chance that the angle ABP will lie be- tween two values 6^ and 6^, which correspond to the values r^ and r^ of -4P, r^ and r^ being the radii of the spherical shell. Now the chance that ABP will lie between 6 and d + B0 = area of zone described by P about B fixed, while has all values from ^ to + B6 ^ surface of sphere generated by P about B fixed, 27rb sin^ x b s[n0 , . /, ,/, 4:7rb' ^ Hence the chance required 1 f ^ . ^ 1, z, ■/,^ Ifa' + b'-r;' a' + b'-r __ I oivi /-/ — I ona H nr\a H \ -— i — i^ - 1 sin ^ = - (cos 0, — cos'^, 2 / „ 2 ^ ' ' ''' 2\ 2ab 2ab 4ab ' The chance that P will lie in an element V of space about P^ = chance of falling in a spherical shell about A as centre of ^, . 1 -, ,. volume of element thickness or, radius r x ^ ;^-^j — n- ' volume ot shell _ 2rSr V _ V 1 1 4rtZ' 47rr''*^r STvab ' r r' 4. Determine the condition to which the vertices of a system of cones which envelope an ellipsoid must be subject, in order that the centres of the ellipses of contact may be equidistant from the centre of the ellipsoid. 1840.] GEOMETRY OF THREE DIMENSIONS. 205 Let ^, 7;, ^, be the coordinates of the vertex of any one of the cones ; then if tlie ellipse be refeiTcd to its centre and axes, the equation to the plane of contact will be 1:^ + ^ + ^ = 1 (1). The centre of the ellipse of contact will be the intersection of (1) with the straight line joining the centre of the ellipse with the vertex of the cone ; its equations are X ^ z Hence if h, A-, I, be the coordinates of the centre of the ellipse, In order that the centres of the ellipses may be equidistant from the centre of the ellipsoid, we must have Ji^ + I? + T^ = constant, p^ suppose ; the equation to the locus of the vertices. 5. Determine the form of the termination of a honeycomb cell on this principle, that if a sphere which will just pass through the hexagonal transverse section be dropped into the cell, the unoccupied space at the extremity of the cell shall be the least possible.* Let ahe (fig. 86) represent half of one of the rhomboidal plates, three of which close each hexagonal cell. ABC re- present the eighth part of a sphere. Then, by the general principle of Envelopes (see Cambridge and Dublin Mathematical Journal^ vol. iii. p. 181), the volume in question is least when the point of contact d is the centre of the rhomboid: or we must have ac = lad. Let OA = a, 07) = r ; .'. a = r cos .30° = r - . * For this solution we are indebted to Mr. Goodwin, 206 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. Let Oac = J .-. a = Od = ad tan0 = ^ac tan^ = ^r cosec^ tan0 ; . 3* , 1 COS(f> ' .'. COS 9 = ,-T . 3* Again, hd = ED sin 60 = r — , and ad = ^r cosecc^ ; .'. t&nbad = ^ = 3* sm(f> = 3* (1 - i)4 = 2*. These are the angles required. 6. The tangent plane to a surface S cuts an ellipsoid, and the locus of the vertex of the cone which touches the ellipsoid in the curve of intersection is another surface S'. Prove that S and S' are reciprocal, that is, that 8 may be generated from 8' in the same mamier as ;S^' has been generated from 8. Take the axes of the ellipsoid as coordinate axes, and let the equation to 8 be ^ = 0. ° • That to its tangent plane at any point xt/z, is , V d8 I , 68 , , d8 This may be put under the form d8 d8 d8 dS dS d8~ dx "^ dy dz If ^, 77, ^, be the coordmates of the vertex of the cone touch- ing the ellipsoid in the cm-ve of intersection with this plane, we have ^ ^ dx a'^"^ dS dS' dx ^ dy dz with similar expressions for t] and ^. 1849.] GEOMETRY OF THREE DIMENSIONS. 207 Again, if with [xyz) a point of S as vertex we describe a cone touching the ellipsoid, the equation to the plane of contact will be «l , ^ , ^ _ 1 d' ^ h' "^ 6' ~ ' 1^, 77, f being its current coordinates. To find the locus of ultimate intersection of these planes, eliminate .r, ?/, z between the differential of the preceding equation, and of this gives ^^ + ^ = 0, h dy Multiplying these equations in order by a?, y, 2, and adding, ^'eget js dS dS ^ ^ + ^;^ + ^^ + '^='' dS ^ dx dx " dy dz with similar expressions for 77, ^. Hence the locus of f , 97, ^ is 5". That is, the locus of the vertex of the cone touching the ellipsoid in its curve of intersection with a tangent plane to 8 is the same as the envelope of the plane of contact when a cone is drawn from a point of 8 as vertex, circumscribing the ellipsoid. This holds for all surfaces, therefore for 8' . But from the mode of generation of 8\ it is easy to see that the envelope of the planes of contact of cones drawn from its points as vertices is 8\ therefore, by what has been proved, the locus of the vertices of the cones touching the ellipsoid in its curves of intersection with the tangent planes to 8' is 8^ that 208 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. is, S may be generated from S' in the same manner as S' was from S, or S and S' are reciprocal. 1850. 1. If from a point be drawn any two lines to the polar plane of in a surface of the second order and meet the plane in A and i?, and if the central conjugate plane of OA meet OB in C, and the central conjugate plane of OB meet OA in Z), CB is parallel to AB. Take that diameter of the ellipsoid which passes through 0, and two diameters conjugate to it, as axes. Let the equation to the ellipsoid be H 2 2 X y z 1- — -I = 1 a b c Let ^ be the distance of from the centre, then the equation to its polar plane is a " = T- Let the coordinates of ^ be — , ?/j s^ ; of ^, -^ , y^, z^. Then the equations to AB are a'" y — V, z — z^ ,. f 3/l - 3^2 ^1 - ^2 The equations to OA will be a" X ^ t y z :75 ^ = f" = ^ = ^ suppose (2), I . therefore the equation to its central conjugate plane is 'a"^ J\ X w,?/ z^z l-f)5=' + ¥ + iP = ° <-'>■ Similarly, the equations to OB are '2 a ^ — -y ^^ = ^ = i = .-, (4), 1850.] GEOMETRY OF THREE DIMENSIONS. 209 therefore that to its central conjugate plane is At the point C, the intersection of (3) with (4), we have 1-^ Similarly, it will be seen that at i), the intersection of (2) and (5), we have a '2 / '•! , the values of .r at C and D are each equal ^o -g + (^ - I) »',? therefore the equations to CD are «^ = T + h--^ri' I VI J '' y,-y^ ^1 - ^2 ' by comparing which equations with (1) we see that CD is parallel to^i?. 2. A plane moves so as always to cut off from an ellipsoid the same volume ; shew that it will in every position touch a similar and concentric ellipsoid. If a plane be drawn touching the interior of two similar and concentric ellipsoids, the point of contact will be the centre of its elliptic section made by the exterior one. Now conceive this plane to receive a small twist about any diameter : it will still remain in contact with the interior ellipsoid, and whatever portion is taken from the volmne intercepted between it and the exterior ellipsoid on one side, will be added to it on the other, therefore that volume will be unaltered. Hence con- versely, it follows that if a plane move so as always to cut off p 210 SOLUTIONS OF SENATF.-IIOUSE PROBLEMS. [I8o0. from an ellipsoid the same volume, the sm*tacc which it always touches will be a similar and concentric ellipsoid. 3. If F{x^ y^ c) = be the equation of a system of curves, where c is a variable parameter, and (f) (.r, y) = the equation of the envelope of the system ; shew that ^ (.x, y) = is the equation of a cylmder whose intersection with the surface jP(.r, y, s) = is the locus of points which in sections parallel to the planes of yx^ zx^ have their tangents parallel to the axis of z. Ex. The cone whose equation is a? + y'^ + s^ = [Ix + my + nzj is cut by planes parallel to the planes of yz and zx ; find the loci of the extremities of the diameters of the sections which are con- jugate to the vertical diameter. (a). The equation (f>{x,y)=0 (1) results from the elimination of c between the equations F{x,y,c) = 0, and -^- = 0. dc It will therefore be also obtained by eliminating s between F{x,y,z)=^0 (2), and -^ = 0. dz Hence, where the sm-faces represented by (1) and (2) inter- sect, we have f- (3). Now the equation to a tangent plane to (1), parallel to the plane of yz. is , ,dF , ,dF ^ 1850.] GEOMETRY OF THREE DIMENSIONS. 211 dF If tlicrcforc -^- = o, this becomes az which is evidently parallel to the axis of z. Hence at the curve of intersection of the cylinder (1; and the surface (2), the tangent in a section parallel to the jdane of yz is parallel to the axis of z. Similarly it may be shewn that the tangent in a section parallel to the plane of xz is parallel to the axis of z. (/3). In the example, the tangent at the required points are parallel to the vertical diameter, tliat is to the axis of z^ hence we get the locus required by eliminating z between F{x^ y, z) = oc' + / 4- z' - [Ix + my + nzf = 0, dF and -T- = 2z — 2n [Ix + my + nz) = 0. The latter equation gives Ix + my 1 - n^ Hence a.'" +f= [Jx + my; (l + ,^^ - n' ^^^^1' _ {Ix + myY is the equation to a cylinder, whose intersection with the given surface is the required locus. 4. If .r, ?/, 2-, be the coordinates of any point P on the surface f{x, y, z) = 0, x\ y\ z of a point F on the surface /(a?', y\ z') = 0, and for any position of P, x' = Ix^ y = my^ z = nz ; and if the surfaces be such that when we take any two points P, Q on the first and two corresponding points P', Q' on the second, PQ is equal to P ^ ; find the fonn of the sm^aces. Let f , 17, ^ be the coordinates of Q. Then ?f , W17, nf are those of Q\ ... PQ- ={x- I^Y +{y- mriY + [z - nXf, PC/ = (I - Ixf + (7; - my? + (^-. nz)% P2 212 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. and these are equal ; hence we get (^ - I^y + {y- mvT +{z- n^Y = (^ - Ixf + (77 - m^/f + {^- nz)\ oi'{l-r')x'+{l-ni')f+{\-7{')z'={\-r')^'+{l-m')v' + {l-n')^\ for all values of x, 3/, z ; |, ■»;, ^, consistent with the equation to the surface /(.r, y, z) = 0. We must therefore have (1 - r) x' + (1 - m') y' + (1 - '?'") z' = constant, a' suppose, which determines the form of the required surfaces, which are evidently central smfaces of the second order, of which the axes of coordinates are principal axes. 5. It is not possible to fill any given space with a number of regular polyhedrons of the same kind except cubes, but this may be done by means of tetrahedrons and octahedrons which have equal faces, by using twice as many of the fonner as of the latter. Consider two octahedi'a so placed that two of their edges shall coincide, and the squares of which they are sides be in the same plane. Let AB (fig. 87) be either of these edges, G a vertex of one octahedi'on, not lying in the plane of the squares, D the corresponding vertex of the other. Then CD = a side of the square = AB = CA = CB = BD = AD^ by definition of a regular octahedron. Hence CADB is a regular tetrahedi'on. Hence if we have a number of octahedra, so placed that one plane shall contain a square section of each, and each edge of each such section coincide with one edge of each of the adjacent sections, an equal number of tetrahedra will fill up the vacant space above the plane, and therefore by using twice as many tetraliedi-a as octahedi-a, we fill up the space above and below. 6. Prove that the tangent plane at any point of the surface [axY + iJyyY + [czY = 2 {bcyz + cazx + ahxy)^ intersects the surface ayz + hzx + cxy — in two straight lines at right angles to one another. 1850.] GEOMETRY OF THREE DIMENSIONS. 21.3 The equation {axY + [hjY + {cz)'' = 2 [hcyz + cazx + ahxy) may be put into the fomi {ax)^ + [hyY- + [czY- = (1). Also ayz + hzx + cxy = may be wiitten « ^ c „ ,_. - + -4 - = 2 . x y z The equation to the tangent plane to (1) at any point [xyz] is Let [l^m^n^^ iO^h^^) ^® ^^^ direction-cosines of the lines m which (2) meets (3), then the condition of these being at right angles to one another, is IJ^ + m^m.^ + n^n^ = (4). Now where (2) meets (3), we have, writing x^y^z^ for xyz in (2), «(ir-(r-(ir-(-)'i(iy^(i)'a' a quadratic m -- whose roots are — ^ , — ^ . Hence ©'^.-(D'^'-e/^-" (^'- njWjj VC2; similarly -^-^ = f — ) : n^n^ \czj ' .-. IJ^ + ?n,?«, + »,?i, =c (aa-)* + (%)* + [czf = by (1). Hence the tangent plane at any point of (1) cuts (3) in two straight lines at right angles to one another. 7. A certain territory is bomided by two meridian circles, and by two parallels of latitude which differ in longitude and latitude respectively by one degree, and is known to lio within certain limits of latitude : find the probable supci*ficial area. 214 SOLUTIONS OF SENATE-HOUSE TROBLEMS. [1851. First, to find tlic ohaiiec that tlic centre of the territory which lies between known limits a, /9 of latitude lies in the zone between paraHels of latitude J and I + hi. area of zone breadth hi This chance = area of zone breadth (a— /3 27r?-cos/ rhl I 'Inrr cos^ rhl cosZ hi (/• the radius of the earth,) sin a — sin /3 ' Then the probable supei-ficies of the territory co%lhl j li sina — sin/S ' A being the area of the territory when its centre lies in the zone between the parallels I and I + hi; 1 W+30 = — - {sin (l + 30') - sin [l - 30')}, 180 27rr'''sin30' cos/. 180 Therefore the probable superficies, 27rr'sin30' /3 180(sina — sm/3) 7rr'sin30' cos' Ihl, 180(sina-sin/3) [a - y8 + 1 (sin2a - sin2^)}. 1851. 1. A line passing through a fixed point and having the sum of its inclinations to tAVO fixed lines through the same point constant, generates a cone of the second order. Any section perpendicular to either of the fixed lines has for a focus its intersection with the fixed line. 1851.] GEOMETRY OF THREE DIMENSIONS. 215 (a). Let OP be the moving line, OS^ OH the fixed lines. Take the lines bisecting the angle 80H and its interior angle respectively, as axes of x and y. Describe a spherical surface about 0, cutting OP, 08, OH in P, 8, H-, join SH, SP, HP hy arcs of great circles, then by the conditions of the problem SP + HP = constant, 2a suppose. Bisect ^7/ in X, join OX, let SX = HX = ^. Draw PM an arc of a great circle, perpendicular to SH, let XM — 0, PM= (f). Then, by Napier's rules cos SP = cos 8M.C0S MP, = C0S(/S+ d) COS(j>, COS HP = cos HM.cos MP, = cos(/S— 6) coacf). XT OT> TTT> o SP+HP SP-HP JNow cos oP + coaiiP = 2 cos - — cos ; .'. cosp cost/ cos(^ = cos a cos , . jrj, Qp o • SP'rHP . S P-HP and cos HP — cos bP = 2 sm sm ; ' a ' a M ' ' 8P-HP .'. smp sma cos 9 = snia sm , therefore adding squares ,, , fcoB^ 13 cos' sm'8mr^0\ , , cos'r 2Ax{z^-Ax'- By'-...) = (1), y,-y + 2By{z^-Ax'-Bf-...) = (2). Again, since the point (xyz) is always at the same distance fi'om the origin, we have X + y + z = rt , or x' + y'+ {Ax' + Bf+...Y = a' (3). The elimination of x, y between (1), (2), (3), would give the equation to the surface. But since the point [xyz] is always indefinitely near to the origin, .r, y, z arc always indefinitely 218 SOLUTIONS OF SENxVTE-HOUSE PROBLEMS. [1851. small, and \vc may neglect their powers higher thau the second. Hence our equations become .-r, — X + 2Az^x = 0, y,-y + 2i?2!,7/ = 0, Eliminating a?, y between these, we get < + y^ - .2 (2^2, -1)"^ ' i^lBz^-X)' ' the required equation to the surface. This surface is evidently of the fourth order, and symmetrical with respect to the planes of yz and zx. Its section, by any plane parallel to the plane of xy is an ellipse, which becomes a circle when the distance z^ of the cutting plane from that of xn = ^ . When z, = — r , the equation becomes x, = 0, "^ A + B ^ 2>1 ' ' shewing that the section is there a straight line parallel to the axis of y, and similarly when ^1 = ^5 the section is a straight line parallel to the axis of x. The points where these lines meet the axis of s, are the centres of curvature of the principal sections for —7 , -^ are the principal radii of curvature at the origin. When ^1 > ^ (supposing A > i5), the area of the section continually increases, as manifestly ought to be the case, since the normals altogether diverge after ^1 > ^ • 4. A plane is drawn through the axis of 3/, such that its trace upon the plane of zx touches the two circles in which the plane of zx meets the surface generated by the revolution romid the axis of z of the circle [x — aY + z^ = c^ (c < a) ; find the equation to the curve of intersection of the plane and surface, and from this equation trace the curve. 1851.] GEOMETRY OF THREE DIMENSIONS. 219 The equation to the phinc will bo z X (0- c [d'-c^f The equation to tlie surface, generated by the revolution round the axis of z of the given circle, is \{x' + yy--aY^z' = 6\ which rationalized becomes [x' + if + z'-^d'-6y = lc{'{x'^y'') (2). To obtain the curve of intersection of (1) and (2), we must turn the planes of ^^ and xy round the axis of _y till (1) coincides with the plane of xy^ and then put z = 0. This is effected by ^^'I'iting [a^_(^\i.j,-c.^ tor X. a ex + (a^ — c^)^ z . ^ — for z. a or since z is to be put = 0, ^^ — x for x. and — for z : this reduces the equation to (a;« + f + a' - 6y = 4 [{d' - &) x' + a'y }, or [x' + y' + d' - c'Y = -i{a'- c') {x' + /) + 4o\y'-', which may be reduced to x' + y' - a^ + c^ = ± 2cy, or x' + [y± c)' = d% shewing that the curve is composed of two circles, the radius of of each of which is a, and whose centres lie on the axis of y, on opposite sides of the origin, and at a distance from it = c. 5. Prove (one of) the two following properties : (1). If [A]^ {D) be two given spheres not intersecting each other, then every sphere which cuts [A] and (/>) in given angles will touch two fixed spheres. 220 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. (2). If (/I), [B] be two given spheres cutting one another, then every sphere which cuts {A) and {B) in given angles will cut orthogonally a fixed sphere. (I). Take the line joining the centres of the spheres for axis of .c, and its middle point for origin : let a, — a be the abscissae of these centres, r,, r^ the radii of [A) and [B] ; x, i/j z the coordinates of the centre of a sphere which cuts {A) and {B) in given angles a, /S ; r its radius. Then we must have [x — ay + / + z^ = r'^ + 1^ — 2rjr cos a, [x + (if + if + z' = r^l; + r' - 2?y cosyS. Let (&, 0, 0) be the coordinates of the centre of a sphere which this moveable sphere always touches. Adding and subtracting the above equations, we find ic' + a' + f + z' = \[r'^ + r/) + r' - v{i,\ cosa + ?'._, cosyS), and — Ixb = - - \rf — r^ — 2r [r^ cosa — r^ cos/3)], y^ - d' =^F - a\ adding these three equations, we have {x —ljf-\- f + ^^ = ?•' — r\ r^cos a + )\ cos /3 + - [ii\ cos a — r, cos /3) \ , + i{<^ + ^-.^ +^ (^-/-O} + ^"^ - «^ (!)• It is evident that the moveable sphere will touch the sphere whose centre is at a distance from the origin if the right hand member of equation (1) be a perfect square, or if jr^cosa + r^cosyS + - (r^cosa— r^cos^S) I = 2 |rj^+ r./+ - [rf— r./) \ a quadi*atic for the detennination of J, shewing that there are two spheres which the moveable sphere always touches. (2). Also it is evident that the moveable sphere will always cut orthogonally the sphere, the abscissa of whose centre is Z», if the right-hand member of equation (1) assume the form 1851.] GEOMETRY OF THREE DIMENSIONS. 221 This it will do if r^ cosa + r^ cosyS + - {i\ cosa — r^ cos/3) = ; h »• COS/8 - r, cosa or — = ~ ' ' a r^ cosyS — r^cosa ' which dctennines the centre of the sphere. ( 222 DIFFKIIENTIAT. EQUATIONS. 1848. cos 37 Assuming that sinx H is a particular integral of the equation S + (-l)^;« (■). find the complete integral of the equation S-(-l>=^" (^)- We see by substitution that not only cos a; y = smr» H , ^ X ' is a particular integral of equation (1), but also since ^ X Hence the complete solution of (2) is . / . cos£c\ -r, ( sin a? y =■ A\ sma; H — j + i? I cosa- where A and B are arbitrary constants. Now assume as the integral of equation (2), . / . cosa-N „ / sin.r\ y ~ A\ sma; ^ J + i> ( cosa; 1 , where A and B are now functions of x which have to be detennined. By the usual assumptions of the method of variable para- meters, we find dy . f sina; cosa?\ -r. / . cosa^ sina;\ -^ = A\ cosa; s— ] - B [ sma; H -^- , ax \ X X j \ X X J ^ , dA I . cosa;\ dB ( sina;\ ^ ,^. and T- sma; H + -^ cosa; = (3). ax \ X J ax \ X J ^ ' 1849.] DIFFERENTIAL EQUATIONS. 223 ., dA ( sin a? cos.rN dB ( . cosa; 8111,0? Also ^— cosa:: 5- ?- sin^c -\ ; ax \ X X j ax \ x x = ^' (4). From the equations (3) and (4), wc proceed to find -y- , -7- , dA {[ sinicV'' cos a; / siiiicN / . cosoj' -5— \ cosic ^ cosa? 4- sma? ^ ax y\ X j X \ X J \ x sin a; / . cosicXl - -w r"^ + — j[ • 2 f sina; = X cosa; V a; dA [^ 1 1\ „ / sina;\ or -7- 14—, 5 = a? cosa; , dx \ X' X I \ £C / ' dA „ / sina;\ , „ ,„, dB „ / . cosa!\ and irom (3) — =. — x \ sina; 4 j ; .'. A = a^'^'sinx — 3/a; sina%?a;, = a;^ sina? 4- 3a; cosa; — 3 sina? 4- C, and B = x^ cosa? — 3/a; cosa;^, = x^ cosa; — 3a; sina; — 3 cosa? 4- D ; therefore the complete integral of (2) is / . cosa;\ , „ . ^ „ . ^. .•. y = [ sma? 4- 1 (a? sma; 4- 3a; cosa? — 3 sina? 4- C j, / sina?\ ... „ , „ ^, 4- ( cosa? ] (a? cosa? — 3a? sma? — 3 cosa? 4- i/j, 2 ^J . cos.-z;\ ^ / sina;^ = a;^ 4- 6 [ sma? H \ -^ D{ cosa? j , C and D being arbitrary constants. 1849. A curve is defined by this property, that the radius of cur- vature at any point in a given multiple of the portion of the 224 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. nonnal intercepted between tlic point and tlie axis of abscissa? ; prove that the length of any portion of tlie curve may be ex- pressed in finite terais of the ordinates of its extremities. The lengths of the radius of curvature and normal are respectively dx) hence the dilFerential equation to the curve is ^"^'^LL = „Ji^r^V = ny \ 1 -I- d'^y "^ \ \dxj _ A dx"" 1 (Fx dy' 1 or y I = — . Lt — -t, B' .^-Ii (^^W — dy - ^^^ ' " dy ~\ [dyj ] dy ' and cot^ -7- = — : dy ny ' .". log cos^ = ^ogCy ; or cos = (GuY n and s = -^ C^y " + 0', C, C being arbitrary constants. 1850.J DIFFERENTIAL EQUATIONS. 225 lleucc the lengtli of any portion of tlie curve is known in terms of the ordinates of its extremities. 1850. 1. lff{x — a, y — h^ z — c) be homogeneous with respect to X — a^ y — h^ z — Cj then y (a; — a, y — b, z — c) =0 is the equa- tion of a cone whose vertex is (a, J, c) ; if the cone pass into a cylinder by «, J, c becoming infinite, shew algebraically that the limiting fonn of the above equation is (j) [nix + ny -^-pz -j- q^ mx -f ny +p'z + q) = 0. Let the axis of the cylinder, to which, as its limiting form, the cone tends as a, Z», c, are indefinitely increased, be parallel to the intersection of the planes mx + ny + pz = 0, m'x + n'y + p'z = 0. Then we have ma + nb + pc = a finite quantity, a suppose, m'n -{ 71 b + pic = a' Hence when «, Z>, c, become infinite we get, neglecting a, a! in comparison with a, b^ c, -.^^=^^ = -.^-^ 0). np — np pm — pm mil — mn Now since f is a homogeneous fimction, we have, if n be its degree, = 0, if iV if if if if ,„> Hence dividing each term of the left-hand member of (2) by the coiTcsponding member of (1), and observing that when a, 6, c, become infinite, the right-hand member will vanish after the division, i¥ - "» ^ + (i^»'' -2>"i) ^ + (»'i«' - »«'«) '£ = 0...(3) : Q 226 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1850. whence, by Lagrange's method, we get dx dy dz np — i^p pm — p'm mn — m'n ' whence mdx + ndy + J)*^^ — ^j m'dx + ndy + p'dz = ; .". mx + vy -^ pz -\- q =0, m'x 4 ny + p'z + 5'' = 0, q^ q being constants. Therefore the integral of (3) is (f> [mx + ny + p^ + q, m'x + n'y + p'z •+ q) = 0, the limiting fonn of f{x — a, y — h^ 2; — c) = 0, when a, />, 0, are indefinitely increased. 2. Prove the following formulae : W- - = '«(^ + ^') (1X5:7 + 93113X5 +•■•)• cosl(,>-2.-) «-.l ^ cs{{n-2r)b-s] ^ ^^^^^^ ^ ' sm(a— c>)sm(a— cj... sm(6— a) sm(o— c) . . . ' where s is the smn of the n quantities a, 5, c,... and r is any integer between 1 and n — \ inclusive. (1). We have Vl.3.5.7 "^9.11.13.15 "^*" = ^{(l?7"3:5) + (9j[5"rL13)+- = {(!-}) - 3 (^-i)} + {{h-h) - 3 (,^1-1^)1 +••• = 2(j-H^-U..-)-(l + ^-^-|+-) (!)• 1850.] DIFFERENTIAL EQUATIONS. 227 Again, generally 1 x' X* x'' 1 , n+x l+J + -5+T'--=2-x^'^[l- Let ic* = — 1, then x = cos|7r H — ^sin^Tr, 1 + ic _ 1 + cos^TT H — * sin^TT ' ' 1 — X 1 — COS^TT - — 4 sin^TT 2 cos'''|7r H — * 2 sin^TT cos^tt 2 sin^^TT * 2 sin^TT cos^tt , 1 COSiTT -\ * siniTT ■ = COtiTT . f J 1- Sin^TT * COS ^77 = — * COt^TT = £-*^"coti7r; ^^^ (r=^) = - H-^ + log COt^TT, and 2x 2* + - 4 2* Oj i 9* , /l+a;\ 2* 42* , , ; ^^Ir^j = 4 (- ^TT + log COt^TT) = ^i (log COt^TT + ^TT) - -i — ( log cot '^ - ^TT Therefore, equating real and imaginary parts, 1 - -l + l i 2,2* (^^^ ^^■^^'^ + i^), 1 3 ~ HA - ... = 2.2* (^"S COt^TT -i^); 1 -J- ' 1 1 57 + ... = TT 2^4' AH ( 1 +.. •)= TT TT 2 2.2* ' ^^ I1.3.5.7 TT 2 (2 + 2*) ' 7r = 96 (2 + 2*) f— L- + ^ +. ^ ^ Vl.3.5.7 9.11.13.15 228 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. (2). In general, Let y = wx, 0) being one of the imaginary cube roots of unity, then £"" = ^ + -]- "^TT "^■" Similarly, e'"'^ = l + ^ + ^V.., ^^=^+f+i4+- Now 1 + ft) + ft)^ = 0, But e'"" + £"'•'■ = e^-^^-** )^ + e(-5--*i )^ = £ - 2 cos -- 2 y ' 1.2.3 1.2..3.4.5.6 ^ V "r -r ; = ^('£" + 2£-^-'cOS^ This problem may also be solved by putting the series = w, we shall then get the differential equation -^ — t« = 0, the in- tegration of which, when the arbitrary constants are properly determined, will give the required value of u. (3). If r lie between 1 and r?, we may assume cos (?i + 1 — 2r) ;r _ A B sin(ic— a) sin(ic — 6)... sin (a? — a) sin(a; — J) "*' ^, J5, ... being quantities independent of iP, .•. cos(»i+l— 2r)a;=-<4sin(ir— J)sin(a;— c)...+jBsin(a7— a)sin(a;— c)... + . . . identically.* * We may justify the above assumption by expanding both sides of this equation in terms of sinx and cosx, and dividing by cos"''j; ; the left-hand side 1851.] DIFFERENTIAL EQUATlOiNS. 229 Putting X = rt, we get cos {n+ l — 2r) a = A ain {a — b) sin (a — c) . . ., ^ cos(n+ 1 — 2r) a , > 8in(« — oj sin(a — cj... t3- -1 1 7? cos(w+l-2r)6 , . bmiilarly B= • /, x • /z .-^ — (2 , •^ 8in(6-a) 8m(6 — cj... cos (« + 1 — 2r) a? cos (n + 1 — 2r) a sin(ic — a) 8in(cc — 5)... 8in(a — J) 8in(a — c)...sin(a; — a) cos (n + 1 — 2r) 5 sin [b — a) sin (& — c) . . .sin [x — 5) cos (n + 1 — 2r) a cos (n + 1 — 2r) & '* sin(a— &)8in(a— c)...sin(a-a;) 8in(5— a)sin(5— c)...sin(6— a;) + . ^°«(''+!--2'-)-^ ,0 (8). 8in (aj — «) s,in[x — 0) . . . In a similar manner it may be shewn that sin [n + 1 — 27-) a sin [n + l — 2r) ^ sin(a— &)sin(a— c)...sin(rt— x) sin(5— a) 8in(5— c)...sin(6— ic) , _ sin (m + 1 - 2r) a; _^ sin(a; — a) sin(a; — J)... (3) coss + (4) sins, where s = a + b +...+ x gives cos{(n + 1 - 2r) a- s} cos{(«+ 1 — 2r) & — s} _ sm [a — 6) sin (a — c) . . . sin {b — a) sin [b — c)... '" ^ ' the required result proved for the w + 1 quantities «, J,... a^. of the equation becomes /(tanx).(seca:)^''''", /being of n — 2r + 1 dimensions, or /(tanx) (1 + tan*^;)'''', which is therefore of n — 1 dimensions, and the equation becomes one of n — I dimensions in tanx ; it may therefore be identically satisfied by the n quantities A, B . . . . We here suppose » + 1 — 2r positive: if however 2r > n + 1, we may A\Tite for 2r, 2m + 2 — 2s, where 2a > 1 < n + 1 ; cos (n + l - 2r) x then becomes cos (n f 1 — 2s) x, in which n + 1 - 2i is always positive. 230 SOLUTIONS OF HENATE-IIUUSE PROBLEMS. [1851. 3. Given f[x) ^f{y) =f{x+y)[\ -f[x)f{y)], find the form oi f{;x). Since f[x) +f{y) =f{x + y) [1 -f{x)f{y)} (1), put X = y = 0^ then 2/(o)=/(0){i-7(om; therefore either /(O) = 0, or 1 -/(0)> = 2, giving /(0)=±(-l)i Taking this latter value, and putting y = in equation (1), f{x)±{-lY-=f{x){l + {-l)if{x)}, which gives f{x) = + (— 1)^ for all values of x ; therefore the given equation is satisfied hj f{x) = ± (— 1)*. Again, if /"(O) = 0, in equation (1) wi'ite y = — x, then f{x)+f{-x)=f{0){l-f{x)f{-x)], .■.f{-x)=-f{x). Differentiating (1) with respect to y, considering x constant, therefore, putting y = - x^ /(-^)=/(0){i+yW1-^l; or putting — x = z^ /(^) =/'(o) {i+TRi'l. Now/'(0) = some constant, C suppose, .-. ^^ = a{i+./>)l'^ Whence f{z) = tanCs, or f[x) = tanCic, which determines the foim of /(ic), C being an arbitrary constant. 1851.] DIFFERENTIAL EQUATIONS. 231 1851. 1. Let P (fig. 98) be any point in a curve and 8 a given point in a straight line A8\ draw SU perpendicular to SP^ and let the tangent at P meet SU^ 8 A respectively in U and T\ find the nature of the curve when 8U bears a constant ratio to 8T. Let the curve be referred to P as pole and 8A as prime radius; then 8U: 8T:: sm8TU: Bm8UT :: 8m{d +8PU): cos 8PU : : sin ^ + cos ^ tan 8PU : 1 ; .'. sin^ + cos^ tan 8PU= constant, e suppose. Now tan 8PU= r -=- , dr ' .'. rcosu -— = e — fund, ar ' cos^ de _ 1 e — m\d dr r ' and log(e — sin &) = log - , • a ^ or sma = e — . r Transforming this equation to rectangular coordinates, y = e (a;' + /)*-c, or eV + (e' - 1) / - 2ci/ - c' = ; shewing that the curve is an ellipse or hyperbola according as e is > or < 1. 2. The equation c — a cos^ cos0 — h ainO Hin sin cos 0) (^Z0 = 0, and (1) may be considered as the complete integral of this equation, or of acos^ sin^ — Jsin^ COS0 asin^ cos^ - Jcos^ sin^ •••\ n c being considered an arbitrary constant. Now (acos^sin^— J8in^cos0)'''+c''' = (acos^sin^— Jsin^cos^)^ + (acos^cos^ + Jcos^sin<^)'* by (1) = a* cos^ + b^ sm^(f>j .'. a cos<9 sin^ - J sin^ cos^ = [a^ cos''^^ + ¥ sm^ - c'"')* ; and similarly, a sin 6 cos — b cos ^ sin0 = (a''' cos'^^ + b^ sin^^ — c''')*, therefore equation (2) becomes [d^ cos^<^ + h^ sin"'^ - c^- [a sin ^ cos ^ — b cos^ sin)* ( 233 ) DEFINITE INTEGRALS. 1849. 1. Shew that ^ 7, ax = ir. +xy Putting X = tan'' 6, we get f ^l[££^ dx = 4. r sin'^^ log tan^t^^. Now generally 1 f[x) dx = I f{a — x) dx^ •) •'0 .-. [ sm'^ log tsinOdd = ( " cos'^ log cot^(/^ •^ •'0 = - 1 cos'^logtan^o?^. Adding these equal quantities and dividing by 2, r^log^ db = - 2 [ 'co82^ log tan^ J6>. Now, integrating by parts, - /co82^ log tSinOdO = - ^ sin 2^ log tan^ + 6 + G, .-. - I co82^ log tan^c?^ = i^r, r" a** logic , 2. Shew that I log .r log \^-^) ^-^ = 7r« (l«^g ^ - 1) • 234 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. Let I logic log f — 2 — j dx = w, dii f lofiTiC then -^ = 2a \ ., ^ — r, dx: da J^ x' + a' ' put X = a tan 0, then dti ^ = 2 T" log (a tan ^) ^7^ (1). Now generally, I /(a?) <;/.c =1 f{a—x) dx, ■Jo ■^ .♦. ^ = 2 riog(acot6')rZ^ (2). (1) + (2) gives, dividing by 2, J /•if a it da = 2 1 log a I •' = Trloga, .*. u = 7ra (loga — 1) -I C. And when a = 0, ?« = 0, .-. c = o, and u = ira (loga — 1), which was to be shewn. 3. Prove that jg (l + 2ecos04e) ^ ^ -^ ~ 1 - e" the upper or lower sign being taken according as e is less or greater than unity. We have r 2e + (1 + e") COS0 ^ _ sinO j (1 + 2e cos + e'f ~ 1 + 2ecos0+e' ' r 2e+ (1+e') COS0, ,^ , ^^ _ , ,. .. ,, . ^^ sS— 2N2 loge (1 + 2e cos^ + e') (?0 J^ (l+2e co80 + e')' *' ^ ' = , — 7i n ^z log 1 + 2e cos + c') + 2e ,- — j^ sr, dB l + 2ecos0-fe' ^^ ^ /, H-2ecos0 + ey 1850.] DEFINITE INTEGRALS. 235 sin 6/ r cos 9 = ... + -_ 1 £ 1 + 2e cosO + e^ j„ 1 + 2t; cost/ + e' I —-^ Ti 7. , between the limits, j„ 1 + 2e COS0 + e" ' W 7fl 1 + «' r" ^0 + e' + 2e COS0 I + e' r dB 2e Jo 1 + e' + 2e cos0 2e J0 And io 1 + e' + 2e cos^ T. tan~^ tan^TT = ^ if f < 1 1 - 1; e' - 1 e + therefore the definite integral becomes in the two cases I + e' TT IT 2e' , 2 + — :^ — t •, — TT = TT - — 73 5\ and — TT - 2e 1-e' 2e 2e(l-e") 2e(l-e') = + TT 1 - e 1850. 1. Shew that la ' / 3 I tan X log (tan a?) dx = Jtt'*', •' (;8). I -L— — ^ (7a: = Tratans or Trocots, and that sm2£ according as s is < or > ^tt. (a). We have, a being < 1, ,x n-\ (See Gregory's Examples^ p. 477.) 236 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. Diftercntiatiug with respect to a, / J o Z" , , TT COSaTT Logzaz = — 1 + z ° sm aTT Putting = tannic, the limits of x will be 0, ^tt, hence . r*"^ 2<.-i 1 /.. \ 7 tt'^ cos aTT 4 I tan X log (tana;) ax = ^-r, , -' whence, puttmg a = §, 27r TT cos 4 I tan^a; logftancc) £?a: = ^ = ~^ 5 sin- / tan X log (tana;) c?a; = ^ tt''*. (/8). Putting a; = a siu^, we have ^o {a'-x')i ^^ _ ^ p- cos'-'^ — sin^ -I—- — a; " -r-— sint/ sm2s sm2£ therefore wTiting — 6 for 6^ = a —————— rfcr. - ^— — + sin smzs Adding these equals and dividing by 2, J-. « . 8m2ej_w 1 . o^ — « ^ . ..„ sm sin2£ sm 2£ = _^ r /"i _ cos- 2s sin2£ i_i^ V (1- sin"''2£sin-'6') ' Tra cos'''2e /'^'^ sec'^^ r" sec J_w 1+cos'i « . „ I ^r 5^r— ^ f?^. 8in2e sin2e J i H-cos''^2£tan*^ f sec^ ^ Now j 1 ^ co8''2£ tan'' 6/ ^^ "" ^^^^^ tan"' (cos 2£ tan^) + C; /•'^ sec'-'^ •*• / ■r~, 27r~z — 27i "" = 7rscc2e; ./_j^l + cos'' 2s tan'^ ' 1850.] DEFINITE INTEGRALS. 237 r" {a'-x')i J I 1 co32e\ .*. / ^ ax = irax -r—- ; — — J _„ a \sm2E 8in2£/ sin2£ = Tra tan s. The above investigation holds if s < ^tt. If e > ^tt, let e = ^TT — e' (s' being < ^tt), then sin 2s = sin2e', and I ' dx = 7ra tans', sin 2 s' = ira cots ; .'. I 1 ' dx = TTci tans or ira cote, J-a « ^-— -re sin 2 s according as s < or > ^tt. 1851. 1. If 3/ be a function of a: defined by the equation a^" = {y — nx)"'^^ [y + nxf~^^ shew that f -^ = r -.-^ = -i^ log^^tJlf. }^y + l3x J^/3y + n'x n + ^ ^ a Since a'-"' = (^ - na;)"^'' (y + nx)"-^ ; therefore, taking the logarithmic differential, / r,\ dy — ndx , „> ^?/ + ndx ^ ' y — nx ' y -\- nx W[n + ^)[y^-nx) + [n - ^){y - vx-)] dy 2 2 2 y — « ic + n [[n — ^){y — nx) — (w + ^)[y + wa;)} <^] y — w'^oj dx _ dy ^ ' ' y + ^x /% + Ti'^aj ' •• J y + ^x~ J ^y + n'x 238 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. between proper limits of the variables. Now when a? = 0, y = a, r ^^ _ r dy " Ja y + ^x~ Ja ^y + n'x^ d{y + 71X) ^y + ri^x + w^ + n^x ' (w + /8)(^ + waj) ' y + ??a; the required result. w + yS log 2. Detennine the value of the definite integral " x'-'{\-xY-'dx [x + ayp ' f Let 1/ X + a 1 + a ' then when a; = 0, ^ = 0, and when a; = 1, ?/ = 1, X = ^ ; 1 + a - y^ . 1 _ ^ = OL+^Kizl) 1 + a - y ^ I + a X -\- a = a t) =^^'-yy' 1 + a- y^ n-x^ \x + . ii-a^r' ^ {i-yr\ ^ i + a-y " [x + ay ofi 1 + a Also 7 r- = ,, ^ , , {x-VaY (1+a)"' cfe _ d'y dy X y 1 + a — ?/ ' — 1 + ^ 7 1851]. DEFINITE INTEGRALS. 239 .-. ^7^^"^l^' da: = ,,/ , y'-' (1 -yf-'dy ; = 1 r(«lI(/3) a^(l + a)* r(a + /3) ' the required value.* * See Gregory's Examples, p. 471. ( 240 ) CALCULUS OF FINITE DIFFERENCES. 1851. 1. PT^ pt^ (fig. 90) are two tangents to a curve drawn at the exti'emities of any chord PS}) passing through the pole S^ and meeting a given line SA in J", f, respectively ; it is required to prove that the cui-ve in which the sum of the reciprocals of ST and St is constant has for its equation ^ = 1 + ecos^ +/(sin^)'^, where y (sin ^)'^ denotes any rational function of (sin^)'^. Let the angle SPT = 0, PST = 6, SP=r, then r sin {0 + (f)) ST^ sin, DA, (fig. 92), rigidly connected, form the sides of a quadi'i lateral figm'e, such that the angle ^ is a right angle, and the points B^ (7, i>, are equidistant from each other: when the whole is suspended at the angle vl, detennine the position of equilibrimn. Let ^, y be the coordinates of the centre of gravity of the system referred to AB^ AD as coordinate axes; and let AB = 2ff, AD = 2^», and the angle ABD = a ; .'. {2a + 2b + 4:{a' + h'y\ x = 2a.a+2(a''+Z»"04[2a+(a' + Z»')4[cos(120°-a)-cos(120°+a)}] = 2a" + 2 {d' + hy [2a + («'' 4 Irf- 3* sin a} = 2a' + 2(«'' + ^»''')*(2« + 3i.^'). Similarly, {2a + 2Z/ + 4 {pi' + W)"^] y = 2h' + 2 [d' + h'f {2b + 3*a). Hence, if be the inclination of AB to the vertical, ^l^ b^-\-{a'' + by{2h + SKa] ^ a'+ {a^ + hy{2a + SKb)' 3. A string of given length is attached to the extremities of the arms of a straight lever without weight, and passes round a small pulley which supports a weight : find the position of equilibrium in which the lever is inclined to the vertical, and prove that the equilibrium is unstable. The inclination of the lever to the horizon will be deter- mined in this case in the method to be shewn in the next problem but one, the point G being now the fulcinim, and P vertically below G instead of above it. To determine whether the equilibrium is stable or unstable, let the lever be turned through a small angle ; then the weight will assume the lowest position it can, and the normal at this point to the ellipse mentioned in the above problem will be vertical. 1848.] STATICS. 247 Hence it is evident that the vertical, through the weight in its displaced position, will intersect the lever on that side of the fulcrum which is lowered in the above arbitrary dis- placement : hence the system will tend further from its position of rest, and the equilibrium is unstable. 4. Two equal strings, of length ?, are attached to the fixed points Aj B^ and C, Z>, respectively, which, if joined, would form a horizontal rectangle ; a sphere, whose diameter equals AB^ is laid symmetrically upon the strings: find the position of equilibrium and the tension of either string, supposing l> AC+^irAB. Shew also how the problem is to be solved when this condition is not fulfilled. The centre of the sphere must lie in the vertical line through the point of intersection of the diagonals of the parallelogram ABCD^ and each string must lie wholly in the plane through its points of support and the centre of the sphere. Let AB=2a, AC=2b^ and the depth of the centre of the sphere = z ; or z = {i{l-TraY-b'}i (1). Let 2' equal the tension of either string ; .-. 4:T jj^ 7^ J = weight of the sphere (2) : from equations (1) and (2) 7" is known. If I = 2b + TTff, = and T = cc ^ the centre of the sphere being in the horizontal plane ABCI). If I <2b + ira^ we must suppose the part of the string not in contact with the sphere to become rigid, so as to support the sphere above the horizontal plane ABCD. Each string must still lie wholly in the plane through its points of support and the centre of the sphere. 248 SOLUTIONS OF SKNATE-IIOUSE PROBLEMS. [1849. 1849. 1. Two unequal weights, connected by a straight rod with- out weight, are suspended by a string fastened at the extremities of the rod, and passing over a fixed point : detemiine the po- sition of equilibrium. Let G (fig. 93) be the centre of gravity of W and W\ the two weights, P the pulley : PG must be vertical, and bisect the angle WP]V'. Let TFPTr= 2/, WW'=2a: then PG is the normal of the ellipse, which has WW for foci and 2l for axis-major ; hence WP : WG : : T^PT^ : WW, «^' ^^'^-wTW''^^ W and IV'P = jjj; :^, 21 ; W+ W ' hence the sides of the triangle WPW are known, and thence its angles : therefore Z WPG = I WPW is known, and WGP = TT — WPG — PWG is known, which is the inclination of TFTF' to the vertical. 2. A smooth body, in the form of a sphere, is divided Into hemispheres, and placed with the plane of division vertical upon a smooth horizontal plane : a string, loaded at its extremities with two equal weights, hangs upon the sphere, passing over its highest point, and cutting the plane of division at right angles : find the least weight which will preserve the equilibrium. Determine whether the equilibrium is stable or unstable. Let a = radius of the sphere ; X = distance of the centre of gravity of the hemisphere from the plane of division ; W = weight of the sphere ; to = weight required. We may consider the string to become rigidly attached to the sphere without disturbing the equilibrium : we then have 1849.] STATICS. 249 each system of a hemisphere and weight attached prevented from turning about the line of intersection of the horizontal plane and plane of division, by the tension iv, at the highest point of the hemisphere. Hence, taking moments about this line, w.2a = Wx -f ioa\ .-. w = -W=lW (1). To consider whether the equilibrium is stable or unstable. If we give either hemisphere a small angular displace- ment {B) about the above line, the weight to rises through a space a^, and the centre of gravity of W falls through a space x.d. Hence the common centre of gravity of the hemisphere and weight rises through a space toae - WxO = 0^ by (1), and the equilibrium is therefore neuter. 3. A slightly elastic string, attached to two points in the circumference of the base of a right cone, at opposite extremities of a diameter, is just long enough to reach over the vertex without stretching. The cone is suspended by it from its middle point: find approximately the increase of its length. Let 2? = mistretched length of the string ; h = height of the cone ; a = radius of its base ; z = the depth through which the cone falls ; 2 (? + \) = the stretched length of the string. Then, by the principle that " tension varies as extension", if T be the tension of the string, T = E J , E & constant weight ; 250 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. h -\- z and 2 T j — — = W the weight of the cone. t -J- A, Also, (h + zY = {l + \y + a'; I I + X or, omitting V and the higher powers of \, W f and A, = 2E {r-dy 4. An equilateral triangle, without weight, has three miequal particles placed at its angular points; the system is suspended from a fixed point by three equal strings at right angles to each other, and fastened to the comers of the triangle : find the inclmation of the plane of the triangle to the horizon. Let J, y, 2, be the coordinates of the centre of gravity of the three weights refen-ed to the strings as axes : ^, y, 2, will be subject to the condition X -]ry -\- z = 1^ if I be the length of the strings. Let 6 be the angle between the nonnal to the plane of the triangle and the line joining the centre of gravity with the origin, which is vertical ; this angle will be the required in- clination of the plane to the horizon. The direction-cosines of these lines are -,. —,. —,. and j=-^ — z=r. — =:^vi , T=i — 4 — ^svi ? f^qr^T^^' respectively; l_ ^ + y + 2 •• ''''^''- ^i' [x' + f + zy - 1 / ~ '6^' {x'+f+zy 1849.] STATICS. 251 5. A piece of string is fastened at its extremities to two fixed points : detennine from mechanical considerations the form which must be assumed by the string in order that the surface generated by its revokition about the Ihie joining the fixed points may be the greatest possible. By Guldinus' property of the centre of gravity, that curve will by its revokition generate the greatest surface whose centre of gravity is furthest from the axis, i.e. is lowest, when the axis is made horizontal and the plane of the curve vertical. Now we know the centre of gravity will assmue the lowest possible position when the string is in equilibrium under the action of gravity : hence the curve required is the common catenary. 6. It is required to support a smooth heavy body, in the form of an ellipsoid, in such a manner, that a given radius in the body shall be vertical, by means of supports at three points : shew that if /, ?«, «, be the direction-cosines of the radius, and the equation of the ellipsoid 2 'i V, X y z {• - — I — = 1 then the three points in question must be on the curve of intersection of the ellipsoid with the cone ll/^ (^. - ^) + ''^-^ (^ - ^.) + ^nxif ( i - 1) = 0. We will assmne the normals at the three points to meet hi some point of the vertical radius.* The equation to the normal at x\, ?/,, 2,, is ^ yjL ?i ' d' h' c' * This is an ass'omption : for the ellipsoid will be supported if two of the normals meet in a point not in the vertical radius, provided the resultant of the corresponding reactions meet the vertical radius in the same pouit as the third normal docs. 252 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. and this line passes through the points Ir^ mr^ nr ; //• — .r, mr — y^ nr — z^ ^ ^7 ii ^ suppose . a^ h' e witli shnilar equations for the coordinates of the other points of support. These equations may be written />• - ^ . = X,, mr -fs = y„ nr 7, s = z'. whence, eliminating r and s by cross-multiplication, fmz^ ny.\ (nx, Iz\ /7y, mx\ or, di'opping the suffix, we have as the equation to the cone on which the three points of support must lie. 1850. 1. A right cone is cut obliquely, and then placed with its section on a horizontal plane : prove that, when the angle of the cone is less than sin"^^, there will be two sections for which the equilibrium is neutral, and for Intennediate sections the cone will fall over. Let ABC (fig. 94) be the section of the cone through its axis, by the plane of the paper, to which the cutting plane is supposed perpendicular. Let the trace BP of the cutting plane make an angle 6 with BC: draw PD perpendicular to BP^ and draw AEF through F, the bisection of BP. Let 2a be the angle of the cone; then /.ABP=7r — a — 0, BDP =e + OL, and APD^^O-a. 1850.] STATICS. 253 Also, let AE = n.EF^ then _AjE _ APs'mAPE _ 2APsm{e-a) " " EF ~ FF smBFB ~ BP _ 2 cos(^ + a) slnf^-a) ~ sm2a ' or n sin 2a = sin 2^ — sin 2a; .-. sin 2^ = («+l) sin 2a. If ?i = 3, E will be the centre of gravity of the part cut off, which will therefore stand on its base in neutral equilibrium, and sin2^ = 4 sin 2a. Hence, if sin 2a < j^, there will be two values of 6, each acute, such that the corresponding cutting planes shall give neutral equilibriiun. For intermediate sections, sin 2^ > 4 sin 2a, and therefore ?i > 4 ; hence the centre of gravity will lie outside the vertical line PD, and the section will fall over. 2. The three corners of a triangle are kept on a circle by three lings capable of sliding along the circle, and the circle is inclined to the horizon at a given angle : find the positions of equilibiium. It is evident that, as the triangle is moved about, its centre of gravity describes a circle about the centre of the circle, the positions of equilibrium are those in which the centre of gravity is at the lowest and highest points respectively of this circle. The corresponding positions of the triangle are easily found. 3. A smooth cylinder is supported in a position of equi- librium by a string which is wound m times round it, and then has its extremities attached to two points A and B in the same horizontal line. The position of equilibrium being that in which the coils are separate, shew how it is determined. 254 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. and how to find the length of the string in contact with the cylinder. Produce the straight lines of the string to meet, as they must do, in a point : project these produced parts and the curved part of the string upon the axis of the cylinder; these projections must be equal ; but the inclination to the axis of the straight and curved parts of the string is the same ; hence the produced parts of the string must be equal in length to the part in contact with the cylinder. Hence the straight parts of the string occupy the same position as they would do if the string, instead of supporting the cylinder, ran under an indefinitely small pulley supporting a weight. This con- sideration determines the inclination of the straight part of the strmg to the vertical. Let 6 = the inclination of the axis of the cylinder to AB', (fi — the inclination of the string to the axis ; Q) = the angular distance from the lowest generating line of the cylinder of the points where the string leaves the cylinder ; 2a = the distance AB- 2l = the length of the string. Then, if we project the line AB and the string upon the axis of the cylinder, we have a cos = 1 cos cji ( 1) . Again, if we project AB and the straight parts of the string produced to meet as above on the plane of either extremity of the cylinder, the line AB would be projected into a line of length 2a sin^, and the string into two lines, each touching the circular end of the cylinder, and of length /sini/r: and these lines touching the circular end of the cylinder, they make with each other the angle tt — 2ft). Hence a sin 9 , . coseo = ^— ^ — : (2). / sni9 ^ ' 185L] STATICS. 255 Also the produced parts of the string each equals half the part m contact with the cylinder = {7mrr + &)?■) cosec<^. Hence, from the ahove projected triangle, (mTrr + cor) cosec d> tan&) = ^^ r = (mTT+ct)) cosec ^ (3). From (1), (2), and (3), we may determine 6 and 0; or the position of the axis of the cylinder and a : whence the lengtli of the part of the string in contact is knoAyn. Another condition is, that the centre of the cylinder must be symmetrically situated with respect to A and B. 1851. 1. A right cylinder upon an elliptic base (the semiaxes of which are a and h) rests with its axis horizontal between two smooth planes inclined at right angles to each other : de- termine the position of equilibrium, (1) when the inclination of one of the planes is greater than tan"* -r , (2) when the inclmation of both planes is less than tan * b a Since the locus of intersection of tangents to an ellipse at right angles to each other is a circle, the locus of the centre of gravity of the cylinder, as the cylinder is turned about in a vertical plane, is a circular arc ; and the centre of gravity is at the extremities of this arc when the axes of the cylinder are parallel to the planes. Also these extremities are the lowest points of the arc when the inclination of both the planes is less than tan~* j- ; but if one of them be greater than tan"* j , one extremity is the highest point of the arc and the other the lowest: hence, in this case, the position of equilibrium is that in which the major axis is parallel to the plane whose inclination is least ; and in the former case there are two positions of equi- librium, viz. when each axis of the cylinder is parallel to either plane. 256 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. 2. A^ Bj Cj are three rough points iii a vertical plane ; P, Q, Bj are the greatest weights which can be severally sup- ported by a weight PF, when connected with it by strings passing over A, B^ C, over A^ B^ and over B^ C, respectively: shew that the coefficient of friction at B = - loe;, -^^r? • We may consider each of the rough points A^ B^ C, as cy- linders of indefinitely small radius : hence, by a known theorem relating to strings passing over rough smfaces, if 6 be the angle through which the string is bent at any of the points whose coefficient of friction is ^, and T^^ T^ be the tensions of the strings on the two sides of the point, if all possible friction is being exerted, we have Let fjbj,^ fjb^, ficj be the friction at A^ B^ and C; a, 7, the inclinations to the horizon of BC and AB respectively: then, by the question, p^ £^^(4-7,^gM^(7-;_3Mcli''+').T.]7 (1)^ Q = e^^Afr-^'.sMBfi^+^J.Tr (2), i? = £'-,(*'-«). £"c'^'+'MF (3); .-. (2) X (3) - (1) gives 1 ,_ QB TT /^B=-lOg3p^^. ( ^57 ) DYNAMICS OF A PARTICLE. 1848. 1. If a and na be the respective distances of a satellite and of the Siin from a planet, ^ and mp the periodic times of the satellite and planet, which are supposed to describe circles round the planet and Sim respectively : shew that the orbit of the satellite will always be concave towards the Sun, pro- vided n be greater than m^. Let the angular velocities of the satellite and planet respec- tively in their orbits be called eo and mw; then it is plain that the rectangular coordinates of the satellite referred to the Sun as origin and axes rightly chosen, are X — na coswi + a coswwi, y = na sinw^ + a sinwio)^. Now, if the path of the satellite pass at any time t from being concave to convex towards the Sun, we have at that time M~ ' dt de dt df ~ ' .'. (n smoit -'r m mima>t) (n sinw^-f ni^ sinmcot) + {n cosQ>t + m coBmcot) [n coBwt + nf cosnicot) = ; .•. n^ + m^ + mn[m + 1) cos(7n —\)(i>t = 0. In order that this equation may not give a possible value of f, we must have n^ + m^ > mn (m + 1) ; .*. T^ — m[m-\-\) n > — ??«"', or or w > m 258 SOLUTIONS or BENATE-HOUSE PT^OBLEMS. [1848. which 19 tlicrcfore tho condition to be fulfilled, in order that tlie path of the satellite may be always concave towards the Sun.* 2. A body of given elasticity is projected with a given velocity, and rebounds n times at a horizontal plane passing througli the point of projection: detcnninc the direction of projection, so that the angle between the direction of projection and the direction of the ball inunediately after the last impact may be the greatest possible. Let a, Kj, a^ ••• ^ni ^^ ^^^^ angles of the first projection, and after the successive impacts; .'. tana„ = e tana,,_j = e^ tana,,,^ = ... = e"tana, if e is the modulus of elasticity ; , . (1 — e") tana .-. tan (a - aj = ; , J. , : " 1 + e tan a we have to determine a, so that this shall be a maximum. Taking the logarithmic diiFerential of this expression with respect to tana, we have 1 2e" tana _ tana 1 + e"tan''a ~ ' .*. 1 — e tan' a = 0, and tana = -r . 3. If a body be projected with a given velocity about a centre of force which cc . ,. ^ ,., , shew that the axis-minor of (dist.)^ ' * Since the above condition assigns an inferior limit to the value of m (n remaining constant), it manifestly precludes the possibility of a motion of the satellite about the Sun in a direction opposite to that of the planet i.e. a retrograde motion as seen from the Sun, -which would clearly require m to be greater than when its path is merely alternately concave and convex and not looped. 1848.] DYNAMICS OF A PARTICLt:. 259 the orbit described will vary as the perpendicular from the centre of force upon the direction of projection; and detennine the locus of the centre of the orbit described. Let r be the distance, a the angle of projection : then h^ = SY.IIZ^ the product of the perpendiculars from the foci on the direction of projection, = SP sma. HP sma, = r (2a — r) sin'^a. And a is constant since the velocity of projection is so ; .'. b cc sina, X r sin a, GC the perpendicular from S upon the direction of pro- jection. Also, if p, = angular distance of the apse, c = distance of projection. 1 = 1 a 1 - e 1 COS(f> c 1 — -7-" = 1 a 1 - P a C0S + >■ = 2.. {^-^)-\* * For this solution we arc indebted to Mr. Gaskin. 262 SOLUTIONS OF SENATE-HOUSE PllOBLEMS. [1848. 6. Force varvinsr as y^. — -., . shew that, when the hatus- • ° (dist.) ' ' rectum is given, an angle = 2 tan"' 5^ , measured from the nearer apse, will be described very nearly In the same time, whether the body moves ui an elliptic or an hyperbolic orbit, whose eccentricities are 1 — a and 1 + a respectively, a being small. We have dt ~7' and r = Z (1 + (1 + a) cos^}"\ in the ellipse and hyperbola respectively, where I is the common latus-rectum : also h^ = fxl is the same in both cases ; hence § = X 'a + (!+«) cos^r = ^-(2cos'i6' + aeos^)-'' ~ h 4 V cos'^i^j r 4 1 /, /, cos^ \ , = -y sec ho [1 + 2a — rr-p^ very nearly ; 4 A ^ \ ~ cos' ^6 J -^ -^ ' .-. t = ^ Jisec^e) {1 ± 2a (sec'''i6' - 2)} d tan^^ = ^T /{I + tan'-'i^ ± 2a (tari*|6' - 1)} d tan 1(9. Hence, if T be the time of describing an angle /3 from the nearer apse, T=^^ [tan 1/3 + 1 tan^iyS ± 2a (^ tan^i/3 - tanlyS)}. Hence the difference of times of describing this arc in the two cases 2«7^ = f|l (itan^'iyS-tani^), which vanishes if /3 = 2 tan~'5% and the proposition is true. DYNAMICS OF A PARTICLE. 263 1849. 1. If the equation for detennining the apsidal distances in a central orbit contain the factor {u — a)'\ shew that a will be a root of the equation (}>{u) - AV = 0, where ^[u] is tlio central force. The differential equation of the orbit will be nnrl dd (1). Multiply by 2 -^ , and integrate ; (duS" _ C [u).du , The general condition for an apse is, that -7^ = ; and there- fore the equation for detennining the apsidal distances is If this equation contain the factor [u — ay, let us suppose that ^/^W^-""=/Wl'.(«-a)'; tlien -^={u-a) .f{u), d'^u du d (du ^^^ de'~ dd'du'Kde = f{n)[f{u)^f'{u).{a-a)]{u-a)', hence m = a is a root of the equation dd' ' that is, a root of the equation (^[u) - //V = 0. 264 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. 2. A body moves from rest at a distance a towards a centre of force, the force varj^ing inversely as the distance : shew that the time of describing the space between ySa and ^a will be a maximum if yS = — j— . We have here d'^x _ fx, ~de~~x' {^\ =2/* log-, \dtj ° 03 smce a? = a, when -r- = : ' dt ' _ 1 dx J {F{x) + C] suppose. Now, let T be the time of describing the space between /8a and /S''af then In order that this may be a maximum, we must have d^ ^' .-. n^"-'F'{/3"a)-F'{^a) =0. 1 But F'{x) = therefore the above condition becomes ^^Z i_-o log^.) or H*/3"-^- 1 = 0; 1849.] DYNAMICS OF A PARTICLE. 265 .-.-3 = 4:. the required expression. 3. A particle is attached to the extremity of a fine string, which is partially wound round a cylinder of diameter c; if the unwound portion of the string be kept stretched, and the particle be projected perpendicularly to its length with a ve- locity F, prove that the string will be wound up after the lapse P of the time -^ , where I is the length of string unwound at the time of projection. Let r be the length of the string unwound at the time t after projection, - the arc which has become covered with string in that time: then, since the only force on the particle, viz. the tension of the string, is always pei'pendicular to its instantaneous direction of motion, the velocity of the particle is uniform: da ^ ^ .'. r -rr = a, constant : at and at the time of projection r-= V' dt ' dd j^ dt f> Also, r = I — -6; c ^\d0 ^i^)i=^ andi(^-l^) =C-lVt: and when < = 0, ^ = 0; .-. 6' = ^/'; J -'-6] =P-cVf.. 206 SOLUTIONS OK SENATE-HOUSE PROBLEMS. [1850. Let T bo the time when the string is all wound up ; .-. t = T, when 1 -^d = 0; r , rp •• Vc' 4. A particle describes an ellipse about a centre of force in the focus 8 (tig. 96) ; about S as centre a circle is described, which 'is cut by the radius vector SP in the point Q ; from Q a line is drawn perpendicular to the direction of the particle's motion, which meets the major-axis in R: prove that R is constant in position, and that QR is proportional to the particle's velocity throughout the motion. From P draw the normal PG^ and from S the perpendicular ^91^ upon the tangent; also draw GL perpendicular to /S'P, PL is half the latus-rectum. Now QR is parallel to PG ; .-. SR = ^.SQ ^e.SQ, by the property of the ellipse ; therefore 8R is constant. Agam, QR = — . 8Q = ^^ . bQ PL.8Q PL.8Q 1 ~ 8PcosP8Y 8Y 8Y' and velocity cc -qy-] .'. QR Qc velocity. Q. E. D. 1850. 1. A heavy particle is fastened by two equal strings of given length to two points in a horizontal line, and then whirled round in a vertical plane ; the velocity is such that, if one of the strings break when the particle is either at its lowest point or half-way between its highest and lowest points, the particle 1850.] DYNAMICS OF A PAUTICLE. 267 will still continue to describe a circle: find the least distance between the point to which the strings are fastened that this may be possible. Let I be the length of either string, 6 be the inclination of the strings to the horizon when the distance between the points is the least possible. Let V be the corresponding ve- locity of the particle at its lowest point before the string breaks ; then [V^ — 2gl &\\\d)^ will be its velocity when the strings are horizontal. Now, if one of the strings break when the particle is at its lowest point, it will proceed in a horizontal circle about the vertical line thi'ough the point of support of the unbroken string, if the velocity be such as to produce a centrifugal force just sufficient to keep the string at the same inclination to the horizon, or, resolving the forces perpendicular to the length of the string, if ^ ^ .8in0 = q cos ^, Z cos ^ ^ ' ,cos'^ . . or l^ = gl -T—^ (1). ^ sm^ ^ ' If one of the strings break when the particle is half-way between its lowest and highest points, it will proceed to de- scribe a vertical circle about the point of support of the other string, provided the velocity (F" — 2^'^ sin^)* be great enough to carry the particle over the highest point of the circle, i.e. if V""- 2glsm6 > ^gl. Now, 6 is supposed to have received its greatest possible value, and therefore, from (1), F its least possible value ; hence V -2gl%n\d = Sgl, or al ^^ — 2ql sin 6 = 3gl: •^ smU .-. 1 - sin'^ - 2 sin'^^ = 3 sin^, or sin* ^ + sin ^ = ^, and 8in^ = - i + (i + ^)* 2 268 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. And the least distance between the points of support = 2?cos^ = (§)Ml + (21)*}W. 2. If P be the perimeter of a closed curve described about a centre of force, t the time of a revolution, h twice the area described in a unit of time, and p the radius of curvature We have at the time <, prove that P = h I — . . -'or p= rvdt J -/ -/i dt ' t'de V dO , r -7- .rar dr P rdr between proper limits. [r'-fY Now let (/-/)i=/(^) •'• / / t!_ 2u between the above limits = f{0)d0 •' = /(27r)-/(0) = 0; • P- [ P^P -I: ■PJ jyr dr dr dp 1850.J DYNAMICS OF A PARTICLE. 269 r —r dr clr P r'de _ rwdd ~ Jo P •' (1 p ''dt 3. If any number of bodies be projected from a given point with the same velocity in one plane, and describe ellipses round a central force which varies inversely as the square of the dis- tance; find the law of force tending to the same centre, under the action of which a body will describe the curve which is the locus of the centres of the different ellipses. Let /Lt be the absolute force, V the velocity, and c the dis- tance of projection. Then, if a be the axis-major, i = ?-^' is). a c fj. Also the equation to the orbit is 1 _ 1 l-ecos(^-a) ^ r~ a 1-e' *' and om' object is to find the relation between e and a ; for if Pj (f> be the polar coordinates of the centre of the ellipse, p = ae, , 9„ be the angles C'P, CJR^ CQ respectively make \A\\\ the apsidal distance, then the time in which the particle will reach P from the apse = ltan-(|tan^,), where a and 5 are now the semiaxes. Let (/<,, /t,), (^.^, l\) be the coordinates of the points P, Q re- ferred to the axes of the figure, the equations to PZ7, Q U will be K K and -^,x + -^,y=\, and the equation to CZ7 through their point of intersection and this line passes through the origin ; therefore \ = - 1, and the equation to C?7 becomes a"- therefore time from PXo R = -i |ta»"' (j tan<^) -tan-^ (f *^^^i) _ 1 a tan^ — tan^, 1 + r? tant^, tan© 6 ' 1851.] DYNAMICS OF A TAKTICLE. 277 DYNAMICS OF A rAKTICLE, K K - K a^ «^ " h - K ah h' ?/ /** /^(/^-/^J 1" + , a rt* ' /^.(^\-^J 6* ^1^2 - ^l^a ah av;' ^4 ">0 1 ^A \K ^4., ah' » 1 d' h' an expression which only changes sign when the suffixes 1 and 2 are interchanged ; it therefore = -J jtan-^ [^ tan 6',^ - tan"^ [^ tan<^^ = time from B to Q. Q. E. D. 6. Two bodies A and 5 revolve in the same conic section roimd the same centre of force in the focus ; shew that if A and B be at opposite extremities of any focal chord, B will appear (to a spectator on A) to move with a constant velocity pei'pen- dicular to SP^ or with a constant velocity perpendicular to the transverse axis, according as A and B describe the conic section in the same or opposite directions. Let PT, QT (fig. 98) be the tangents at the points P and Q at the extremity of the focal chord BSQ'j draw Sl\ SY' per- pendiculars from S upon those tangents. First J suppose the bodies A and B to be moving in the same direction about S^ then the relative velocity of A and B per- pendicular to FSQ = l\smSPV+ V^^mSQY', 278 SOLUTIONS OF SENATE-HOUSE FHUBLEMS. [1851. it' \\ aiul r, be the velocities of Ji and B _J_SY h SY' ~ SY' SP'^ SY' SF ~ \SP^ SF h = , ,7 : IS constant. ^ lat. rectum Secondly^ suppose A and B to be moving in opposite direc- tions about S^ then their relative velocity pei'pendicular to the transverse axis = \\co^PTS- V^cosQT'S - cosSPY- oD> -^^ on^^' ' - cosSQY' SP sin SPY ' e SF sin SQ, Y' ' e = - I — cotSFY--^ cotSQY''] ~e [sF = - {u cot<^ — u' cot0') suppose. Let ASP =6, then 2 7 2 If = - (1 +e cos^) [I the latus-rectmu), A ^A. \ du and. cot m = ^^ : ^ u do' du 2e . .•. i( cot 9 = — Jn— T sine', and writing tt + ^ for ^, and neglecting the change of sign, since cotd)' = — ^77 and not , -j^ as above, we find ^ u do u do ' ?i cot 9 = -y sma, whence we see that the above expression gives the relative velocity perpendicular to the transverse axis equal to zero. 1851.] DYNAMICS OF A PARTICLE. 279 7. A body m moves with a uniform velocity v = — — /i* in a circular tube whose radius is a, and attracts a body m' within the same tube with a force = , ,. ,„ ; shew that if m and m be originally situated in the opposite extremities of a diameter and m at rest, the two bodies will meet one another 2aa at the end of the time b sin a * Let P, Q (fig. 99) be the positions of m and m at the time t after the beginning of motion, and let PCA, QCB be their angular distances from their original positions ^, J5 at the extremities of the diameter AGB\ let PGA = 6^ PCQ = (f)y V also QCB = - t: join PQ ; then, for the motion of m', a fl cos|0 (2a)' mi^^' V Also, d = TT — (b t ; a •*• de " df ' and 2 '-^ ^ = _ iL .^^^ ^ . dt df 4«* 3in'^<^ dt ' dt J 4a* Vsin*^)^ V* cos'^'a sin''^<^ + C Now, when < = 0, -^ = , d) = 7r: ' dt a v^ v" cos'^a ,, ^, ••• a' — ^('+^); .-. C = SGC'OL- 1, 280 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. (It a b1ii^(/) d^ — r« = ; vertical motion of the same, or ao)^ sin 6 — aco^ cos ^ = ; and horizontal motion of the other point, or aco^ cos 6 + aa>^ sin — rco = 0. Hence, aco^ = rco cos^, ao)^ = rco sin^; .'. a -7- = r cosd.f— r sin^.w", dco (where /=-£), 1848.] RIGID DYNAMICS. 289 a —jJi = r {iiT\u.f+ r cosc/.eu , dt It dto Also, X = r cos^, y = r sin^; d^'x .". -j^ = — r sin O.f— r cos6.q)'\ — ^ = r cos 0.f — r sill 6.q)\ Hence the equations of motion become - Mr sm0.f- Mr cos^.w' = R, + R\ (1), Mr cose.f- Mr sin^.o)' = E + R\ (2), = i? + ir, - Mg (3), MFr cos e.f- Mh\ sin O.ui' = - Ra,' - R! p,' sin ^ (4), MWr sin B.f + MU'r cos 6. ai' = R/i' + R'ci' cos 6 (5) , Mk'rf = R'/i' sin - R\d' cos ^ ... (6), (4) cos^ + (5) sln^ + (6), 2Mk\f=: [R^ + R',) d' sin 6* - [R^ + JS;) a^ cos (9 = -Ma'rf, by (1) and (2); .••/=o, and CD is constant. Hence (4) cos^ + (5) sin^, = i?^ s\n0 - R^ cos^, and (1) sin^ — (2) cos^, = R\ sin^ - R\ COS0. Hence there is no action perpendicular to the plane tln-oiigh the radius vector and the axis of the cylinder. Let R and R' be the horizontal pressures in that plane on the base and at U 290 SOLUTIONS OF SBNATE-iroUSE PROIU.EMS. [1848. the other point of contact. Then (1) cos^ + (2) sin^ and (5) co8^ — (4) sin^ give 11 + R = - Mrc^' (7), and R + E = mKvcS' (8): a' (3), (7), and (8) are the only equations for determining i?, R\B^, and B\. Tlie indetcmiinatcness of the problem arises from the cir- cmustancc, that there are more pressures on the sphere than are necessary to produce the motion required. Thus, there are the two vertical forces B^ and B'^ to support the weight, the two radial forces B and B' to curve the path of the centre of the sphere, and the two, B and B\^ to oppose the tendency of the sphere to rotate about a horizontal axis perpendicular to the radius-vector. Hence the above equations contain the sums of couples of these quantities. Considerations of elasticity, which prevents all such ambiguities in nature, would remove them from the solution of the problem. G. A uniform bent lever, whose arms are at right angles to each other, is capable of being enclosed in the interior of a smooth spherical surface ; determine the position of equi- librium. Find also the time of a small oscillation when the position of equilibrium is slightly disturbed. Since the reactions of the sphere all pass through the centre, it is plain that the resultant force of gravity upon the lever must also pass through the centre of the sphere ; hence, its centre of gravity must lie vertically under the centre of the sphere. Let C (fig. 101) be the angle of the lever ACB, join AB-, bisect AB, AC, BC, in 0, D, and E, and in ED take the point a, such that EG : ED :: AC : AC + BC: join OE, OG, OD: G will be the centre of gravity of the lever, and OD, OE, will be pei-pendicular to AC, BC: also will be 1848.] RTfJID DYNAMICS. 291 the centre of the sphere. Hence we must have 00 vertical. But EG: GD'.:CD'.EC::OE:OD; therefore OG bisects the right angle 0Z>, and^C, 5C, are equally inclined to the horizon. ^Vhen the lever is slightly disturbed in its own (vertical) plane from its position of equilibrium, it will manifestly oscillate as if it were attached to an axis through 0, and the sphere removed. Hence we have to find the radius (/>■) of gyration of the straight lines AC^ BC, about an axis through 0, per- pendicular to the plane of BC^ CA. Let A^j, k^, be the radii for AC^ BCj respectively : then k;' = 0D' + IAD"" = EC + ^AB' = ¥ + !«•■', if ^ C = 2a, BC = 2h. Similarly, K = a' + W, and l>^-J<±JK^ll±^^^±4^±Vl^l(a + hf. Again, to find 0G{= Z), the distance of the centre of gravity of the lever from 0. We have ._ j^^, &'mOEG _ a j-,^sinOED smEOG ^ TTTh smEOG 2-ah ~ a + V therefore the time of a small oscillation _ 2^' {a + hf 7. A section of the surtace of a circular right cone (whose axis is horizontal and vertical angle 60°) is formed by a plane pei'pendicular to the slant side, so as to contain the vertex ; U2 292 SOLUTIONS OF si:nati:- HOUSE i'kouuems. [1848. shew tliat when the surface so cut off makes small oscillations 21a about the axis, the length (jf the isochronous pendulum = -— (whether the elliptic base be included in the surface or not), a being the length of the pei-pcndicular drawn from the vertex upon the elliptic base. Let k be the radius of gj^ration of the section about the axis ; r being the distance of the element hS of the surface from the axis. Let hS be projected upon a plane pei-pendicular to the axis, and hS' be the corresponding elementary sm-face ; .-. hS' = S>Scos30°; , _ sa^v = the square of the radius of gyration of the elliptic projection on a plane perpendicular to the axis. Similarly, the radius of gyration of the elliptic base equals the radius of gyration of its projection on the same plane, which is the same as the projection of the whole section. To find this radius we must first find the axes of the elliptic base BC (fig. 102). Since the vertical angle BAC=SO°^ we have, if AC=aj and if h equals the semi-axis minor, (2^)"''= CF.BE=a.2a', .'. ¥ = K. We may now also find Oo, the distance of o, the centre of the base from 0, the point where the axis pierces the base. We have Oo^oC-OC 3i 1 1849.] RIGID DYNAMICS. 293 •*. 06' = jW, and Oo" = lOd'-- - W,2 or Od = ia. Let a', &', be the semi-axes of the ellipse BDj which is the projection upon the plane BE of the conical surface, as well as of the elliptic base BCj .-. 2a' = BD= CF+ \[BE- CF) = a + \a = la^ and a = |a, and h' = semi-axis minor of BG 1 Hence the radius'^ of g)T*ation of BD about Oo' and the length of the simple pendulum of BD about the axis 17 «■'' ^ , = 4X6 W"-^' = T5« + i« — 2I/T — TB<^* Smce the length of the simple pendulum for the conical surface is the same as for the elliptic base, it is the same for the conical sm^face alone and taken with the base. 1849. 1. If a miifonn inextenslble string, in the form of any continuous curve, be subjected to an impulsive tension at its extremities, the tension at any point will vary directly as the velocity communicated to that point in the direction of the radius of absolute curvature, and inversely as the curvature. 294 iSOLUTlONS OF SENATE-HOUSE I'UOBLEMS. [1849. Let T be the tension at any point, then T -r- ^ "^ ji ^ y > are the tensions in dircetions of the axes ; and since the tensions arc impulsive, wc have difference of tensions at the extremities of any small arc QC velocity communicated to the arc ; or -^ al + I —r« as ec ^- (1). as as at ^ ' Similarly, |ir+r§*cc| (2), %'^-^'i'^-% (^)- dx d'^x dy d^y dz d'^z rp ^ It ~d? 'dt ~dl It Is' "^ urx\' Td^ /^v * d'^x dx d^y dy d^z dz smce -jT 1-+-?^ T^ + TiF -7-=0. ds ds ds ds ds ds Now, the direction-cosines of the radius of cui'vature are d'^x d'^y d'^z d£^ ds ds^ d''x\' (d'yV fd''zy] i ' Also, if p he the radius of cui'vature, 1 (fd'xy (dW fd' Ai p~ Wds'J "^ [ds'J "^ Us'J ^ ' ,'. T Gc velocity communicated to [xyz] in the direction of the radius of absolute curvature, and inversely as the curvatm*e. 2. The nut of a screw rests upon a smooth horizontal plane, over a hole cut so as to allow a free passage for the screw, and the screw descends through the nut by its own weight: detcnnine the motion. 18-19.] KTGID DYNAMICS. 295 At time t let P be the whole action between the screw and nut perpendicular to the thread of the screw, which makes an angle a suppose with the horizon. Then the whole vertical force on the screw = Mg—Fcosa^ moment of the whole horizontal force = Pa sina, on the nut = — Fa sina. Hence, if y = depth of any point of the screw below a fixed plane, ft), &>', the angular velocities of the screw and nut, d'^j/ Pcosa df 7 2 ^Ca dt rrid(o' ^ dt ~ M' The geometrical condition is, that each two cori'esponding points of the screw and nut in contact have the same motion perpendicular to the thread; dy , . .*. rto) sma 7" cosa = aco sma. at Differentiating this equation and substituting from the above. u M Pa sina M 1 Pa sina doi d'^y _ . di (O a sma -j cosa -^ = a sma —r- : dt dt' dt PiC sin'^'a P cosV Pd^ sin'^a whence P is constant, and its value known : by substitution of this value we determine the three required parts of the motion, which thus appear to be uniformly jWcelerated. 3. The centre of a rough sphere is fixed ; if another sphere be placed on the top of it and just displaced, determine the motion of both spheres. Let 0, o, (fig. 103) be the centres of the spheres at the time t ; 0(7, oc, the two radii which in the beginning of motion were 296 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. vertical, so that C, c, coincided : then, calling the different parts and angles, as m the figure, we have for the equations of motion, for the lower sphere, MU'^=Fa (1). for the upper one, M'^ = B smcf> - FcoH (2), M'^ = B cos<^ + i^sinc^ - M[g (3), M'k"^^ = Fb (4);- and for the geometrical condition we must express the circum- stance that the spheres roll without sliding ; ... a{(})-e) = b{e'-) (5). Also we have X = [a + b) sin^, 7/ = [a + b) coscf). Takmg (1) ^ — (4) «, we have MFb~-MTa'^^0- dt' df ' whence we find ., M¥b . ^ M'V'a ^ suppose. Hence (5) becomes a{(l>-0) = nbO - b<^y a ^ + ^ J. or a = 7 d> ; and & = n r d>. a + no Now the expression for the vis viva gives us «.(|)V>ri(.+.)'(f)Vr(f)] = 23/-,,(.+.-,); 1849.] RIGID DYNAMIC'S. 297 and substituting the above values of d and 6\ we get = 2M'g{a-\-h-y) = 2M'g[a + h) (l-cos(/)) which may be written shortly dt \m^ (1 - cos^) = 7>i'''sin"^^<^ dt 1 f/<^ sin^0 ' and mt +(7 = 2 log tan \^. The constant C may be determined by supposing ^ to have a very small value a, when i = ; whence ^ , tani(f> mt = 2 log \-^ , ^ tanja ' and tan^^ = tan^a £*"", which determines + (^)(^'^ T = g cos^ — -^cos2^, by (1) and (2). Hence we find, by substituting the value of -j^ from (3), -^{co82^+|j8in"''(^+ ) (A), _ Ra %md 306 SOLUTIONS OF SENATE-HOUSE PIIOBLEMS. [1849. Also we have the geometrical condition, that P must have no motion in the direction CP after impact ; whence V co3^ - v sin^ - r^PO amGPC = 0, or V cos^ — v sin6 — wa sin^ = 0. Finding r, v', from equations (A) in terms of w, and substi- tuting in this last equations, we get fV- — cot^] cos^ - — sin^ - ^'x sln^ = 0, or V cos^ — ( — ^— ^ + a smO ) •cj = .'. CT = V sinO COS0 h a sm y a ich I is to be a maximum by the variation of 0. N( DW, a 4 = 37r' a, k' = W -«'^; .'. in Va sin0 COS0 - a' cos'^0 Va sin 2^ «' - a' (1 + COS 20) Taking the logarithmic differential, a' sin 20 = cot20 - «'-a'(l + cos20)' or (a' - a') cot 20 - a'' cosec20 = 0, or cos20 = a — a 16 97r' - 16 ' which determines the position of the point P. 11. A imiform solid cylinder is revolving with a given angular velocity about its centre of gravity, which is fixed ; the 1841).] KKilD DYNAMICS. 307 cylinder then receives a blow of given intensity in a direction perpendicular to the plane in which the axis moves: determine the subsequent motion. Since any section of the cylinder through its axis is a prin- cipal section, the blow takes place in a principal plane, and therefore only generates a velocity about the axis (that of X suppose) perpendicular to the axis of previous rotation (that of j/), and the axis of the cylinder (that of 2). Hence, if A be the moment of inertia about the axes of X and y^ C that about the axis of z ; and w,, eo^, Wg, be the angular velocities about these respective axes at any time t after impact, we have as equations of motion, A'^-{C-A)a>.^^o.^ = 0, C^-{A-A)co^ii — (cost/— cos^J. Now, to connect w and 0, we have two expressions for the rlB motion of the centre of the base, viz. h sina.w and h cosa . -r- ; dO ^ .-. 0) = ^ cota; fd()\^ 3 qh sin^a , ^ n\ •■• [it) "2 'ks^f'^''^^ -"><); .*. -jY = - J \n sin0 , I the length of the side ; or if the motion be very small, d''0 3 «7/sin'''a „ ^ k' Therefore the time of a small oscillation = 47r (3^Z)^ sina 13. An indefinitely great number of indefinitely thin cylin- drical shells, just fitting one within another, are revolving with different angular velocities, but in the same sense, about their common axis; also the angular velocity of each shell is pro- portional to a positive power (the n"^) of its radius, and that of the outermost shell is tu. Prove that if the system of shells 1849.] RIGID DYNAMICS. 309 be suddenly united into a solid cylinder, the cylinder will revolve about its axis with the angular velocity . If the bodies composing the Solar system were suddenly to become rigidly connected, explain what the nature of the sub- sequent motion would be. If 0)^ be the angular velocity about the axis of any particle, at a distance r from the axis, the area described by it in the time t n n+!j 7' r a a if a be the radius of the outer shell. Hence the sum of the areas described by all the particles in the same cylindrical shell, of thickness Sr, = TTO) -%- t . Br. a Hence the sum of all areas described by all particles in time t t r — ' « Jo a = rrmt « + 4 If w be the angular velocity after uniting, the sum of the areas described by all the areas in an equal time = TT'ot I r^hr •' = TTWC . — 4 By the principle of the conservation of areas, the two above sums of areas must be equal, or a' or TT-art — = TTtOt 1 4 n + 4 ' 4(i) « + 4 310 soLLTioNS OF sENATK-iiuL'sE rKOBLi::>:s. [1850. In the Solar system the areas of the orbits vary as the square of the mean distances, and the periodic times squared as the mean distances cubed, and therefore the mean angular velocities vary as the square roots of the mean distances. Hence, if all the bodies of the system were suddenly to become rigidly connected, the bodies nearer the sun would have their motions suddenly accelerated, and those furthest from it would be suddenly re- tarded ; after which all would proceed with a common unifonn and, so to speak, an average angular motion. 1850. 1. A parallelogram, whose centre is fixed, is rotating about one of its principal axes in its plane ; find how it must be stiiick that, after the blow, it may rotate with the same angular ve- locity about the other. Since the effect of a blow upon the velocity of rotation of the body about a principal- axis depends only upon the moment of the blow about that axis, it is plain that the blow must in this case be perpendicular to the plane of the parallelogram, and Its moments about the two principal axes in its plane must be equal and be due to the velocity of rotation already existing about one of them, and in a direction to destroy it. Let A, B^ be the moments of inertia about these principal axes, 0) the velocity of rotation about A before the blow (/) IS given, x, «/, the coordinates of the point of application of the impulse ; Am y Bm which equations determine the point of application of the im- pulse. 2. Three equal smooth spheres (radius r) are placed together on a horizontal plane, and kept in contact by a string passed round them in the plane of their centres. A cone of given weight [W] and vertical angle (2a), is placed between them 1850.] nroiD dynamics. 311 80 that its axis is vertical : find the tension of the string ; and if the string be suddenly cut, find when the cone will strike the plane. If R be the pressure between any sphere and the cone, we have for the equililibrimn of the cone 3i2sma = TF; and for that of any sphere, i?cosa = 2 TcosItt „ FTcota the required tension. At the time t after the string is cut, let y be the height of the vertex of the cone above the plane, x the distance of any sphere fi'om the axis of the cone : the equation of vis viva gives us, if W be the weight of any sphere, where y^ equals the height of the vertex of the cone in the position of equilibrium. We have also to express the condition, that the motion of the point of any sphere in contact with the cone in the direc- tion perpendicular to the generating line of the cone through the point of its contact with that sphere, is equal to the motion of that point in the same direction : or dx dii . T- cosa = — J- sma : dt dt ' dx dy ^ .'. -J- = r tana. dt dt Hence the above equation becomes ( W+ 3 W tan'^a) (^)' = 2 Wg {y, - y) : or differentiating, ^ Wg_ de ~ Tr+3ir'tan''a = — /' suppose, 312 SOLUTIONS OF SENATE-HOUSE FKOliLEMS. [1850. a constant retarding force. And the cone starts from rest; therefore the time of describing the space y^,, To find y^. From fig. 107 it is evident that OiV, the per- pendicular on the axis of the cone from 0, the centre of any sphere in the position of equilibrium, is the distance of any angular point of any equilateral triangle of side It from its centre ; /--» AT TT 2 .-. 6>A = r sec - = p r ; .*. OT = ON seca = ^ ?' seca, and PT = ( -5 seca - 1 J r ; CT = PT coseca 2 -J seca coseca — coseca j r ; .-. CN=^ CT-NT=CT- OTsma I r^ (seca coseca - tana) — coseca^ r = f-i cot a — coseca] r, and y^ = r — CN = f 1 -f coseea — —^ cota j r-; 1(1+ coseca - -j cota) ( W-\- 3 W tan'a) r j 3. Two equal particles of mass m are fixed at the ex- tremities of the axis of a prolate spheroid, of which the mass is il/, the eccentricity of the generating ellipse being e. The spheroid is struck by a couple and then left to move freely; 1850.] RIGID DYNAMICS. 313 shew that throughout the motion it will constantly have contact with a single plane, if m = ■^o^e^ The spheroid will manifestly have contact with a fixed plane parallel to the invariable plane, if it be similar to the momenta! spheroid of the system consisting of the spheroid with the two masses [m) at its poles. Let -4, jB, be the moments of inertia of this system about its axis of figure and an axis through Ita centre perpendicular to its axis of figure. Hence, if a, 5, be the semiaxes of the generating ellipse, the condition of contact with a single plane is Ad'=^Bh' (1). Now A = lMb\ B==\M{a^ + h')-\-2ma\ and condition (1) becomes or 2ma' = lM{a'-b')- = me\ 4. A small arc of a hoop is removed and replaced by two small straight lines, tangents to the circle at the ends of the arc, their mass being so disposed that the centre of gravity remains still at the centre of the hoop. If the hoop be now rolled along a horizontal plane, sufficiently rough to prevent sliding, with an angular velocity a> not great enough to make it leap, shew that motion will never cease unless . f o)''* d^ + F cos'^a \2ga ' (1 — cosg) cos a be a whole number ; where a is the radius of the hoop, k its radius of gyration, and 2a the angle subtended by the arc removed. 314 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. The motion of rolling on the circular rim will be unifoiTn : hence the angular velocity at the time of impinging on the apex of the tangent will be to. To determine the motion immediately after this impact. See fig. 1 08. Let F^ F' be the impulsive actions between the apex and the plane along the plane and pei'pendicular to it; v^, v^ the velocities of the centre of gravity in the same directions, and ■C7 the angular velocity after impact. The equations for finding y^, V . and -nr, are F F' MFzj = Mk^m + Fa - F'a tana. Also, to express the condition that the apex must be at rest after impact, we have v^ — aw = 0, Vy — a tana.-cT = 0. Hence we have Jc^z: = Jc^oi + a^ (o) - w) — (a tana)^.^, a'(l+tan''a) + k' = nco suppose. ;Next, to consider the continuous motion of turning about the apex : let 6 be the inclination to the horizon of the radius to the apex at the interval t after impact. Then, taking the equation of moments about the apex, M{/c^ + c^ sec'^a) -yy = — Mga sec a cos^ ; 'd6\ (da\ 2qa sec a . ^ ^ \dt J F + a" sec^a 2qa seca , . ^ > 7 2, 2 2- sm e - cos a k -f a sec a ^ ' 1850.] RIGID DYNAMICS. 31,') \dtj k~ + a' sec'a ^ i \ n a 1 A'^^^' In passing from a motion of rotation about the axis to a motion of rolling on the circular rim, there is no impulsive motion ; hence the angular motion at the time of the second impact of the apex on the plane is w or noa. Similarly, the angular velocity at the time of the >•'*' impact is ifca. Now, if the hoop ever comes to rest it must be by just balancing on the apex : suppose this happens when it is rolling over the apex for the ?«"' time; then equation (1) shews that we must have = w ft) - ^-^ — ^ — 1 -cosa), /c + a^ sec'a ^ '^ o 1 if 2(7a sec a ,, or 2m log?i = log ijj^-^, r^-^z l-cosa '^ [[k + a sec a) w ft)" («^ -f P cos''^a) lof ^ I2qa (1 - cosa) cosa or m = ^\ - 1 2log|n-^^^tan^a| by inverting both the quantities under the logarithmic sign. Hence, if this expression for m be a whole number, the hoop will come to rest as it is rolling over the apex for the ?»"• time : if it be not a whole number it will never come to rest. 5. Two similar homogeneous cords are similarly stretched, and one of them loaded at its middle point with a small weight /x; shew that the fundamental note of the loaded cord will be lower than that of the other, and that if t denote the time of vibration of the loaded cord for any possible note, the values of t are given by the equation tan H fiVl = A Uxl< wlicrc / is the length of the cord, • the mass of a unit of length. 316 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. and T its tension. Under what initial circumstances will both strings sound the same note? See Duhamel in the Journ. de VEcole Polytech. tom. xvii. It is evident that when the loaded cord is sounding its fun- damental note, its two halves meet in a salient angle pointing from the line joining its extremities. Thus the two halves are portions of the trochoid which a longer cord would assume in vi- brating, and will vibrate in the same maimer as if they actually were parts of such a trochoid : hence the loaded cord is virtually longer than the unloaded one, and capable of a deeper note. Let ic, 3/, be the coordinates of any particle of the string at the time f, referred to the line joining the extremities of the string, and a line through its middle point perpendicular to it and in the plane of vibration, as axes of x and y. The equation of transversal vibrations will be (Poisson, Mecanique^ n°. 490) , . rmr ,. , > mir y = h sm -^— [\L — x] cos ^— at^ A A< where h is the greatest value of y for any particle, a = ( - ) , and \ a quantity to be determined by the circumstance that the middle point of the string is attached to the weight [i. Let y be the ordinate of this weight at the same time t : the equation of motion of ^l is Now tan d'y = KIL = hrriTT rmr I rmr — 2t . — — cos — - - cos — - at] A, A 2 A ••• y = 2t h\ 1 rmr I rmr — . — . —:. cos — — - cos -T— at. fi rmr a A 2 A y = Jt=o = J . rmr I rmr h sm — - - cos — - at : A ' mir ' d^ = 2c \ . , T — . — , since a = - . fx. ymr £ 1850.J RIGID DYNAMICS. 317 Now, let t represent the time of vibration ; _ 2X, 1 _ 2\ /£ m ' a m ' \T and the above equation becomes, by eliminating \, the required equation for the determmation of all possible values of «. The value of t answering to the fundamental note is its greatest value; it is plainly such that T-l;) "2- Now suppose fi indefinitely small, or the weight removed, the value t' of t then answering to the fundamental note is TT? /S\4 _ TT T-[t) -2' and therefore t > t\ or the fundamental note is lower for the loaded than for the unloaded cord, as shewn above. The two cords will evidently sound the same note when the middle point of each is made a node ; in which case the note will be that due to a length which is any submultiple of \l. 6. If a body hang by a string, and through any point of the string a series of horizontal lines be di-awn, with any one of which the body may be rigidly connected and perform small oscillations about it, the time of oscillation will be a maximum about a line at right angles to the one about which it is a minimum : prove this, and shew how to find the position of these two lines, and the time of oscillation about any other, in tenns of the times about these two and the angle which it makes with them. If t be the time of the body's oscillation about any one of these lines about which its moment of inertia is Q^ we have the relation « = ^^ (j4)'' 318 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. where M is the body's mass, h the depth of its centre of gravity below the horizontal line in question. Now the general ex- pression for Q is Q = ^hn{x' + ^f + z'); therefore if we make the horizontal plane through the lines the plane of xy^ 'S.hmz^ is the same for all the lines, and the relative magnitudes of Q for the different lines will depend upon But this expression will be unaltered if we project every particle of the body upon the plane of xy. Hence, as in the case of a plane lamina, we shall have two axes in the plane at right angles to each other, for one of Avlilch Q will be a maximum and for the other a minimum : hence the first part of the proposition is tnie. Also, as far as the part ^hn (.c^ + y"^) of Q is concerned, we shall have the usual relation Q = Q, co^'9 + Q,^ sin'6', where Q^^ Q^ are the maximum and minimum moments, and 6 the angle which the axis of Q makes with that of Q^ : hence the true relation between Q^ Q^^ and Q^ is Q - 28m.z' = {Q^- ^Bmz^) cos'0 + {Q,- ^Smz') sm-*^, or Q = Q^ cos'^^ + Q^ sin^^, as before. Hence, if #, t^^ t^^ be the times about the lines about which Qi Qit Qii ^1'® the moments of inertia, t = {t'^ cod'd + f.; sln'^)i To find the positions of the lines of greatest and least moments in the given horizontal plane. Let the direction-cosines of this plane referred to the prin- cipal axes thi'ough the point where the string pierces the plane be ?, m^ n] and let Q be the moment about the line whose direction-cosines referred to the same axes are a, /3, 7 ; then we are to have, if ^, B^ C, be the principal moments, Q = Ad^ + B^^ + C;| — r) coseca cosec/S sm 7 + ?-cosS (5), X = (r^ — r) coseca cosec/S C0S7 coss. Now the equation of vis viva gives us 322 .SOLUTIONS OF SliNATE-IIOlSE PKOBLEMS. [1851. where «„ is the height of p where tlic solid starts from rest ; .', {cosec^rt cosec'''/9 cos'^y cos'^'s + (coseca cosec/9 sin 7 — cosS)"'' Pi-:' ,._ , (dry l'< v^o 1 (dry + —^ cotp coseca} I ^-1 = 2/y [z^^ — 7\^ coseca cosec/3 sin7 + r (coseca cosec)8 sin7 — cos8)}, an equation which, when integrated, will give lis r, and there- fore also X and 2, at any time t. From equation (5) it appears that as z will necessarily be diminished by the force of gravity, if the solid starts from rest it will roll toward C or from it, according as coseca cosec/3 sin7 > or < cos 8. 3. Shew that the diiference of the moments of inertia of a body round two axes in a given plane which are equally inclined to a fixed line in the same plane, is proportional to the sine of the angle between those axes. Let ^j, Q^ be the maximum and minimum moments about lines in that plane, Q^ Q the moments about any two lines in the plane making angle 0, 0' with the axis of the moment Q^ ; then, by Problem 6, 1850, <2 = ()^cos'0 + 1 ; a sphere (density p'), whose radius is also «, on being put into the cylinder, is supported in such a position that it touches the free surface at its vertex : shew that p \ n The equation of equilibrium is volume of fluid displaced = — .^Tra" (1). r Fig. 110 shews the position of the sphere in the fluid, the dotted line representing the continuation of the section by the plane of the paper of the free surface. Let F, the vertex of the free surface, be the origin of coordinates, the axis of the cylinder that of 2, and r the distance of any point from it. Then the equations to the free surface and that of the sphere are 2 2a CO ^ n ^ and r'"' = 2az — s'"* : therefore if z' be the height above V of the circle of intersection, or z' = 2(7 f 1 Hence the volume of the fluid displaced = irj^{2az-z')dz--.- TT -I « ( 1 — -] z"^ — !«'■ 3' = W(l-l){a(l-i)-Sa(.-l 1849.] therefore from f 1 ' IIYDUOHTATrcS. ^.|W = ..a^(l-^,, 325 or - = ('--)' > V nj 3. If X, F, Z^ be the forces acting at a point [xyz) of a mass of heterogeneous fluid in equilibrium, and Xdx + Ydy + Zdz^ be not a perfect differential, then the pressure and density will be constant thi'oughout the curves of which the differential equations are dx dy dz d]^_d^~ dZ _dX~ dX _dY' dz dy dx dz dy dx Let p^ p be the pressure and density at the point (xyz) ; ]) + dp the pressiu'e at the point [x + dx^ y + dy^ z + dz)^ then dp = p {Xdx ^-Ydy + Zdz) (1). In order that this equation may hold, and therefore equi- librium be possible, we must have the right-hand side of this equation a perfect derivative of tkree independent variables, or we must have dp Y dpZ dz dpZ dx dpX dy dpX dpY dx ' or ldY_ _dZ\ ^ ^dp _ ydp^ '^ \dz dy) dy dz dZ dX\ __ Y ^P ydp dx dz J ^ dz dx (A). dX (' \ dy dx dY \ _ Y^ _ x^^ dx dy. 326 SOLUTIONS OF SKNATE-HOUSE PliOHLEMS. [1849. ^lultiplying these equations in order by A', Y^ Z, and adding, we get as the condition wliicli the forces X, F, Z^ must satisfy in order that they may be able to produce equilibrium. Now let dx^ dfy, dz, in the expression (1) for f^>, be such that dx fh/ dz , . d_Y_dZ^(lz'_t(X~clX_dY ^'' dz dy dx dz dy dx then dp will be the variation of p as we pass from one point to the adjacent point of any of the curves of which these are the dif- ferential equations. Now, combinmg (l) and (3), we get by (2) dp = 0, wherefore p is constant along the curves whose differential equations are (3). Also from equations (A) we may put equation (3) in the fonn dx dy dz dy dz dz dx dx dy dy =. r(x'^ - Zf \ dz dx dz = r{Y^i-xf\: V dx dyj and multiplying these equations by y- , y- , -^ , respectively, we get ' ^ dp = -fdx + -^ dy + ~dz = 0: dx dy dz or p is also constant along these curves. HYDROSTATICS, 327 1850. 1. Three equal cylinders arc placed in contact upon a horizontal plane, sufficiently rough to prevent sliding: find how much water must be poured into the space between the cylinders, in order to disturb the equilibrium. Let h be the depth of the water poured in when each cylinder is on the pomt of turning about a tangent line to its base, in which case the water will run out between the cylinders. Now the moment of the fluid pressures upon each cylinder about the tangent line to the base about which the cylinder would begin to turn, is the same as the moment of the fluid pressures on a vertical rectangle of height h and breadth equal to the radius [r] of each cylinder about its base* = /' I gp [h — z) zdz Now this must equal the moment of the weight [Mg] of each cylinder about the same line or Mgr^ is the required height. 2. All space being supposed filled with an elastic fluid whose volume at a given density is known, the particles of which are attracted to a given point by a force varying as the distance : find the pressure on a circular disc placed with its centre at the centre of force. Let fi = absolute force of attraction at distance unity ; the attractions A', Y, Z^ parallel to the axes at the point [xyz] are • For this is the natiire of the sectiuii of each cylinder supposed of a height /(, made by a plane through the lines of its contact with the other cylinders. SOLUTIONS OF SENATE-HOUSE PKOULEMS. [1850. — /i.r, — /i^, — t^^i tlic centre of force being origin: hence dji = p [Xdx + Ydy + Zdz) = — fip [xdx + ydy + zdz) = — yuprdr if r'- = x^- + ?/■' + s'' ; and p=kp\ .'. -^ — — ukrdr ; To detennine C, wc have p = kp = CkB-^'^"" • BM = mass contained between two con- secutive spheres having C for cen- tre, radii r and r + Sr = 47rpr'S?- = A7rCk.rh'^'""-'Br', . 3/ = whole mass, and therefore known, ATvCk Let ^jfikr^ = 2, and r'Wr = z^dz • .-. M=4.7rCk-=— / sW^: and /%^s-v. = r(f) = ir(i) and M is known ; hence G is also known. Hence, if P be the pressure on the annulus (radius a) we have 8P = 27rrBr.j) 1850.J HYDROSTATICS. 329 from r = 0) /^'^ 3. A hollow cylinder is filled with inelastic fluid and made to revolve about a vertical axis attached to the centre of its upper plane face with a velocity sufficient to retain it at the same inclination to the axis. Find at what point of the face a hole might be bored without loss of any fluid. Let ft) be the angular velocity of rotation : then, if the fluid were contained in an open vessel, the latus-rectum (Z) of the generating parabola of the free surface would be -^. Now since it is supposed by the question that there is a point in the upper plane face where the pressure of the fluid is zero, it is manifest that the face touches the above free smiace at this point. This point will evidently lie in the diameter of the face most inclined to the horizon, at a distance r suppose from the centre of the face. Let a be the inclination of the face to the vertical, li the distance of the vertex of the supposed free smiiaee above the centre of the face, the equation to the free surface is and for a;, y we may write r cos a, r sin a, .-. r^ sin'^a = Z (r cos a — h) : the roots of this equation are equal, , , cosa .-. r = \l ^-^ . sm a g cos a to sm a 4. A mass of inelastic fluid is contained between three co- ordinate planes, each of which attracts with a force which varies 330 SOLUTIONS OF SKNATE-HOUSE PROBLEMS. [1850. as tlic distance, and the absolute forces of attraction ytt,, yu,.^, fx^^ are in harmonic progression. Half an ellipsoid is fixed with its plane surface against one of the coordinate planes, and its surface touching the other planes ; its axes being parallel to the coordinate axes and proportional to //-,"*, yu,./*, fi'^. If there be not sufficient fluid quite to cover the ellipsoid, the uncovered part will be bounded by a semicircle. The attractions X, F, Z^ parallel to the axes arc - yu-jir, .'. flp = Xdx + Y(hj + Zch (if p = unity) = - {H'.sccIx + fi_^ydy + fi^zdz) ; therefore the equation to the free surface is fM^x^ + fi^^y'^ + fi^z^ = a constant = C suppose (1). The equation to the ellipsoid, if it be bisected by xz^ is ti,[x-af + ^.y + fJi,{z-cy = C (2). (1) — (2) gives for the plane of intersection 2fjb^ax + 2/Zg02; = a constant = 2 ( C')^ A suppose ; ••• M'l^x + /*3*^ = -4 (3), since a — 1 , , <- — i — (3) may be put in the fonn fj,/' = A' - 2Afi^^x + fi^x^ ; subtracting this equation from (1) gives 2/A^a?" + fJ^,f = 2Afiix - A^ + C (4), the equation to the projection on {xy) of the curve of intersection. Let = — — — = -^ since /a,, /w,^, ^u,.^, arc in hannonic progression. tan0 = I'-i V, 1850.] HYDUOSTATICS. 331 This equation, taken with (4), shews that tlie axes of the projection on xij of the curve of intersection parallel to x and y respectively, are in the ratio of cos0 : 1; hence the curve of intersection must be circular, evidently a semicircle, whose diameter lies in xz^ and its plane pei'peudicular to xz. 5. A rectangular vessel is filled with fluid of twice its weight, and placed with its open end downwards upon a hori- zontal plane, which is then made to revolve round each side of the base successively, one of these sides being greater and the other less than three times its height: find when the fluid will begin to escape in each case, supposing the centres of gravity of the vessel and the fluid to coincide. If the vessel had a base instead of being opened at the lower end, the moment of the fluid pressure on its inside about any side of the base would be the same as that of its weight acting at its centre of gravity : hence, when the vessel is open at the lower end, the moment of the fluid pressm-es about a side of the base will be that of the weight acting at its centre of gravity, minus the moment of the fluid pressures on the plane on which the vessel rests. To find this moment, M suppose. Let the horizontal plane be supposed to have been turned through an angle a, and let r be the distance of any point in it from a horizontal line in it, at the same height as the highest edge of the vessel : the dis- tance of the edge about which the vessel is being turned will be, if a be this edge, h the other edge, and h the height of the vessel, h -f h cot a. Hence M = i f/pr sin a . adr [h + h cot a — r) J Acota pb*hcoti = gpa sina I [{b + h cota) r — /•"} dr J A cota = gpabsma{^{b+hcota) {b + 2hcota)-^[b''+3bhcota+3h''Qot''a)] = gpab sina {\l>^ + \1d} cota). Let Whc the weight of the vessel, and therefore 2 W that of the fluid: then, when the water begins to flow out, the ni(»ment 332 SOLUTIONS OF SENATE-HOUSE TKOBLEMS. [1850. about tlic edge ct of all the forees on this vessel, including its weight, is zero, or 3 W{^b cos a — ^h sin a) — M = 0, or, since W = gpabh, and b = Snh suppose, — [Stik cosa — h sina) — S7ih sina {^.3nh + ^h cota) = 0, or 3« cosa — sina — n {n sina + cosa) = ; 2n .'. tana = -2 r; which gives the value of a when the two values of n are substituted, one >, the other < 1. In both cases, however, a is < 45°. ( 333 ) HYDRODYNAMICS. 1848. 1. A cylindrical vessel, with its axis vertical, is filled with fluid, which issues from a great number of small orifices pierced in the side : find the surface which touches all the streams of spouting fluid. This surface is evidently a surface of revolution, having the axis of the cylinder for axis. Its generating curve is the line which touches all the parabolic jets of water from the different orifices in the same generating line of the cylinder. These jets have all this generating line for axis, and a common directrix in the plane in which they lie, viz. the horizontal line at the level of the surface : for the velocity of efflux is that due to the distance from this line. Hence, making the common axis and directrix axes of x and y respectively, the equation to the jet whose point of efliux is at a depth h is f = Ah{x-h). To find the line which this cuiwc always touches, differentiate with respect to /*, considering x^ y constant ; .-. = a; - 27*, and eliminating ^, f = 'lx.\x, or y = x\ the equation a straight line through the origin, inclined to the vertical at an angle of 45°. Hence the surface required is a right-angled cone placed on the cylinder in an inverted position. 2. A closed vessel is filled with water, containing in it a piece of cork which is free to move ; if the vessel be suddenly 334 soi.rTioNs uF senate-house prohlems. [1849. moved forwards by a blow, shew that the eork will shoot for- wards relatively to the water. Suppose, for an instant, the eork removed, and Its plaee occupied by solidified water ; when the blow is stnick this mass of solidified water will instantaneously receive a velocity V equal to that of the surrounding water, and the impulse on it will be MV^ if M be its mass. But when the cork is in the place of this solidified water, the impulsive actions on it of the suiTOund- ing' fluid will be the same as they were on the solidified water, and therefore the impulse on it will be the same. But the cork is lighter than the same volume of solidified water, and therefore the same impulse will impart a greater velocity to it, or the cork will move forward relatively to the water. 3. A closed vessel is filled with water which is at rest, and the vessel is then moved in any mamier: apply the principle of the consei'vation of areas to prove that, if the vessel have any motion of rotation, no finite portion of the w^ater can remain at rest relatively to the vessel. The principle of consei'vation of areas about any axis must apply to the whole mass of water. But if any portion of the water remain at rest relatively to the vessel, we may suppose it to become solidified and rigidly attached to the vessel without altering the motion of any particle of the water : but in this case it is evident that the principle of the couseixation of areas about any axis must also apply to the part of the -water not solidified ; consequently it must also apply to the solidified poi- tion of the water which, since the water is originally at rest, can therefore have no motion of rotation, which is absurd if. the vessel have any motion of rotation. Therefore, if the vessel have any motion of rotation there cannot be any finite portion of the water which remains at rest relatively to it. 1849. 1. Supposing the effect of friction in the case of aerial vibrations in a tube of uniform bore to be the production of 1849.] HYDRODYNAMICS. 336 a retarding force on each particle equal to / x velocity, prove that the equation of motion will be satisfied by taking cs'^^' sm — ■{ a f 1 — Y— ^ j t — xl as the type of the vibrations. Let X be the coordinate of any particle at rest, oc + ^ its coordinate when displaced at time t ; then the equation of motion will be df ~ clx' J dt ' Now, if we assume P = cr^^' sin —- (nat — x), where 7?'^ = 1 — '. ., ., , ^ \ ^ '^' 16(1-77 ' we have -~ +/f = cz'^-^' ■! — T— cos — {7iat - ^) + ^ sin — {nat — x)[', ,„ 47r''^fl^ . 27r . , = — c£ -•" ■; sm — (wa^ — ic), by substitution of the value of n. Ai '2 ^"1 47r'''(t''^ ,,, . 27r , , Also a -T-T, = ^-5- «?£ *-^^ sm ^{nat — x)\ dx- X X ^ ' ' ' df '^•^ dt dx' ^^ d''^ d"^ d^ and the equation of motion is satisfied. 2. Steam is nishlng from a boiler through a conical pipe, the diameters of the extremities of which are D and d respec- tively : prove that if V and v be the coiTesponding velocities of the steam, r = T — £ =* 336 SOLUTIONS OF SEXATE-IIOUSE PROBLEMS. [1851. where k is the pressure divided by the density, and supposed constant. The motion may be supposed to be that of a fluid diverging from a centre, the centre being the vertex of the cone, of whicli tlic pipe fonns a portion. Let J) and p be the pressure and density at the distance ?• from the centre of motion at the time t, when the velocity at that point is u ; then, since the motion is wholly radial, its equation is 1 dp du du . . . . - , . ~ 'j~ — ~ ~ji ~ ~ '^zr y smce the motion is steady ... (1). Also 2) = hp (2). The consideration of continuity gives the equation ?//)r^ = constant, or ^ipr^ = constant (3). From (1) and (2), k dp du p dr dr ^ or k \o^p = C — ^ii\ Let P, p be the pressures at the two extremities of the pipe, But from (3), ^•log P w- or P _ P ~ £ 2* . P vd' P~ VB'' . V = ' d' 185L L If a regular homogeneous tetrahedron be placed in any position whatever in a fluid whose density varies as the depth, shew that when the resistance of the fluid is neglected, the 1851.] HYDRODYNAMICS. 337 7A4 tetrahedron will make vertical oscillations in the time 2-77 h being the depth of the centre of gravity of the tetrahedron in the position of equilibrimn. We shall first shew that the centres of gravity of the tetra- hedron and of the displaced fluid are in the same vertical, what- ever position the tetrahedron occupies in the fluid. Let the centre of gravity of the tetrahedron be at a depth z below the sui*face, and be taken as origin of rectangular co- ordinates [xyz)^ the latter vertically downwards. Let the den- sity at a depth z below the surface be fiz : the density at the point xyz will be ix[z' + z) = c + fiz suppose. Let r, y, i, be the coordinates of the centre of gravity of the displaced fluid ; .-. (mass) X = JfJ{c + fiz) xdxdydz : but JJJxdxdydz = 0, because the centre of gravity of the solid is origin, and JJfxzdxdydz = 0, because every system of rectangular axes through the centre of gravity of a regular solid is 2i 2)'>'incipal system ; .'. ^ = 0, and similarly y = : hence the centre of gravity of the fluid displaced lies in the vertical line through that of the tetrahedron. Thus, in whatever position the tetrahedi'on be originally placed, its centre of gravity will move m a vertical line, and make finite oscillations in that line. The force acting downwards on the solid at any time = the weight of the solid — weight of fluid displaced = g(TV -g fjj{c + fiz) dx dy dz (if V be the volume of the tetrahedi'on, o- its density) = go- V — gc F, J5. ""' -df+''^'=^^ 338 SOLUTIONS OF SENATE-HOUSE rilOBLEMS. [1851. since the centre of gravity is the origin of coordinates, and therefore fjjz dx dy dz = 0. Hence the equation of oscillating motion is since c = /xz'j and the time of an oscillation Now A, the depth of the centre of gravity in the position, is d'^z the value of z in the above equation, when —p^ = ; or A = — , and the time of an oscillation This proposition is equally true of any homogeneous regular solid. ( 339 GEOMETRICAI. OPTICS. 1848. 1. Compare the brightuess of the Earth as seen from Venus with the brightness of Venus as seen from the Earth, supposing the sizes and reflecting powers of the two bodies equal. Let Sy E, Vy (fig. Ill) be the respective positions of the Sun, and the centres of the Earth and Venus, at the time when their brightness is to be compared. From E di-aw the straight lines JSa, Eb perpendiciilar to ES and EV m the plane of the ecliptic, and similarly T c, Vd per- pendicular to VS and VE: the part of the Earth seen from Venus will be contained between planes pei'pendicular to the ecliptic through Ea^ Eb ; and the part of Venus seen from the Earth between the planes Vc^ Vd. Let Q be the quantity of light that falls upon a imit of the sm*face of Venus which has the Sun in its zenith. To find the quantity of light reflected to the Earth from ajiy element 8S of the sm-face of Venus. Let the latitude and longitude of the element B8, referred to the Sun as origin, and plane of the ecliptic as plane of longitude, be 6 and (f). The quantity of light reflected to the Earth from SS will = QBS X cosine z. d. of Sun x cosine z. D. of Earth = QBS.cos9cos(li.cosecos{V-(f>), F= l SVE, = Q . r'^ Bd cos^ B(f) . cos^O cos = ^Qr'Jcofi'e{cosV.-^sm{V- 2<^) + Oj <7^: z2 340 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. from = {V— ^7r]] to \ and if /A be the refractive index of the ray, 342 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. Now, If and (f) be the coordinates of the rays of refractive Index fji whicli come out parallel to their consecutives, 0', ' may be foiuid in terms of /x. from the equations d6' '^ d" dd' ~ ' d^ d^ ^' _ dS' dii>' ' dB' ' those consecutive rays being supposed to come out parallel whicli before Incidence lie In a plane defined by the particular value of -^ . du Now, supposing and to retain the particular meaning assigned to them above, since & and <^' are now fimctlons of /t, dB (dd dO d'\ dO' dO dfi [dO'"^ dcf>"de') dfi'^ dfi' d(f> _ /d(f) d(f) d<}>'\ dO' # dfi \dd' d(f)' ' dd' J dfi dfx ' The first tenns In these expressions for -7- , -^ correspond to the variation of the direction of emergence due to the va- riation of the direction of Incidence ; the second tenns coiTcspond to the variation of the direction of emergence due to the va- riation of /A In the differently coloured rays. But we have seen that the first tenns are each equal to zero ; hence the whole variation of 6 and <^ is due to the vaiiatlon of fj, In the dif- ferently coloured rays, or we may obtain -7- , ■— by differen- tiating as if the differently coloured rays which severally come out parallel to their consecutives started from P in the same direction. In the application to the case of the rainbow all the incident rays are parallel ; we may, however, differentiate for -^ , -^ as if all the angles of incidence of rays which come out parallel to their consecutives were equal. 1848.] GEOMETRICAL OPTICS. 343 Here each ray which comes to the eye moves in the same plane; and since the rays incident upon the raindrop are all parallel, the angular coordinate after emergence will be i), the deviation. Hence, to find -y- we differentiate i), consider- dD „, ^, d' ing constant; and s,in, considering fi constant, and put dD ^ d^ = '- This equation will give us a value of ^, which substituted in D will determine the amount of deviation of the rays of the re- fractive index /t, by which the corresponding part of the rainbow is seen : let this value be D = 2mir -f i/r : then, if a^ be < tt, the deviation at the first refraction will be towards the eye, and the red rays will appear on the inside of the arch : if i/r be > tt, the deviation at the first refraction will be from the eye, and the red rays will appear on the outside of the arch. 4. Every diameter {d) of the extreme boundary of a sphe- rical reflector subtends a right angle at C the centre of the sphere: supposing parallel rays (inclined at an angle a to the ^44 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. axis of the reflector) to be incident upon every point of the extreme boundary, shew that the section of the reflected pencil made by a plane passing through C and peqiendicular to the axis, will be an hyperbola, whose axes are d and J cot a. Let CO (fig. 112) be the axis of the mirror AOB-^ let Or, parallel to the direction of incident rays, be taken for axis of x ; Cy, pei-pendicular to it in the plane OCx^ for axis of y, and Cz pei-j^endicular to Cb, Cy for axis of z. Let Cx pierce the miiTor at the point o: Join P, any point in the bomidaiy of the miiTor by arcs of great circles, with 0, o : call Po, PoA, and : OF will be 45°. The ray reflected from P will pass through the point D of Co, such that the perpendicular from D upon CF bisects CP, t. e. through the point {^a sec 6^ 0, o) : also the coordinates of P are a cos^, a smd cos^, a ski 6 s'm(f>. Hence the equation of the reflected ray is X — ia sec6 ii z , . a 1 a = ^ a ± = — • a • — T = ^ SUppOSe...(l). aeost/ — ^a secc^ asmc^cosip asmc^sm9 ^^ ^ Also, from the triangle OPo, cos45° = cos^ cosa — sin^ sina cos(^ (2). Our object will be to eliminate ^, <^ from these equations, and find the relation between r\ and z when we have written. a; = ?; sin a, y = 7] cosar the equation so fonned will evidently be the equation of the curve in which the plane through C cuts the surface formed by the refracted rays. From (1), X cos — la = a\ cos^d — \a\ ; .-. aV cos'a - a\x cosB + ^x^ = ^a^X{\ - 1) + Ix' ; .-. a\ cos e = ^x+ {^d'\{\ - 1) + lx'}i (3)- Hence equation (2) becomes, multiplying by aX, -^a\ = [^x + {^a^X{X- 1) + ^x^] cosa - y sina, 1849.] GEOMETRICAL OPTICS. 345 or {|rt'''\(X— 1) + ^x"]* cosa = -^ a\ + y sina — ^x cosa, it" or {^a''X(\— 1) + \7f sin* a]* cosa = ^ «^ + ^ sina cosa ...(4) : hence, squaring, \ci\ (A. — 1) cos'^a = \d'''}^ + -^ ciXt] sina cosa ; .'. aX sin'''a + a cos'^'a = — %^i) sina cosa, and aX = — 2-7; cot a — a cot'^a. Again, from equations (1), d'X' Qm^e = f + z' : adding this equation to (3)'^, «V = ^x' + f + z' + K^(X- 1) + {ia'X(X- 1) + \x']^-x^ or by (4), ia'^X (X + 1 ) = ^77^ + i77^cos'"'a + z^ ^ ( tu: «^ + \'r] sina cosa ) « sina ; cosa \2* y ' or retaining only the second powers of t] and z^ Iff cot'a = \'i]\\ + cos'a) + z"* - tf + ^t^'' sin'a + &c., or T\' cot'^a — z'' = &c.; shewing that the required locus is a hyperbola, the ratio of whose axes is cot a. Now the axis in the plane OCx is evidently d: hence the axes are d and d cot a. 1849. 1. A ray of light is incident upon one of two reflectors inclined to each other at an angle - , in a direction parallel to a line which is at right angles to their intersection, and bisects the angle between them : supposing the intensity of a ray reflected at an angle (j> to be to that of the incident ray as e cos(f> to 1, shew that the intensity of the ray after it has suffered n reflexions will be to that of the incident ray as e" to 2"-'. 346 SOLUTIONS OF SENATE-HOUSE PROBLEAIS. [1849. From the problem on p. 60 it appears that, if n be even, the successive angles of reflexion are complcmentaiy of Itt S tt 7} — 1 it n — 1 tt Stt Itt 2'w' 2'n '" 2 'w' ~~2 'n"'2'n^ 2n' Ilcncc the intensity of the ray after n reflexions : that of the incident ray -"( . 1 TT . 3 TT .71—1 TtV , sm - - . sm - - ... sm — - — . - : 1. 2 n 2 71 2 nj Similarly, when n is odd, this ratio is „ / . 1 TT . 3 TT . n ttV . e sm - — . sm - - . . . sm - . - : 1. V 2 w 2 n 2 nJ Both these ratios may be expressed by the general formula „.l7r .37r .2?i — Itt, e siij - - . sm - - . . . sin — : 1 : 2 71 2 n 2 w ' and in Hymers' Tlieory of Equations^ Art. 22, Ex. 20, it appears, by making ^ = 0, that . 1 TT . 3 TT . 2« - 1 TT 1 sm - - sm - - . . . sm — - — - = -^r=-, , 2 n 2 71 2 w 2 " whether n be odd or even : hence each of the above ratios = e" : 2"-\ 2. A transparent medimn is bounded by two parallel planes ; the refractive index is constant thi^oughout any plane parallel to the bounding planes, but varies continuously in the direction of the nonual to those planes: shew how to find the path of a ray of light tlu'ough such a medium, and prove that in passing through a section for which the refractive index is a maximmn or a mmimmn, the path will in general have a point of contraiy flexure. It is evident that the path of any ray will lie in one plane : in this plane take two lines, one perpendicular to the bounding planes, the other parallel to them as axes of x and y re- spectively. 1850.] GEOMETRICAL OPTICS. 347 At the point [xy] let the inchnation of the path to the axis of X be 0, and let the refractive index be yit: at an adjacent point [x + Sa-, ?/ + Sy), let these be ^ + 8^, fi-\- Bfi: then, by the law of refraction, Bin = — sm{(f> + S(f)) + — 1 (sin^ + COS080), or = cos ^B(f) ■+ — sin ~ + - -f- = 0'. ax fi ax .'. logging + log/i = constant = logO, .*. sin© = — , 01* 1 + I ^- I — I Vy / ) dxV _ ffM dy) " \G dy f//.^^ ^-* which, since /i is a known function of a*, is a diflferential equation for the determination of the path. 7 When /x. is a maximmn or minimum, we have -y- = 0, and as the ray passes through such a section, its path usually suffers inflexion. 1850. 1. If a string be wrapped round a glass prism, whose section is an equilateral triangle, so as to be always inclined at the same 348 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. angle to the axis of the prism, the portion of the string seen by internal reflexion will appear to be parallel to the portions seen directly. Let AB (fig. 113) be a portion of the string seen directly, BC a portion seen by internal reflexion at the surface AC oi the prism : and first, let the eye be In such a position that the small pencil of rays by which any point of BG very near B is seen, shall pass through some point of the surface ABB' very near to B\ a point in the edge AB'. Then, if the eye be at a con- siderable distance from the prism so that the axes of small visual pencils may be considered parallel to each other, any point of BC very near C will be seen by a small pencil which passes through the surface ABB' at some point very near C a point in the edge BC. Let B'U^ G'E be the directions of these small pencils upon emergence. Now the plane CC'E is parallel to the plane BB'E] and if we draw a plane through A parallel to these, the plane BB'E will be equidistant from the other two, since the prism is equilateral, and AB^ BC equally inclined to its axis; hence BC = AB'^ and B'C is parallel to AB: hence BC appears parallel to AB. Now let the eye be moved in any manner without approaching too near the prism ; it may easily be seen that the locus of the points where the rays from BC to the eye cross the plane ABB' is parallel to B' C", and therefore to AB. Hence BC will always appear parallel to AB; and the same may be shewn of any other portion of the string seen by internal reflexion. 2. A rectangular box, at the bottom of which is a plane mirror, contains an unknown quantity of water ; from the angle at which a ray of light must enter through one of two small holes in the lid in order that after refraction and reflexion It may emerge at the other, determine the height of the- water in the box. Let a be the height of the box, x the depth of the water, 2h the distance between the small holes in the lid j ^, (f)' the angles of Incidence and refraction when the ray enters the water. 1850.J GEOMETRICAL Ol'TICS. 349 they will also be the angle of refraction and incidence as it emerges from it : hence we have h = [a — x) tan + ic tan 0', sin0 = /isin^'; a tan 6 — h tan^ — tan^' a tan (j) — b , sin cb ' tan-, Ps^ Pt, perpendicular to these lines respectively. Then Pr + Pt= OP (sill POr + sin POf) = 2 OPsin^iPOr + POf) co^{POr - POt) = 2 0PsmPOs costOs = 2Pscos6^. Hence, if Xj y, the coordinates of P, be substituted in Mj, u^^ a^, we shall have u^ + u^ = 2a^ cos^j, and the same may be shewn wherever the point P is taken ; hence generally, u^ 4- ti^ = 2a^ cos 0^:'\ so u^ + u^ = 2a^cose^ \ ■?/, 4- u = 2a cos^ » ' 1 n n J From equations (A) 6^^ ^.2V ^„ must be found, and thence Wj, «*2V ^n from equations (B), and the path of the ray will then be fully determined. If there be four reflectors we find, by adding the P' and S'^ of equations (A), O, + e^ + 0, + 0, = 27r-a^-^,: .similarly, by adding the 2"<* and 4}^^ ^, + ^. + ^3 + ^4 = 2vr - a.^ - a^ : 1851.] GEOMETRICAL OPTICS. 353 hence we must have a, + a, = a^ -f- a^ = TT, or the quadrilateral must be inscrlbable in a circle in order that the problem may be possible. If however this condition is satisfied, still the problem is indetemiinatc ; for if wc treat equations (B) as above, we find ttj cos^j + rtgcos^j = a^ cos6^ + a^cos^^ (1). This is an identical equation ; It will therefore lead us, by means of equating the coefficicjits of x aiid i/ on the two sides of the equation, to three conditions between ^,, 6,^^ 6^, 6^^ and constants : between these constants there ai'e also two relations arising from the circumstance that the quadrilateral may be inscribed in a circle, and the three conditions between ^,, 6^^ 6^^ 6^^ and constants amount to only one equation independent of equa- tions (A). This condition, together with equations (A), will detci*mine ^1? ^2i ^35 ^4- ^^^ since we have derived the condition (1) from equation (B), it shews that those equations are equivalent to only three independent equations, and ii^^ u^, u^, u^ are therefore indetemiinatc. The dii-ection only of the different parts of the path of the ray can be determined; with these directions any position will satisfy the problem. Similarly, if there be any even number of reflectors, we may, from equations (A), deduce the condition a, + 0(3 + ... + a„_, = (x^ + a^ + ... + a„ or the siuns of the alternate angles must be equal : this condition is satisfied if the polygon can be inscribed in a circle, and the problem is then possible. The problem is still indeterminate, for from equations (B) we may deduce the condition a,cos^,+a3Cos^34-...+ a„_,cos^,,_,=rtjjCos^2+a^cos^_, +...+ «„co8^,, : this Identical equation will lead, as before, to one equation of condition independent of equations (A) between 6^^ 0^, ... 0^^ and constants, which, with equations (A), serves to detennine AA 354 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. O^^O.^y.. 6^. Still equations (B) are equivalent to only n — \ independent equations, and ?ij, ?<2, ... w„ are therefore inde- terminable. Ill fact, we have shewn in Problem 2, page 59, that ' if a ray of light, after being reflected any number of times in one plane, at any number of plane surfaces, return on its fonner course, the same will be time of any ray parallel to the former which is reflected at the same surfaces in the same order, provided the number of reflexions be even.' ( 355 ) ASTRONOMY. 1848. 1. If there had been uo stars, how might the absoUite periodic times of the Earth and planets have been determined, even if the eqnator had coincided with the ecliptic ? We might first have detenuined the synodic time ( T) of the Earth and any superior planet by observing the interval between successive conjunctions. Let jE, P be the periodic times of the Earth and the planet, 27r 27r , . , , . . -'. 'Y ~ 'p ~ relative angular velocity of the Earth and planet - yr, or E=t{i-^ We might then have observed the elongation from the Sun of P and the other planets at their points of station : this gives the ratio of the distances of the Earth and each planet from the Sun, and therefore, by Kepler's law that the squares of the periods are as the cubes of the mean distances, it would give E approximately the ratio of the periods or -p for each of the planets; whence from above E^ and the periods of all the planets, would be known. 2. A star map is laid down on the gnomonic projection, the plane of projection being parallel to the equator: give a gra- phical solution of the problem, to determine the time at a AA2 35G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. kno\NTi place l>v observing wlicn two stars laid down in the map arc in the same vertical plane. Since the place is known, we may draw about the centre of the map the circle described by the projection of the place on accomit of the Earth's daily rotation. Let a line through the two stars intersect this circle in two points A and B. Then it is plain from the method of projection, that when the projection of the place is at A or B, the two stars arc in the same vertical. Let the differences of longitude of the Sim in its place among the stars on the day of observation, and the points A and B, be obseiwed ; this longitude, converted into hours at the rate of 15 degrees to an hour, w^ill give the interval since last noon or the true solar time. 3. Shew that at the equinoxes the extremity of the shadow of the style of a vertical south dial will trace upon the dial-plate a horizontal straight line at a distance acosec? from the upper extremity of the style, a being the length of the style and / the latitude of the place. On the day of the equinox the Sun appears to move in a great circle of the celestial sphere. We may consider the extremity of the style as the centre of that sphere, or that the Sun moves in a plane through the extremity. The nonual to this plane lies in the vertical plane thi'ough the north and south points, therefore its intersection with the dial-plate will be a horizontal straight line ; this is the line traced out by the ex- tremity of the shadow of the style. Let the plane of the paper be the vertical plane containing the style AB (fig. 117). Draw AG vertical and BC perpen- dicular to AB: then, since AB is parallel to the Earth's axis, and the Sxm is in the equator, C is the extremity of the shadow at noon, and ^ is the distance of the horizontal line from A : it = ABcosecACB = a co&ecl. ASTRONOMY. 357 1850. 1. If a rod be fixed into a vertical wall which faces the south and the shadow of it be cast upon the wall by the Sun, find the curve upon which the shadow of the end of the rod will be situated every day at mean noon, the Sun being supposed to move imifonnly in the ecliptic with his mean motion. The mean Smi is situated on the equator at the same distance from "Y* as the true Sun ; it is mean noon when this mean Sim is due south. Let S, S' (fig. 118) be the true and mean Suns, then nr >S' = "V >S", and if we draw SD an arc of a great circle perpen- dicular to the equator, and call 'Y' S, tp D, SD^ i, a, S, re- spectively, we have S'I)=yS-'rD = L-a. Now let E be the extremity of the rod, its shadow when the Sun is in v , P the position of its shadow when the Sun is at yS, and S' on the meridian, i.e. due south. Then if we draw ON horizontal, NF vertical in the plane of the wall, and join EO, EN, EP] OEN== S'D, NEP = SB, EPN=l - S, where I = latitude of the place. Call On, x, JVP, y, EO, d, .'. X = dtoxiS'd, ^ {d' + xy-smSD ,, , -r . , tanZ — tana or X = dtaiiiL — a) = d — — ; — -^ ^ ' I + tana tsuiL , tana — cos w tana . , = d 5 , smce cosw taniy = tana cos...(2). sm/cotw— cos/sma ^ '' "We have now to eliminate a between (1) and (2). 358 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850. From (2) sin 7 cot G) , (d'^ + x'^)^ — ; cost = -^^ , sma y • 7 V or sin a = sin 6 cotco -r^. rrd- {(.r + xy + ycosl' From (1) (coso) cos^rt + sin^a) x = d{l — cosw) sina cosa, or (coso) + (1 — COSO)) sm'aYaf = (f (1 — cosw)'' sin^a (1 - sin^a) ; .-. [cosw {{d^ + xy+ycosl]'^ + (1 — cosw) sin'"*? cot^ co.^^]'"' x^ = (1 - COSO))'' cof 0) sin^? d'f [{{d^ + x^ + y cos?]'' - sm7 cof eo.?/''], the equation to the required cm've. 2. Suppose that dm'ing the day of the equinox, a man walks in a horizontal plane towards the Sim at a miiform rate ; prove that the equation of the path described by hun is ny \ _ sin? ^^ ^^^^^ _^ ^ „,,^, sin -^ + 0=^ £«'-' + e ,, Vasecf / 2 V / where x and y are the coordinates of his position at any time, measured along and at right angles to his meridian at noon ; I is his latitude, and a is the space he walks over while the Earth revolves through an angle n. Deduce the particular cases of his being at the pole and at the equator. Let 0) be the angular velocity of rotation of the Earth about its axis ; the angle apparently described by the Sun in the time t will be ft)^, since the Sim is in the equator, it being the day of the equinox. If a be the Sun's azimuth at time <, dy -~ = tana. ax Now I and mf are the sides of a right-angled triangle, sup- posing the man to start at noon, of which the angle opposite tot is a y 1850.] ASTRONOMY. 359 t&ncot .•. tana = -v— ,- , dy tanwf dx sin? ' ■V dy _ tanw^ *°^ ds ~ (sm^Z + tau'^a)«)i ' cfe sinZ (/5 (sin' ? + tan' «»«)** Also, * = - G)^ n . n - tan -s sin - s ay a a ^ fsin^Z + tan'-sV fsin'^Z cos' - s + 1 - cos'' - sV V a J \ a a J . n . n sua — s sm - s a , a — sec6 4 ; [ 1 — cos''? cos''- s] f sec' ? — cos' - s j .*. y = -secZ-jcos'M cos - scosZj — 4 ; •.• y = when s = 0, or cos -s= sec? cosf — —i+l] (1). a \asecl J ^ sm I cos - s . . dx a [sm? 1 — sm -s + sm - sj V a I a sm? cos-« a ( sin'' ? + cos' ? sin'' -s\ n cos -s 7 ^ = tan? [tan' ? + sin'' - s j 360 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. a . w / . . . w \' sin - s + tan"'' I + sin" - s ) a \ a ) .'. X = - tan Hog i ; '.' x = Q when s = ; n tan / / \4 "' .'. sin -s + (tan'/ + sin' - s] = tan? e° '^°' ? a \ a J .-. 8in-5- (tan'/+sin^-5) = - tanZ e""'^"' , a \ a J or (sec'Z-cos'^sV = ^ /ga-^ + g-^ ^ the required equation to the path. We have in this sokition considered I constant ; if, however, the man be at the pole, I will = ^tt, and sec/, tan/ will be susceptible of great changes when / alters but very little ; hence we must consider his motion as indefinitely small compared with that of the Sun, or - indefinitely small : hence the above equation leads us to x = 0, ^ = ; and the man merely stands at the pole looking towards the Sun. If he be on the equator, tan / = 0, and therefore a; = 0, or he walks along the equator. 1851. The declination of the Sun at two obsei'vations S, 8', and the Sun's motion in right ascension and longitude in the in- tei-val between the observations, are equal: shew that if &) be the obliquity, and a, / the Sun's right ascension and longitude at the first observation, cosw = cosS cos 8'; tana = sinS cot 8'; cot/ = sin 8' cotS. Let P, K (fig. 119) be the poles of the equator and ecliptic; >S', S' the two positions of the Sun : then, since the differences 1851.] ASTRONOMY. 361 of the Sun's longitude and right ascension at S and S' are equal, SS'=SPS'. Let the angle P8K=^'^ draw KR^ an arc of a great circle, to meet SP produced at right angles: then, in the triangle SPS\ sin/S-S" ^mPSS' = sin -S'P sin /SP/S", or cos<^ = co88'; .*. ^ = 2', and KR = (f)=:B': also PR = 8, KP = ft), lKPR = 90 - a, lPKR = I Hence, by Napier's rules, cos 0) = cos 8 cos S' (1), sin 8 = tana tan 8', or tana = sin 8 cot 8' (2),. sin 8' = cot? tan 8, or cotZ = sin8' cot8 (3), and (1), (2), (3), are the formulae required to be proved. ( 362 ) DISTURBED MOTION. 1848. Two bodies, P, P (fig. 120), describe round a central body S circular orbits lying in one plane, the orbit of P being within that of P'; prove that the disturbing force of P' on P, when wholly central and additions, will be equal to the disturbing force when P, P' are on opposite sides of /S, provided SP' be a mean proportional between SP and SP + SP. Let Pj be the position of P when the distui'bing force [F^ on P, is wholly central and additious ; . F -J^ ^ • > P^P"''P^P"> (where fi is the absolute force of attraction of P'). Under the condition that the force of attraction of P' on S and Pj, perpendicular to SP^, is the same, i.e. that SPP^ is an equilateral triangle, .-. PP^ = PS, IJ,.SP^ or F = SP" Let F^ be the disturbing force when P is in opposition to P' at P„ P = fi fi if SP" P^P" ' and P, = P, 1 1 ^P SP" {SP^ + SPY~ SP"' or, dropping the suffixes, because SP^ = SP^, {SP+ spy - SP' = ^ {SP+ spy 1849.] DISTURBED MOTION. 363 or SRSP' + 2SF" = SF' + 2SRSP" + SF% or SF'= SF{SP+SF'), or if SF" be a mean proportional between SF and SF + SF. 1849. If, in addition to the force of the Sun on a planet, there be a small force tending towards the Smi, and varying invei*sely as the m^^ power of the distance of the planet from the Smi, prove that the perihelion of the orbit will have a progressive or re- gressive motion, according as m is greater or less than 2. Can you explain this result by reasoning similar to that used in " Airy's Gravitation'''''^ If F be the whole central force on the planet we shall have = jXU + fill , where fi' is very small. The equation of motion is d% F d'^u II u! „,_.. For a first approximation, d^u yU- W + " ~ P " ^' which will be very approximately satisfied by M = «{1 +e cos(c^ — a)], if c be very near unity, and a = ^ ; .-. ^^ m"'-» = ^ a"'-' {1 + {m -2) e cos(c(9 - a)|, omitting higher powers of e. Hence, for a second approximation, -j^ +u- a- — a"' ' -- do fJL fl + u- a- — a"'"' - — a"""' [m -2) e cos {c0 - a) = 0, 364 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. which is satisfied by u = a\l+ — a'"-' + e cos (cd - a)]. if ae{l- c') cos [cO - a) - — a'""' {m -2) e cos(c^ _ «) = 0, A* or l-c' = ^(m-2)a!"-': hence c is < or > 1 according as m is > or < 2. Now, the argument {c6 — a) may be put in the form ^-{a+(l-c)^}; whence it appears that the above equation between u and is the equation to an ellipse, the longitude of whose apse is a + (1 — c) ^; its apse will therefore progress or regress accord- ing as c is < or > 1, i.e. according as m is > or < 2. This result may be explained in a manner similar to that used in Aiiy's Chavitation^ Art. 98. Let P, A be perihelion and aphelion. The disturbing force is towards >S' both at P and A ; it will therefore progress about P and regress about A. To consider which of these effects will be the greater. If the disturbing force at P, A and the other points of the orbit were propor- tional to the inverse square of the distance, its only eflfect would be to alter the magnitude of the central force in a certain ratio without altering its law ; it would therefore have no effect upon the position of the apsides, or its eflfects about P and A would be equal. But if the disturbing force vary inversely as the (distance)"', where m is > 2, the ratio of its intensity at P to its intensity at A will be greater than the ratio of the intensities of the central force at those points ; hence its effect will be greater at P than at -4, or the progression at P will be greater than the regression at A ; i. e. on the whole the perihelion will progress. Similarly, it may be shewn that if m be < 2, the perihelion will regress. ( 365 ) ATTRACTIONS. 1848. 1. A sphere is composed of an immense number of free particles, equally distributed, which gi'avitate to each other without interfering: supposing the particles to have no initial velocity, prove that the mean density about a given particle will vaiy inversely as the cube of its distance from the centre. The attraction upon any particle will be the same as if the matter nearer than itself to the centre were collected there, and attracted with a force varying inversely as the square of the distance. This attracting mass will remain the same for the same particle throughout the motion. Let x^ x-\- ^x be the distances from the centre at the time ^, of two particles situated in the same radius, whose original distances from the centre were «, a + S« ; d'^x fji, •'• ~cie^~ x'' and U- =2/. --- =2^ ,dtj \x aj ax dt ( ci\^ X dx \2fiJ (ax—x^)^ a \^ ( ^a ^a — x 2fjbJ \{ax — xy {ao; — x;y But /A depends upon the mass originally contained within the sphere radius a ; .*. fi^ a^ = Ca^ suppose ; 366 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. In order to find the relation between ^u- at the time t and 8a, we must differentiate this equation, considering x and a variable, and t constant; _ /,, lx\ aSx — xBa •■■ ^ -^ W a' ' or 8x = '- 8a. a Hence the volume of the shell originally contained between the spheres of radii a, a + Sa, i. e. of volume iTrci^Sa^ is now of x^ volume Attx^Sx = 47r — S« a x^. Hence the density of the matter in this shell, Avhich varies inversely as the voliune, varies inversely as its (radius)' : hence the proposition is time. 2. Prove geometrically, or otherwise, that if g be the attrac- tion which a particle m exerts on a point m a closed surface 8, the angle between the direction of g and the normal, doi an element of 8^ JJg cos, Odd) = 4:77771, or = 0, according as m is within or without 8, the attraction of m at the ni distance r being — ^ . Extend this result to the case of a finite mass cut by 8, and thence prove by taking for 8 an elementary parallelopiped, that if V be the potential of any mass for an internal particle, d'V d'V d'V 'd^-^W^w=-^''p' About the particle m as centre describe a sphere of radius unity ; and let a cone having m in in its vertex, and circum- scribing the element dco of the smiace 8, include a portion dco' of the surface of this sphere : then the relation between day and dw' will be dot) cos 6 = r\ do)'. 1848. ATTRACTIONS. 367 Also, let g be tlic attraction which m exerts on a point at distance unity; .■.g = l: ' r hence g coaOdo) = g'doi'^ and fjg cosddo) = g'co' = the whole attraction of m on w', where w' is the whole projection of S on the surface of the sphere: hence, if m be external to S we see, by taking the projection of each element with its proper sign, that to' = ; but if m be within S, cu' = 47r ; and, by the question, g' = m ; •'• !J9 cos^c?&) = 47rw, or = 0, according as m is within S or without it. This equation expresses the value of the sum of the attrac- tions of a particle m on the different points of a closed surface, each resolved in the normal to the surface at the point. Now, suppose the sm'face S to cut from a finite mass the mass M^ the above equation holds for every element of this mass, and therefore for the whole, if the symbols involved be properly modified : we shall, therefore, still have the sum of the attrac- tions on each point of ^S", resolved in the nonnal at that point, = 47rJ/. Again, suppose S to be an elementaiy parallelepiped so small that the density [p] may be supposed uniform throughout it: let V be the potential of a mass for an internal particle whose coordinates are a?, y, z. Let P (fig. 121) be the point a;, ^, s, and the comer of a parallelopipcd whose edges Zx^ Sy, S^, arc parallel to the coordinate axes. The above considerations shew that the sum of the attrac- tions on the faces, each resolved in a direction pei-pendicular to the face, will be due to the matter contained in the parallele- piped: now dV rrr p[x - ^) d^ dy d^ -III dx- JJj{[j--^f+{^-r,Y+{z-^y]i^ 368 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [18-48. the integration extending throughout the parallelepiped ; hence dV . at P, -J— will be positive, since ^ is always greater than x ; at P it will be negative, since ^ is always less than x + 8x. Hence the absolute magnitude of the attractions parallel to the axis of a^ at P and j) will be, respectively, dV ^ dV d'V^ -J— and J =-T- ex. ax ax ax Now we may consider the surfaces il/P, mj) so small, that the attraction on every point of each of them is the same : hence the whole attractions on JWP, mp parallel to the axis of x d^V = -—hxhj 8z. The whole expression for JJg cos Odo) is in this way found to be / d'V d'V d'V\ . . ^ [-d^-df-^)^''^^ ^^' ^^"^^ ••• = ^'^^^ = 4:7rp 8x By Bz ; d'V d'V d'V •'- d^-^df-^d^ = - ^^^- 3. Supposing a mass of homogeneous fluid, which attracts every particle of matter with a force varying as -.. ..; , to be enclosed within a thin spherical shell, find the path described by a heavy body let fall from any point of the surface of the fluid, the resistance varying as the velocity. Prove also that the body will reach the axis and equator of the spheroid after the same intei-vals respectively, from whatever points of the surface it begins to fall. The attractions of the spheroid on any particle within it perpendicular to the axis and equator, vary respectively as the distances {x, y) of the particle from that line and plane, = ixx^ fix suppose. 1848.] ATTKACTIONS. 369 Also, by the question, the resistance ou the particle in motion = kv = k — : hence the resolved parts of it are J ds dx - ds dti , dx , dii dt ds ' dt ds ^ dt^ dt Hence the equations of motion are d X dx ^ + k^ + ;.x = 0, d'?/ , dy Let a, yS be the roots of the equation z^ + kz + fi = 0, and a', /9' those of z"" + kz + yu,' = : the above equations give X = Ae"" + Bt^', y = A's*' + B'zl^'\ A, B, A\ B\ being arbitrary constants to be determined by the circumstance that the body falls from rest from a given position. The circumstance that it falls from rest gives us the condition that ^ = 0, ^ = 0, when t=0: dt ^ dt ^ ' .'. = AoL + ^/3, and = A' a + B'^' ; A B ^ .-. - = - - = C suppose, A' B' „, W ^~V^ suppose ; the equations for the determination of the relation between X and y by the elimination of t. Hence, if f, t' be times of fallmg to the axis and equator respectively, = fit" - ai^\ BB 370 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849. .-. log/3 -\- at = loga + yS^; log a — logyS t = so « = a-y8 ' loga' — logyS' _ x' - ^' • whence it appears that f, t' arc independent of the particle's original position. 1849. 1. Each particle of two indefinite straight lines, lying in the same plane, attracts with a force which varies inversely as the distance. Determine the motion of a body projected in any direction along the plane. We must first find the attraction of either of the lines AB (fig. 122) upon the particle in any position P. From P draw PD, the perpendicular on AB^ and join PQ^ Q being a point at the distance x from D. Let hA be the attraction of an element hx of the line about Q resolved in PD: .: BA^^cobQPB fjbadx .'. A = u tan~^ - a fJiTT. from a; = — 00 to a; = + CO Hence P will be attracted by two constant attractions in constant directions, which are therefore equivalent to a constant attraction in a constant direction, viz. 2yu,7r sin a (2a the angle between the lines), parallel to the internal bisector of the lines. Hence the case is the common case of projectiles, and the path will be parabolic. 2. The attraction of a uniform filament of matter, in the form of a plane curve, upon a particle is replaced by that of a circular filament having the particle for its centre: find the 1850.] ATTRACTIONS. 371 law of density of the circular filament in order that this may be done. Let the cuitc be referred to the particle as pole, and let a be the radius of the circle, yu. the density of the filament In the form of the cun^e, p that of the circular filament at the point [B] ; pa 80 fJ'^s ^ ■' ^^ ,:^ > fia ds r = fia \n^ + [dej 1 ' u 1 7- ? fia ^J' if p be the perpendicular from the ! particle on the tangent at the point (r. ,0) 5 1 1850. A uniform rod is placed witli its middle point against a rough circle, in whose centre resides a force attracting inversely as the square of the distance : if the rod be slightly disturbed from the position of equilibrium, find the time of a small oscillation. Let 6 be the inclination of the rod to the horizon at the time t] ^ the distance of its middle point from the point of contact with the circle. Since the motion is small, we may take the equations of motion about the instantaneous axis of rotation : hence we have where k is the radius of gyration of the rod about its middle point, and L the moment of the attractions on the rod about the point of contact. bb2 372 SOLUTIONS OF SENATE-HOUSE I'KOBLEMS. [I8f)l. To find L. In fig. 120 the moment about D of the attrac- tion on an element at Q (P being the centre of the circle), if /i be the absolute force of the attraction, p the mass of a unit of length of the rod, fipa X Sx {a' + xj' _ , ^M- , 2? being the length of the rod, {(i'^rf \^ ' a^ + r V d'^rj omitting p and higher powers of |^, 2fipal g _ 2/jbpdH „ to the same degree of approximation : and the equation of motion becomes "^^ {d'+ry- or, since M= 2?p, g+ ^^ = 0. '^^ id'-^rfu' Hence the time of a small oscillation r=2 /Lt^a 1851. 1. Two uniform straight rods AB^ CD (fig. 123), mutually- attracting each other with forces varying as the distance, are 1851.] ATTRACTIONS. 373 constrained to move in two grooves ABO, CDO at right angles to each other ; dctenuine the time at which the extremity of one of the rods reaches the point of intersection of the grooves. The attraction towards of each element p'hri of CD upon any element ph^ of AB, at a distance f from 0, will be the same, viz. fip'SrjpB^.^ : hence, if AB = 2a, CB = 2b, the whole attraction towards of CB on AB will r = 2fipp'b J -, ^dl. if a? be the distance from of the middle point of AB^ = Afipp'hax. The equation of motion of AB is therefore d'^x 2pa -y^= - ifipp'baxj or -YY + w"'a: = 0, if n^ = 2fip'b ; .'. x = A cos[nt + B) = x^ cosntj if t = at the beginning of motion, and x^ is the original value of X. Hence, if t be the interval before A arrives at 0, we have a = Xq cos nt, 1 -I a or t = ,- — rfTi cos — . {2fip'b)i x^ 2. If a portion of a thin spherical shell, whose projections upon the three coordinate planes through the centre are A, J5, C\ attract a particle at the centre with a force varying as any function of the distance, shew that the particle will begin to move in the direction of a straight line whose equatioas are X y z A^B^C' Let be the angle which the radius drawn to the element SS of the shell makes with the axis of .r; then, if r = radius 374 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. of sphere, and (/•) be the law of attraction, the attraction of 8S on the particle parallel to the axis of x will be {r)A. So Y=<}>{r)B, Z={r)C. And the equations to the direction of the resultant attraction, which is the direction in which the particle will begin to move, are X y z X^ Y^Z' X y z ( 375 ) PHYSICAL OPTICAS. 1848. A spherical wave of light is incident directly on a lens : find approximately the retardation of the several portions of the • ,. , .111 wave, and prove m this way the common equation ~ / * Suppose the lens to be a positive concavo-convex whose thickness at the middle point is indefinitely small : take this middle point as origin of coordinates and the axis of the lens for axis of x. The retardation of any part or ray of the wave will = (/u. — 1) X length of the path in glass = (/a — 1) p suppose. Let ic, y be the coordinates of the point of incidence of this ray, Q the inclination to the axis of the part of the ray within the lens. The equation to the two surfaces will be V' = ^r^ (1), and 77' = 2s| (2), very nearly ; since | is very small for all rays near the axis. Now (2) is satisfied by the coordinates x — p cos 9^ y- p sin 0, or, as B is veiy small as well as p^ hj x — p and y ; .-. f = 2s[x-p) ='-y-2sp', 1 ny p = \r--sn therefore the retardation of this portion of the wave = (-')(7.-^)f- Hence, if w be the distance from the origin of the centre of the incident wave, the equivalent length in air of the ray 37(> SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848. we are considering from the centre of the wave to the point of emergence, = {(«-^)'+/]- + /*P = u — X ■\- ^ +At \ — very nearly 2m \r h) 2 "^ Consequently this ray upon emergence is in the same phase as the ray incident directly when it has travelled a distance \ V ii\ V— alter emergence. \u r ^\r s)\ 2 ^ Now the geometrical focus upon emergence is the centre of curvature at the vertex of the surface of revolution, which is the locus of all parts of the wave which, after transmission, are in the same phase of vibration. Let V be the radius of this sphere when only the parts of the wave indefinitely near the axis have emerged : the sphere will then pass through the points [x-p, y) or (^1^, 3/j and y being indefinitely small ; u r ^\r syi 2 ' .-. / = 2v 11 ^1 _ l\\ t 4. t u r^ *^\r s/j 2 "*" 25 11/ -N /^l 1 /. - = --f 0^-1 V u \r s 111 1/11 or = -^ , if -, ^ /A - 1) V u J J \r s But evidently v is the distance from the lens of the geome- trical focus upon emergence ; hence this is the usual formvila. 1850.] PHYSICAL OPTICS. 377 1850. A and B being two fixed points, and P such that AF= /m.BP, the locus of P is a circle. Shew from this property how to construct a lens of common glass, such that a direct pencil incident from a determinate point will be refracted without aben'ation. The property enunciated will be found In Prob. 8, p. 157. Let A, B (fig. 124) be the points from which the pencil is to diverge before and after passing through the lens without aberration. Draw the circular arc HCH' such, that If Q be any point In It, BQ = fi.AQ. With centre A describe any circular arc HcH' intersecting HCH' in H^ H': HCH'c Is the section by the plane of the paper of such a lens as Is required. For a ray incident upon the lens from A will sufiier no deviation ; and BQ-{fi.QP+AP) =fj,.AQ-{fi.QP+AP) = {fi- l)AP Is constant: and therefore, by reasoning similar to that in Airy's Tracts^ p. 276, it appears that the pencil diverging from A will, after emergence, diverge from B. 1851. If [6) be the angle which one of the planes of polarization makes with the plane passing through the normal to the front of the \f ave and either optic axis of a blaxal crystal, and y,, v^ be the two velocities of transmission of the wave, shew that (y,cos^)'''+ {v^%meY = h\ Since the planes of polarization respectively bisect the acute and obtuse angles between the two planes through the normal to the front and the optic axes (Griffin's Double Refraction^ Art. 21, p. 12), It follows that the angle between these two planes = 2$. Now, In accordance with the usual notation, the equations to normal to the plane front are ? = J^ = ? (1). I m n 378 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851. Those to the optic axes are 3^ = ^^ (;^^4±(J^ = (2). If therefore tlie equations to the above planes are Ax + Bi/ + Cz = 0, A'x + i?'?/ + C'z = 0, we must have Al + Bm + Cn = 0, A{d'-b-y + C{U'-c')^- = 0; A B C _ Similarly A ^b; C _ , AA' + BB' + CC Also cos2^ = {A + B' + cj {A" + B" + cy- AA' + BB'+CC ~ {{AA'+BB'+CC'y+[BC'-CB'y+{CA'-AC'y+{AB'-BA'YY- ' Now AA'+BB' + GC'= [m' {¥- c') + r {W- c^yd' {a'- b')-m' {d'-b')] rr' = {[r-^m'){b'-c') - {m' + n'){d'-¥)]rr' = {{r+m') b'-{l-n')6'- (I- P) d'+ {m'+n') &'} rr' '.' P + m' + n' = 1 = _ {a' -b' + & - [I'd' + m'V' + d'c^) ] rr' = - ( U— 25^) rr suppose, [BC'-CB'Y + {CA'-AC'f + [AB'-BA'Y = 4 {id'r' 4 m' + m'd') {d' - h')[h' - c') rV = Anf{d'-b'){b'-c')rV' = 4 Vr\"' suppose ; 1851.] PHYSICAL OPTICS. co-e '^' - ^ ^«--^-{(2i^_?7r + 4F}* 2V - U 379 = (1 _ ^i^ _ ,„'-^) 7/c' + .ti^c'a' + {l-r- m'] d'h' = W suppose ; Now Vj"'*, vj'' are the roots of the equation considered as a quadratic in v^ ; .-. v^Jrv'^ = f [h' + c'O + m'^ (c^ + «•-=) + n'' {d' + J'^) = (l_,,i-''_n'-')(J'''+c2)+(l_^--^_f-')(c''+a^)+(l_Z--'-7;i ){d'+l'') = a'^ + ?>■■' + c' - {rd' + m^V' + ?i^c''') and y>; = T^&^c'' + y«'''c'^«''' + d'd'W . . cos2c/ = v;' (1 + cos2^) + v^' (1 - cos 2^) = 2l/\ and (y^ cos^)' + ((',, sin 6)' = 6"^ ( 380 ) CALCULUS OF YAIUATIONS. 1848. A aud B are two given points in the generating line of a surface of revolution whose axis is vertical : supposing a body acted on by gravity, to descend along the surface from A to Bj find its form when the whole pressure upon it between the two given points is the least possible. Find also the form of the surface when the length of the generating line between the point A and B is also given, and point out the difference between the two results. Let y be the depth of any point of AB below A, x its dis- tance from the axis ; the pressure at this point will be P^Mg'^ + M-, as p V being the velocity, and p the radius of curvature at the point : hence we must have [Pds = Mg I f 1 + — . -^ I dx^ a minimum. TT Tr , 2?/ ds Here F = 1 + — -^ p dx _i _ J£^. dV _ 1q i^ 1 dy 1 +^ dp (1+/)'^' ^ dq 1 +2^ 1848.] CALCULUS OF VARIATIONS. 381 the equation between iV, P, Q^ is ^^ dP d'Q ^ or N--Up-^4)=0. ax \ ax) dQ _ 2p ^py But ^-^ = -,-f^^ + dx 1+/ ' (!+/)•'' or , ^sin0 + a'^^cos(<^-6') = O (2); also {^ + xY + / = d' (3). From (1) and (2) \ '^d'co%\<^-e)] \dt or, since sin^ = -, tanrf> = ^ = w, a dx ^^ X-W Also, from (3) " dx^ ' {d'-Tf-y 1851.] CALCIJLUS OF VAKIATIONS. 383 dt 1 + ^•y [[d^-yy+pyY\ d^ 1 c[a-y ^a{(«^-j/?+P3/F+^4>?^^^. or Here V contains only y and p^ .-. V-Pp = C, [{(« -y npy] +f^p^- [{{a^-yy+pyY^jc^^i + - (« -3^ ) -o, {a^-yy{{^-yy+py] + 9. ^.f)i [[(^a^^,jy-+pyr+kYY- = o ; 1- or {C'k-y)p = {d^-y% C = |l - (^)y'; .-. C'k sm" '^ + [p" - ff = x^ 0", the required equation to the curve : the constants C", C" are to be detemilned by the two given positions of the rod which give two points through Avhich the curve must pass. The curve is independent of the angular velocity of projection. ( 384 APPENDIX. The following problem in Astronomy was set in 1848. If a rectangular court be enclosed within a wall of given height, and one of its sides be inclined at an angle of 30° to the meridian, detennine the breadths of the shadows of the walls on a given day at noon, and the portions of the courts and walls which will be enveloped in the shadow, the latitude being 52° 30' north, and the Sun's declination on the given day 7° 30' north. By referring to the problem on p. 64, we see that here = 30° and = latitude — Sun's declination = 45°, r. a = \h^ ^ " 2" ^'' Let ?j, ?2 ^6 the lengths of the walls, whose shadows are respectively of the breadth «, 5, the area of the courts enveloped in shade will be \a + [l^ — a) Z>, or l^a + I J) — ah ; and the shaded parts of the walls the whole of the two walls, and two triangles \ha^ \hh of the other two. The following solution of the problem on p, 148, is due to Mr. Gaskin. Let TP, TQ (fig. 126) be the two given tangents, take the line AB as axis of cc, and let OP' Q be the chord of contact of any conic touching TP, TQ^ and passing through A^ B. Take APPENDIX, 386 as origin, and let '7 — ) ''7' — a a be the equations to TP, TQ respectively ; also let OA = a, OB = /3. Let the equation to OP' Q' he 7/ = mx, then that to the conic will be , Hence, putting ?/ = 0, we get o,. 1, /i + i)l + (_L,_'|!) = o (.), a; V« aj X \aa \J ^ '^ the roots of which equation are - , p 5 whence we see that 1111 a p a a or the line OPQ is divided harmonically in -4, i?, whence is one of the foci of involution of the system of points P, Q^ A^ P, so that the chords of contact of all conies touching PP, TQ and passing tlu'ough -4, P, cut AB in one of the points 0, 0', if 0' be the other focus of involution. Now, in order that (1) may represent a rectangular hyperbola, the sum of the coefficients of x'' and y'' must = ; hence -— - m^ + 777 - 1=0. an 00 But by (2), 1 _ 1 m' a/3 aa \ Combining these equations, we get aa 00 J \aa ap giving two values for «i, equal and of opposite signs^ so that there can be constructed two pair of rectangular hyperbolae CC 386 SOLUTIONS OF SENATE-HOUSE PROBLEMS. whose chords of contact meet in one of the foci of invohition, and are equally inclined to AB. The relation between the four lines Ox, Oy, Op, Oq, in the problem on p. 158, may be expressed thus: Ox and Oy each bisect the lines between 0}), Oq parallel to the other. For let the equations to Op, Oq, refen-ed to Ox, Oy as axes, be 2/ = mx, y = m'x. Then if Oa = a, Ob = b, Od = a', Ob' = b', the equations to ah, db' are - +1 =1 a^b ' X y a b '.+'Tr=l; whence, if (xy) be the point Q of intersection of these lines, \a a J But Q lies on the line Oq, or y = mx, ^b b'^ ■'' («-^'Ka"a')='^^^"^'K^-T')' 11 /I lA ^ a a \b bj The condition that the point of intersection of this line lies on Op or y = mx, is derived from this equation by interchanging a, a!, and writing m for m, 1 1 , ? + m a a ■(M)-- Hence m = — m, which expresses the above relation between Ox, Oy, Op, Oq. The same thing may be proved geometrically by making any one of the points a, b, a, or b', remove to an infinite distance. APPENDIX. 387 The following statical Problems set in 1850 have been omitted. 1. A heavy rod, whose weight is TV, rests upon a fulcrum at its middle point, when loaded at one end with a weight W, the density at any point of the rod at the distance x from a TTtJC certain point in it varies as sin — , a being the length of the rod : find the ratio of W to TF', and determine at which point the density is zero when this ratio is the greatest possible. Let c be the distance from the centre of the rod of the point where the density is zero, p the density at the point x = ^a. The conditions of the problem give r^ . TTic , r*-"-' . irx , p sm — ax ■\- [ p sm — dx = W, ^ Jo ^ pa (^ (tt 7rc\ , /tt ttcX) „_ or^|l-eos(- + -) + l-cos(j--)| = P^'; w W ••• P'^-^ (•)• Also taking moments about the fulcrum which is at the middle point of the rod, r^"'' . . irx . r . , . TTcc Pi (ic — c) sm — ax = p I [c- x) sm — Jc ti '^J^^'a TTX , ax + p \ (c + ic) sm — ax -\- W - /•*'*' . irx J ( [*-"'' . irx J ^ fi"-' . irx J \ ^.., a or pi xsm — ax — pel I sm — ax + I sm — ax ] = W - T-T- f . TTX , ax TTX a^ . TTX JN ow ixsm — ax = cos 1 — 5 sm \- c: J a TT a TT a ^ . TTX , a . ire (a (a icsm — rta; = -sm — \-z-\-c-\-\- -c a IT « 2 V2 a . TTC = — sm — TT a 888 SOLUTIONS OF SENATE-HOUSE PROBLEMS. and I sin — ax + l sin — ax = — , as shewn in (1 a'' . TTC 2 oca ,,,, a r /IN TT W fa' . TTC 2ac\ „, a or from (1 - — - sin \ = W -; ' 2 a Vtt a tt J 2 a sin 2c a This ratio = oo when c = ; it has a maximum value when . TTC . a sm 2c is a mimmum ; a or, clIfFerentlating with respect to c, TTC ^ ^ TT COS 2 = 0, a ire COS — = a TT which determines the value of c. 2. Portions are cut from an ellipsoid by planes which are parallel and equidistant from the centre ; if -ct be the length of a pei-pendicular from the centre upon either plane, and Z, m, ?i, the cosines of the angles which it makes with the axes, shew that the remainder will rest when placed with a section on a horizontal plane, if 1 r m' n' — = or > -^ + -p- + ^ , w a o c a, &, c, being the axes of the ellipsoid; and express the con- dition that ]j such solids, when placed on each other with their sections coincident, and their centres in a line inclined to the vertical, shall not fall over. The one portion will rest with a section upon a horizontal plane if the vertical line drawn through its centre of gravity, APPENDIX. 389 which is the centre of the ellipsoid, fall within the section; i.e. if zj be equal or less than the radius vector (r) of the ellipsoid drawn in the same direction, or since if -^ = or > -^ = or > -, + 77 + -^ . «j T a b c Wlien there are ^; such solids placed on each other as above described, the height of the centre of gravity above the plane will be jjicr ; and, if p be the distance of the foot of the per- pendicular drawn from the centre of gravity on the horizontal plane from the centre of the section, p the same distance when there is but one solid, we have p = i^p : the condition that the p solids shall not fall over is, that pp shall be equal to or less than the radius vector of the section through the foot of the said perpendicular. The equation to the cutting plane is Ix + my + nz = vj (l) ; and if a, /8, 7, be the coordinates of the centre of the section, a, /8, 7, are subject to the conditions (see Gregory's Solid Geometry^ Art. 121) la + myS -f W7 = ■z^, a /9 7 -ra- , s and ciH U'm c'n a'l' + H'm' + c^i' The coordinates of the foot of the perpendicular on (1) are IzTj v/jCT, w-ct; hence the equations to the radius vector of the section through this foot are y - ^ _ g - 7 _ = p,r suppose (3), Izs — o. wa — ^ 7*^ — 7 where r is the distance of the point {xyz) fi'om (a/37). If we substitute these values of ic, y, 2, in the equation ta the ellipsoid x' f _^ g' _ 1 i? + 6^ + c^-*' 390 SOLUTIONS OF SENATE-HOUSE PROBLEMS. we shall find the length of the radius vector r of the section through the foot of the perpendicular, {{1^ - a) fir -\- aY {(yn^ - /3) fir + /3Y _^ {(»^ - 7) Z^^ + tF _ 1 7? ^ h' + 6' '~^' the roots of this equation are equal ; Now, from equations (2), la 1)1/3 7i »' % \^ ---- MttaiA VrMi-4r. liHa 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewals only: Tel. No. 642-3405 Renewals may be made 4 days priod to date due. Renewed books are subject to immediate recall. ^0610,972 8 4 fiCC'D LD MA R 7 73 -2 PM 3 7 AUG 2 19751 9 .^PP 18 1"^^'' EEC CIB.MAR 29 77 MAR 2 8 1984 I ;^ 3 KEC, CIR. HAR ^ 9 84 (N8837sl0)476-A.32 ^"'^^"^^^^^j^f '°""* NEW CLASS cDbmniDflT A Commentary on the Works of Aiistophanes. By W. G. 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