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Wa-tlj 3g?J,( 



Saciiaiti College i/iitatj 

TbQ lAWrmce SeTentfflo Solioo'. 



TRANSFERRED TO 
CABOT SCIENCE LIBRARY 



o 

A COMPANION 



TO 



WEIGLEY'S COLLECTION 



OP 



EXAMPLES AND PROBLEMS 

BEING ILLUSTRATIONS OF MATHEMATICAL PROCESSES 

AND METHODS OF SOLUTION. 



JOHN PLATTS, 

HEAD MASTER OF THE GOYERNMENT COLLEGE, BBKABES, 



AND THE 

Rev. ALFRED WRIGLEY, M.A., F.R.A.S., &c. 

OF ST. JOHN'S COLLEGE, CAMBRIDGE ; 
AND THE BOYAL MIUTARY COLLEGE, ADDIBOOMBB. 



CAMBRIDGE: 
DEIGHTON, BELL, AND CO. 

LONDON: BELL AND DALDY. 
186L 



Mccth 1,%%M 



ri\i.j.::.^^.jro.] Irom 



The Lawrenco Scjent-ific Schooi. 



Q February, G3S. 



PRINTED BY C. J. CLAY, M.A. 
AT THE UNIVEBSITY PBESS. 






PREFACE, 



The object of this work is, not to furnish to Wrigley's 
Collection a Key in which every Example and Problem is 
worked out and solved, but to exemplify Mathematical 
Operations and Methods of Solution by numerous illus- 
trations. A Key, it is judged inexpedient to publish, on 
the ground that it would destroy or impair the value of 
the Collection as an Exercise book. But a work, in which 
principles are illustrated and processes set forth, such as 
would serve to solve every question in the Collection, will, 
it is anticipated, prove convenient for the Tutor and espe- 
cially useful for the Student. 

The Arithmetic and the beginning of the Algebra in 
the Collection are left untouched. 

Any list of Errata^ sent to the Publishers, would be 
gratefully received. 



September 21, 1861. 



CONTENTS. 



The following are the numbers of the Examples and Problems in the Fourth 
and subsequent Editions of the Collection discussed in this work. 



Ex.35 

••• 86 

... 87 

«•• oo 

••• 89 

... 40 

... 41 

... 42 

... 43 
• . • 44 
... 45 
... 40 
... 47 
... 48 
... 49 
... 50 
... 51 
... 52 
... 53 

... 54 
... 55 
>•• 56 
... 57 
... 58 
... 59 
... 60 
... 61 
•i. 62 
... 63 
... 70 
... 71 
... 72 



ALGEBRA. 

O. o. 21. 22. 2^ ......... 

17- 19- 34- * - 

3- 9- 13 

20. 23. 25 

6. la 17. 21. 22 

9. 17. 48. 52. 54. 55. 57. 61. 62. 63. 64. 65. 67. 68. 69. . 

9- lo- *S 

27. 3a 40. 41. 42. 43. 44. 45. 46. 47. 4a 49. 56. 57. 58. 59. 

62. 63. 64. 65. 66. 

16. 18. 



19. 



I. 4. 9. II. . 
la II. 12. 18. 



42. 49- 

I. a 

I. 5. IT. 15. 19. 28. 32. 37. 40. 41. 44* . 

I. a 9 

7. 9. 17 

9* ^4 •.*....... 

4. 10. 14. 20. 23. 26. 28. 30 

43« 64. 9i. lOi. 14. 17. 22. 23. 28a. 30. 3 1 1. 313. 



32. 35- 



38. 39- 43. 45. 



4. a 
4. II, 

2. s. 



12. 



I. 4. 8 






I. 4. 6 

I, c 


• • • • • • • • • • • 

I. 12 


I. 3. c 


I. 4. 5. 9. 11. 13. 14. 16. 18. 19. 20. . 

I. 3. 9. II. 14. 16. 18. . 

I. 5 


• • • 
• • • ■ 


• 

• 


I. 7. 13 


•♦ 




T. 8. 12. 15. 18. 20. 22. ... 


• p • 


■ 



PAGK 
1—3 

3. 4 

4, 5 

6, r 

8, 9 
9—14 
15 

16-27 
27, 28 
29, 30 
31, 32 

32 

33 
34—36 

37 
38, 39 

39 
40—42 

42—47 
48, 49 
50, 51 

52 
53, 54 
54-56 

57 

68, 59 
59—63 
63-66 
65, 66 

66 
66—68 



VI 



CONTENTS. 



THEORY OF EQUATIONS. 



Ex. 1 

... 8 

... 8 

... 4 

... 6 

... 6 



... 7 

.1. o 
... if 

... 10 
... 11 
... 12 
... 18 
... 14 
... 16 
... 16 
... 17 



5- 9- 
I. 3- 
3. 4. 

2. II 

I. II 

I. 2. 



II. 



II. 13. 15. 20. 

15- . . 



14. 17. 19. 



J. 

2. 4. 

I. 6. 

1. S' 

2. 9. 
I. II 
I. 2. 
I. 4. 
I. 4. 



7. 



10. 14. 



12. 13. 15. 



9. 12. 

1. 5. 14. 15. 19. 20. 

2. 6. II. 14. 17. 18. . 



PAGE 

69 

G9. 70 

70, 71 

71, 72 
73, 74 

75 

15 

16 

77 
78—80 

80 

81 
82—84 

84 
84—87 
87—91 
92—95 



GEOMETRY. 

I. 2. 4. 6. 9. 13. 17. 25. 29. 32. 35. 36. 37. 40. 42. 46. 48. 50. 
B3' 55- 57- to. 62. 63. 65. 67. 68. 70. 73. 75. 78. 80. 84. 87. 
89. 91. 94. 95. 98. loi. 104. 105. 107. 109. 113. 115. 117. 

118. 119. 121. 124. 126. 129. 131. 134. 139 



96—118 



MENSURATION. 



Ex. 1 

••• « 

... o 

... 4 

... 6 

... 

... 7 

• . • O 
• • • iJ 

... 10 

... 11 

... 12 
.. 18 



6. 8. 13. . . 
8. II. 13. . 
2. 3. 9. II. 12. 



9. 13. 14. 
4. 6. 7. 14. 



3. 9. 10 



I. 2. 10. 14. 
6. 



I. 7. . . 

4. . . . 

5. 9. 13. 15. 17. 



• • 



• • • • 



• • • • 



7. 9. . 
... 14 2. 5. 6. 
...16 3. 10. 13. 18. 



■ ft 



119 

119, 120 

120, 121 

121, 122 

122, 123 
124 

124—126 

126 

127 

127 

128 
128—130 

131 

132 

133 



PLANE TRIGONOMETRY. 

Ex. 1 I. 4. 6. 9 

... 2 3. 4. II. 12. 13. 15. 18. 21. 32. 34. 40. 42. 44. 48. 

... 8 I. 2. 3 

... 4 I. 2. 3. 7 

... 5 I. 2. 5 

... 6 I. 3. 4. 10 



134 

135—137 

138 
138, 139 

139 
140, 141 



■ • 



■ • • 



• •• 



• • • 



• • • 



• • • 



CONTENTS. Vll 

PAGE 

Ex. 7 I. 3. 7 1*2 

8 I. 6. 9. II . . . . • . . 143 

9 I. 2. 6. 9 lU 

10 3. 5- 8- 10 145, 146 

... 11 I. 2. 5. 6. 7. 10. II. 13 146—148 

... 12 2. 4. 6 148, 149 

... 18 2. 7. 8 150 

... 14 3. 5. 9. 143. 15. 17. 19. 2o 150—152 

... 15 5. 7. 8. 9. 12. 13. 20. 24. 25. 26. 27. 28. ss. 35. . . . 152^158 

... 16 I. 2. 4. 7. 9 159, 160 

... 17 I. 8. 15. 17. 19 160—162 

... 18 I. 3. 17. 18. 20. 23. 26. 31. 36 163—168 

19 I. 3. 5. 7. 10. II. . . . % 168—170 

80 I. 2. 3. 6. 10. II. 14. 17. 20. 22. 23 170—174 

... 21 I. 4. 7. 9 17&-177 

... 22 3- S 177—178 

... 28 I. 5 178 

APPLICATION OF ALGEBRA TO GEOMETRY. 

I. 2. 7. 9. II. 13. 14. 17. 20. 25. 26. 31. 32. 36. 41. 42. 43. 51. 

B3' 54- 55- 5^ 179—190 

ANALYTICAL GEOMETRY AND CONIC SECTIONS. 

Ex. 1 I. 5. 9. 13. 18. 22. 251. 254. 27. 30. 36. 39 191—196 

... 2 I. 6. II. 14. 21. 23. 27. 29 196—199 

... 3 I. 4« 7- 9- "• T3. 16. 19. 22. 24. 29. S3' 35- 4i- 48. 51. 56. 

64. 66. 68 199—209 

... 4 1. 3. 6. 8. 1 1. 13. 15. 18. 21. 26. 28. 29. 32. S3' 38. 42. 43. 

46. 48. 50. 51. 53. 59. 60. 61. 66. 70. 75. 78. 82. 83. 91. 92. 210—225 

5 I. 6. 9. II. 14. 19. 21. 27. 30. 32. S3 21&— 230 

6 I. 3. 5. 7. 13. 16. 18. 21. 22. 25. 28 231-236 

72 236 

8 3. 4. 6. 9. 12. 13. 16. 18. 19. 21. 22 237 — ^242 

9 I. 6. 9 243 245 



STATICS. 

Ex. 1 I. 3. S- 6. 8. 10. II. 15. 17. 20 246—250 

... 2 I. 6. 9. 12. 17. 19 251—263 

... 8 I. 5. 8. 9. IT. 12. 14. 17. 21. 23. 27. 31. 32. 34. 37. 38. 41. 44. 263—264 

... 4 I. 3 265, 266 

... 5 I. 5. 6. 7. 12. 17. 19. 22. 23. 26. 32. 33 266—274 

... 6 I. 5. 7. 8. 10. 12. 18. 22. 24. 26. 28. 29. 32. 34. 35. 36. 38. 39. 42. 275—284 

... 7 I. 2. 5. 9. 12. 13, IS 285—288 

... 8 1 289 

... 9 I. 2. 3. 5. 7. 10 289—292 

... 10 I. 5. 8. II. 14. 17. 20. 23. 27. 30 292—296 

... 11 J. 4. 7 296, 297 

... 12 1. 8. II 297—299 



• • B 



■ • ■ 



VIU CONTENTS. 

PAGE 

Ex. 18 3- 5- 7- • • • • • • • 299, 300 

... 14 I. 5 301 

... 15 I. 3. 4. 8. 9. II 302— ^J06 

DYNAMICS. 

Ex. 1 I. 3. 7. 10. II. 16. 18. 20. 23. 24, 26. 28. 29. 30. 31, 33. 36. 

41. 43. 44. 45. 46 307 — 319 

I. 3. 5. 8. 10. 14. 15. 16. 17. 18. 20. 22. 27. 29. 30. . 320—325 
I. 5. 6. 9. II. 13. 15. 17. 19. 21. 23. 25. 26. 27. 29. 31. 33. 

35. 37 326-335 

I. 2. 4. 5. 7. 9. II. 13. 15. 18. 19. 22. 24. 26. . . . 336—311 
I. 3. 7. 9. II. 14. 17. 18. 21. 25. 27. 3a 31. 33. 36. 37. 39. 

40. 42. 43. 45. 46. 48. 5a 52. 54. 343—357 

I. 2. 3. 7. 10. 13. 16. 19. 21. 23 358—363 

I. 4. 6. 9. 13. 15. 17. 19. 21. 25. 28. 30 364—370 

I. 2. 4. 5. 6. 8. la 12 373 — 377 

HYDBOSTATICS. 

I. 2. 7. 9. 12. 13. 17. 18. 21. 24. 25. 28. 33. 36. 38. 40. 42. 44. 379—386 

I. 2. 4. 7. la 15. 17 , . . . 387 — 391 

I. 2. 4. 5. 6. 7. 9. la II. 13. 15. 17. 19. . . . . 392 — KH 

I. 3 404, 405 

I. 2. 4. 7. 9. la 13. 16. 20, 21. 23. 29. 32. 35. 36. 38. . 405—412 

I. 4. 7. 9. II. 15. 18. . . 412 — 418 

I. 2. 6. 9. la 12. 14. 16. 17. 418—423 

I* 3. 5* 7. 9. II. 14. 16. 18. 19. 21. 22. 23. 25. 26. 27. 3a 423—432 

HYDBODYKAMICS. 

Ex. 9 I. 3. 6. 9. 12. 15. 19. 22. 25. 28. 31. 34. 36. 37. 38. 40. 42. 433—443 

... 10 I. 3. 6. 8. II. 12 444—448 





2 




8 




4 




5 




6 




7 




8 



Ex. 


1 




2 




8 




4 




6 




6 




7 




8 



ERRATA. 

PAQB 

89 (14) Change the sign in X^ thus, •\-'jx*-Sx-4; hence^ the roots are real, 

and lie as follows { - 2, - i } ; { - i, 0} ; { i, 2} ; { i, 2}. 

90 (19) The signs in X^, Xg, Xij should all be changed ; hence there will be 

2 pairs of imaginary roots, and the 3 real roots will be found to lie thus 
{-«,-!}; {I, 2}; {I, 2}. 

91 (20) See the Appendix. 
128 Insert Bx, 12 before (5). 

184 Last line, for d read d' . 

185 Fifth line, for vdd read vdd', 

270 Insert 19. at the beginning of the line — Let the axis, &c. 

297 The problem (7) belongs to the Wheel and Axle, and should precede Ex, 12. 

347 In probs. 25 and 27 for tan a read tan a. 



ALGEBRA. 



I. Simple Equations. 

Ex. 86. 

aoB — l^ iaair — h „ , 
T ^ — ' = c, find x; 

{cKcf + h 



(aa?)* = 6 + ; 



c— I 

s \a 



(V 






8. -(a + a?)* + - (a + a?)*«=T-.» > find x\ 



a * ' 0? 






3 aaj' 



or (a + aj)' = --T-; 

f 
a 05 

J* 



2 SIMPL£ EQUATIONS. 

21. oaj + I = ' — ^ ^^, , find a?, 

a + (oj + a»)* 

Then -^=_^ 4.x. 

Transpose i and square, 

/ax — i V __ g* 



/ oa; 4- I Y _ £ , 
W-i/ "a" ' 



400; X 



{ax - i)» a» • 
oa? - I = + V4a^ = ± riaVa ; 
/, « =a ± aa* + a""'. 

22 a^-{a-{a^-ax )^]^ , . , 

Dividendo and Componendo, 

{a-{a-a^x)^]^ 1-6 

o ( \l + bjj (i + b)*' 
i+a;+(aaj+a;»)* (a + a;)*-** 



. {(a + X)* - «*}' ^ ^ (3 + a;)* + g* 



(g + a;)^ - g* 
(a + a)* + »* 



= a* 



SIMnS XQUATXOKB. 



Componendo and Dividendo, 



. a^/ i+A * i^. , 4tt* 
••«"Vi_o*/ ' (I -a*)'' 



aa' 



8 



n. SntVLTARBOUS EQUATIONS. 



Ex. 86. 



17. 



3« 5y 9 
I I _ I 

5» ^~4> 



(I) 
(2) 



find ^ and y. 



X 



Multiply (i) by 5, {%) by 3, and subtract ; 

15 ' 3^ 5x13 65 6s- 

XT I 13 1 I 13 48^13 

5aj 16 X iia ' 3y 4 ^9^ ^9^ 

64 
^ 35 35 



19. 



ah c 

y X V 

u o c 



y find X and y. 



r+S-^V^'' ^^>- 

Multiply (i) by , {2) by r , and subtract ; 



SIMULTANEOUS EQUATIONS. 

abclab '\- ac --hi^ . .| , abclac — ab^hc) 
•*• «==— Jxi T-i — Trii Bimilarly , y = -TT^^ t-z — rr-f 



24, 



y*-(M-a?)*=:(y-aj)* (i) 

3(!20-aj)*=!i(y^aj)* ....{2) 



find a; and ^. 



Then 1/* = f- 



+ i)(y-a?)*; 



3. 



.-. 2505 =i6y; 
•*. 9a; = 16 (y — a?). 

From(ii), 9(^o-aj) = 4(y.aj) = ^; 

4 

.•. 4 X 20 — 4a? 5= a? ; 

/, aj = — = 10, andy = a5. 
5 

Ex, 87. 

3a?-7yH-4« = i | (i) 

— 5^ + 9y— « = 22> (2), find a?, y, and z. 
x-%y+ = J (3) 

Adding (2) and (3) we get, -4aj + 7^ = 22 (4), 

from (i) 33? - 7y + 45 = I, 
from (3) 4aj — 8y + 45 = o ; 

.-. -x+y =1, or, -4a? + 4y = 4, 

from (4) — 4a; + 7y = a2; 



.-. aj= y-i = 5, 
and « = 2y — a?= 7. 



and y=^6; 



9. 



aa; + « — 4 3y — 6«+i a? — 2 

13 13 4 ' 

3a;-ay + 5 4a!-5y + yg a . gy-gz+g 
5 7 ^ 6 



X 



--y + 3« = a. 



(I) 
(3) 



SIMULTANEOUS EQUATIONS. 5 

From (i), {2,6x + 13^ - 52) + (36y — 7a« + la) = 390? - ys, 

whence 13a: — 36^ + 59^ = 38, 
from (3), 13a?- I i7y + 351^ = 334 ; 

.'. a9a« — 81^=196 (4). 

From (a), 

(irj6a?— 84^+210)— (i^aj-l5qy 4-aio«) = 60+ (io5y—3i5«4-aio), 

whence 6aj — 39^ + 105;? = 60 ; 
or aoj— i3y+ 35« = ao, 
and from (3), 2x— i8y + 54s = 36 ; 

••• i9«-5y=i6; 

.% 19 X 2g2z — 1460^ = 467a, 
and from (4), 19 x 4ig2z - 15393^ = 3734 ; 

••• 79^=948, orysia; 
... j5- —=^4i and 0?= i84-9y — !i7«= i8. 

13. ay = 3(a? + y), xz^8{x + z), 7y« = 9(y + «), find a?,y,«. 

Divide the first by 3a?y, 

then - + -=- (i) 

^ y 3 

similarly ^+j = | {^h 

and ' + - = 2 (3). 

if z g ^^' 

Add and divide by 2, 

x^y^z^2\3^S^9)' 
From this subtract (3), 

then - = -(- + 3--) = - — ; .*. a=-6^. 
a? aVs 8 9/ 144' ^ 

Similarly, 

y iaV3 » 9/ 144 



subtract (2), i=ig-| + ^) = ^; /. y = 27r. 



«.M(-M-9=^' ■■■'-> 



41 • 



6 SIMULTANSOUS EQUATIONS. 



20. Let « s the time A took to do the work, 

y= B 

n J 



« = C 



and let w = the work. 
Then-=^'8daaywork; - = 5'8; -= O's, 



and sinoe B and G w orking together can perfonn m times as 
much per daj as Ay 

— -— + -; .•• — = - + - (i). 

X y z X y z ^ ' 



Simikrly ^ = i + i (4), 

aiid£ = i + i (3). 



From (i) and (a), = ;. 

^ ' ^ ' X y y X 

X y 

a; m + I time taken by A 

or — ss — — rs ; rL ^ 

y n + 1 time taken by £ * 

q. ., , a? _ m+i _ time taken by j1 
"" p + I "" time taken by G ' 

n y _ n + I time taken by B 
« ""2^ -M "~ time taken by G ' 

Agamfrom(i), w = - + - = — — +— -; 

y 2? w + i p + i 

I I w I 

"w + i 2? + i w + i m + i^ 

III 

/. + h = I, 

m+ 1 n+i p + 1 

23. Let the length of the conrse be x yards, and the dura- 
tion of the race be t minutes. 

Then A rmis 05 + 90 yards, whilst B nms a? - 90 yards, and 
A would run zx-i-^x yards, whilst J? would run !W? — -? a?. 



SIMULTANEOUS EQUATIOHS. 

I 

QiCti -4" "■ US 

TT oj + oo A*& rate of rauning 6 12 + 1 

Hence ^- ~ -^s ^ — — — = . 

aj — 90 -Bs I lij — I* 

o 

.*• — = — ; /.«?== 1080 yaxdfl, 
90 I "^ 

. . ^^s time of running _ ^s rate of running 
^*^' B'a "A'a ' 

I 

t O II 



o 

Dividendo, - = — 
3 « 



.-. ^ = ^ = 16^ minutes* 

25. Let X = the No. of bushels of wheat, 

y= • barley, 

«= rye, 

then a? + y + « = ioo (i). 

The value in £'s is therefore 

- + ^+^ = ioox-.. (i), 

5 60 20 6 ^ '^ 

and2«x^ + g+(aj + io)i:=g(aj+y4-^« + io)...(3), 

from (2), i%x + 7y + 9« = icx)o, 
from (i), 9a? + 9y + 90 = 900 ; 

.-. 3a?-2y=ioo (4). 

Again from (3), 

i8« + 7y + lija? + lao = iap+ loy + 20z + 100 ; 
.•. aa?-3y-ij«=-iZo, 
from (i), ^o; + 2;^ + 2s;^ B aoo ; 

.". 40?— y=i8o, or 8a; — rjy = 36o, 
from (4), 3a; — !jy = 100 ; 

.'. 5a? =260, a? = 5:z; 



8 SIMULTANEOUS EQUATIONS. 

.\ y = 4«- 180 = 28, 

« = IOO-(iB+y)=ao. 

Ex. 89. 



whence r^ =— ^ 



or 



••• 3«*+9 = ^«*+i8; 
.'. a* = 9, and ^=±3. 

10. 35:1^ +^±7 = III! fi^d^. 

, 5a^ + y ^ 5i-gg 35-ga? _i6 , 
••5a»-7 9 99' 

Componendo and Dividendo, ^ = ^ . 

7 7' 



17. 



(I + a!)» (i - a) 



i = «, 



21. 



i + o; ^ i-a; "' 
. ^-^■¥0!' , i + x + ai' 

" ~T+^ — "'^ i-x -+«'=«» 

I — 335 1 + 30! 

whence aj = + ff^H^V . 

(i -«)»+! (l + xV-I 



QUADRATIC EQUATIONS. I 

Then 

{i + x)^-i ^ {i'x)^-i ^ (i--a?)^+i ^ (i + a?)^+i ^^. 
(i_a.)*+i (i-a?)*-i (i + a?)*-j (i + a?)*+i 

. (i--a^)^4-i-{(i + a;)*+(i-a?)^} 

(i-a^)*+i + f(i + a;)*+(i-a;)*} _^. 

X ' 

.\ z{{i + x)^+{i-xf}^ax, 

64 (i - aj") = (a'aj» - 8)» = a^a:*- i6a*i»» + 64; 
/. i6ar*(a'-4)=a^ic^; 

22. {(i + a?)' - ax}^ + {(i - a?)» + okk}* = a?. 

Transpose, and square, 
I + 2aJ + ar» - CMC = aj' + I - aa? + a^ + aaj - aa? { ( I - aj) » + oaj j * ; 

/. 4aj - a?" - 2,ax = - 2aj {(i - a?)' + ax}^. 

Divide by a;, and square, 
ar» + 4aa? + 4a* - 8a; - i6a 4- 16 = 4 - 8a? + 4^* + 4ax; 

.'. 3a?» = 4o" + ia-i6a; 

it / . K 4 (« — i) (« — ^) 
... af^'^{a^ + 3-4^)=— f ^J 



.■...,. {(---)j-^3) }^ 



Ex. 40. 

9 a?' — 2a.-|--2- = o; 

/.a; ^^"T-j^ j^ ^Q jQQ? 



10 QUADBATIC EQUATIONS. 



• 
• 


05- 


lo' 
3 


=± 


6 

lO' 

3 


• • 


X 


=»' 


or 


lo' 



17. 170?' + 190? = 1848, findoj. 

By the Hindu method, 

(i) Multiply the equation by 4 x coefficient of a^. 

» 

{7,) Add to both sides the square of the coefficient of x. 
Then 
4 X I7*aj* + 4 X 17 X 19a? + 19* = 4 X 17 X 1848 + 19* = 17,6025 ; 

.-. 1^ X 17a? + 19 = ± 355, 

.= ±3^==9j.,or-ii. 

48. a?+(aj»-aaj + 6*)* = - + 6; 

/. (aj'-aaj + 5*)-a(aj»-6K»4-ft')* = i*-oJ; 

/. (aj*-aaj + 6»)-o(aj»-aa? + &')*+— =S*-a& + ~; 

4 4 

whence o^ — aar + 6* = J", or (a -ft)*. 

(O - a^-aa? + ft» = ft»; 

« . X ^ twu f 
.*. 07 = a, o. 

(2^ a:* - oa? + ft» = a* - aoft 4- 1' ; 

. . a" c;a' - 8aft 
.'. 05* — aa?4- — =s^^ ; 

4 4 






QUADRATIC EQUATIOHB. 



11 



52. 



5lx + (ix-i)*) 



.♦. {«+ (aas -!)*}-{(« -!)»}* = 2 (losr)*; 

5 

1 8a! - /2x\i 

■••T--(f)*' 

a 8 

54. caj = {(I + 0?)*- 1} . {(I - aj)*+ i} ; 

.-. c{(H-a;)-i} = {(i + aj)*-i}.{(i-aj)* + i}....(i); 
.-. c{(i + a?)*+i} = (i-a?)*+i; 
.•• c(i + a?)*-(i-a?)*«i-c; 

/. c*(i + a?) + (i— a?)-«c (i— aj")*s=i-i;c + c"; 

."• !ic (i -(»')*= ^c + (c"-i) a?; 
.". 4c" — 4c'a^ = 4c'+c*aj»-ijc*iB* + a:* + 4c(c*-i)a?; 
.-. aj'{c*4-2c*+ i} + 4c(c'- i)a? = o; 

/. ar = o, and , 



55. 



^+-- ==-+-) — , find X. 
S^ 2,x V3 4/ 3 



^ iiH^^HihiH'l^v 



12 .QUADRATIC £QU ANIONS. 

••V64 S 4af) 3V8 W^9 ' 



"8 «aj "" 3 * 

whence as = 6, or — . 

3 



57. 



/. a:* — i8aj" = 4a? — 48; 
.*. ar* — i^r" 4- 49 = 435* + 4» + 1 ; 
/. 05"- 7 33 ± (aaj + i); 
whence a? = 4, — ii, ^ i ± V7« 

61, 40^ + 12a? (i + aj)*=: 27 (i + a?), find x. 
Then 4^ + 40? x 3 (i + aj)*+ 9 (i + a;) = 36 (i + a?) ; 

/. riaj= (± 6- 3) (1 + a?)*= 3 (i + a?)* or - 9(1 + 0?)*; 

whence a? = 3, -2 |(9±V97), 

62, iiaj»+ (a^+ 9)* =* aj*- 9. 

Then (a;" + 9) + (af+9)* + i = a?*-oj» + -; 

4 4 

whence a.= ±(3±^)*, ±{^-^)\ 

63, (a + a;)* + 4 (a - a?)*- 5 (a* - a?)* = o, 

•••f^=(M)*-**°"' 

• . a? — — ^ Cv* ^~ o. 

^5 



<iUADBATIC EQUATIONS. 13 



64. v + r = 2ar. 



^"^ + * 

(a - xr {a + xy 

Sqnarmg, ^^ ^+:z (a*-a^r+ ^^ ^ = 4«; 

^ ° a — 0? ^ ' a + x ^^ 

{a + xY + (a - a?)* , , .. * 

.-. ^ ^j--^ ^=4^»-2(a--aj»)*; 

.'• «^ + 3«^ = ^«' - ^<^ - (a* - a?*)*. 

.\ (a* - aj*)^ =» a« - 5aaj* = 5a (a" - a:*) - 4a5. 

Assume a^-^af^ y\ 
then y3 ^ ^^y« -- ^ 4^^ 

or y!*(y-o)=4a(y'-o»); 

•". y — a==o, or y = a; 
/. aj = o. 

Again, y» = 4ay + 4a'; 

.*. y = 2a (i ± V^) ; /. a; = + a (± 8 V^ - ii)*. 

66. (I +xf + (I -a?)' = (i -^ a:*)*. 

Dividing by (i +aj) , and transposing, we have 

/I — a3\ $ / I —ayX s _ 

Completing the square, (^)' - (i^)' +i = - J, 

I -a; fi, , / v)^ {i±V^}* 

whence « . iil^l^; = -i±^ = ± i VT^. 



14 QUADRATIC EQUATIONS. 

(i+a?)* (i — oj)3 
mi ^ i-a; + a?* , i + a? + aj" 

" (i-a!)» (i + a;)» *' 

4a;* _ c— a 

•■• (i-.aj»)'~ 3 • 
Componendo, il±^ = ^. 

Whence a. = f/^=lM^ ^ (c-^4±^^WTT)H 

68. (a + ajj* + (a - a?)* = A; 

.-. !za +4** (a» - a?*)* - 2 (a* - a?»)* = A^. 

CoBxpleting the sqwe, 

(a' - af)* - aA» (a» -a?f+h* = a+-, 

whence (o» - tB»)* - A» = ± (a + — ]* ; 
.-.(«'-«•) = |a'± (a + 1)*}'; 

and « = ±ra'-|A»±(a + |)W 

69. aa; (i - a^)* = a (i + a5<), find 35. 
■ Thena(l-a.)*=«(^ + a.). . 



QUADBATIC EQUATIONS, 15 

Square and transpose, 

Subtract 4a' jfrom both sides, and complete the square. 

Whence a = ± - £- i ± (i - a«)* ± {a ± 3 (i - a*)*}*]* 

Ex. 41. 

4 4 
Whence the values of a? are found to be 

10. aj*-aj» + ^ar»-aj + i = o,- 

4 

4 4 

4 
whence •aj'+i— - = ±aj; 

••• ^ = J{3±V=T}, or = i(^i±V=:T5). 



15. 






,\ a? = - 1 IS a root 



16 SIMULTANEOUS EQUATIONS* 

Also I —x + a^'-x^ + a^^a{i+4x + 6af + 4x^ + 001*). 
Whence (a- 1) (a:* + 1) + (4« + i) {(x^ + x)+ {6a - i) oj* = o, 

or (a-i)(« + ^) +(4a+i)(«+^) = -(4a+i), 

whence a: + I=^^±^*. {-(4a+ i)*± 5*1 

= 2p suppose ; 
••. a?* — 2jx» = — I, 

. ._ ( I + 4^) + 5* ( I 4- 4^)^ ± {- 1 o + 6oa ± ^ X 5^ ( 1 4- 4a)^}^ 

4(i-a) 



Ex. 42. 

27. a.*+/=4, (I)] . . . 

8 « > 9 find X and y. 

a'+r = a8,...(2)J 

Dividing {2) hj (i), a? - a;*y* + y = 7, 
squaring (i), aj4-2a?*y*4-y= 16; 

.-. 3aj*/ =9, or x^y^ = 3 (3), 

From (i), x + 2a^y^ + y = 16, 
from (3), 4a#y* = la ; 

.-. a?-aaj*y* + y = 4> 
or »* — y* = ± a : 
but a?* + y* = 4; 

/. a?* = 3, or i; a? = 9, or i; 
y* = i, or 3; y = 1, or 9. 



37. a^ -y4=: 14560, a; -y = 8, find x and y. 

Substitute a? — 8 for y in (i). 
Then ic* - (aj* - 4 x 8aj» + 6 x 8'a^ - 4 x 8aj» + 8^) = 14560 ; 
.-. 4x8iB3-6x8V + 4X 83a;= 14560 + 4096 = 18656; 

.•. a?5 -latic^ 4. 64^?*- 583^=0, 

or a?(a^- iaa? + ii) + 53(^-ii)=0 5 
.-. a?- 11^6, and cc(a?- l) + 53=o. 



a? = ii) 



Whence a? = ii]_ ^^^ ^^ 



38. iB* + y* = 641, xn (a?» +y') = 290, find x and y. 

Then ^?22y^-2ajy«aj*+y^*=64ij 

4X a'(a3r)*+4xax64i (aiy)'+64i'=64i' + 4xaxa90»=i04i'; 
.-. aar'y' = ± 124LlL-ll= aoo, or - 841. 

Hence 05*+^' = 39, or ± 10 V- a, 
a,i_,y> = ai, or ± Vi48a ^ 

.-. a! = ±5, ±a, ± l± 5 '^^ ± ^37°^)*' 
y = ±a, ±5, ± {± 5 ^^=^ T VstS^}*. 

40. a5» + **=3s*»> a»-«**»a, fiiid»aJid«i 
.-. a + » =< ± (5aM)*r »♦-«»« + «»«*-'«»» + ** -±r^- 

Bat a^ +a!»e' +»*«8aj*«»'; 

.-. a» (o^ + «•), or a^^a' = 8a;'«' T -| ; 



18 SIMULTANEOUS EQUATIONS. 

Hence ^z* = ± j ; 

* 

.,^»(_4.y=f_^y=.,(xoo)^ 

Via5/ \icx>o/ ^ ' 

Then a? + « == ± (ioo)»o, 

a?-« = ±{-2(ioo)»}*; 
.-. a;=±(io)»(-5 + -ix5*); « = ± (lo)* (-5- -ix 5*). 

41. a? — y = i, (ic' + y*) (aj» — y*) =5:247, find a; and y. 

■ 

Now aj' + y* = i + i&a:iy, "by squaring (i), 
and a?* — y5=:i4-3a5y(a — y) = i + 3a:y, by cubing (i). 

Hence (i + aay) (i + 3a?y) = %A1 ; 
.\xy^6, or -^; 

.-. a? + y=±5, or ±(-y)*; 

42. aj'y" + ajy^ = 156, !?aj*y* — a'y^ = 144, find a; and y. 

The equations being homogeneous, put y^xz\ 

-^ z^ + z^ 156 ^ i + z^ 13 
Then — r = -^ , or = -^ . 

Hence z= , or — - . 
3 4 



SIMULTANEOUS EQUATIONS. 

Now a:*y" (iZO? - y) = 144, 
and ^ = i-« = «-|^4 or «+^ = ^; 

...a,y. = 3X144 ^^^8 ^,4X244 

4 . 15 



19 



.*. xtf* = 156 — 108 = 48. 

» .3 
.-. y« = |x48 = 3aj 

/. y = a, and a = 3. 



8 



a^y* = ^ X 144 -i 144 = - ^ X 144, 
a* 16 ' 

.*. y=-(ao5-8)J, anda!=|(ao5-8)». 



43. 



y 81 , „, a^^ 

^-rn=(3y + i8)^ 



X xy ^ •' ■ ' y > } find » and y. 
y + 3a^ = 9 + 3(a^)*' 
Clearing (i) of fractions, 

y»-8i = aa^(y + 9); 

••• y+9=o> or y=-9> 

and y — 9 SB aa^, 

from (a), y - 9 = 3a'' (y*- 1) ; 

/. a)' = o, or aj = o, 
and y* — I = - ; whence y = -^ , or — 9. 

But aa^ = y-9 = -^, or -18; 

•••«=(-f)',»r(-9)» 
, -|(f)*,or3(#. 



20 



IJIHULTAKEOUS EQUATIOXa. 



44. 



©'- ® 



6i 



+ 1 



u 



; find X stud y. 



45. 



From (:*), (a2()* ^x 4- Vy) = 7^ 5 
/. V^ (a? + y + a V^) = 78". 
From (i), fc + y«6i + \/icy; 
.'. V^ ((Si + 3 Vfl^) = 78* ; 

whence V«?y = 36, or ' ^j 

and /. X = 81, 16, - (7 4- 36 V^^^), 

y=i6, 81, i (7 - 36 V- 0,2). 
3 

X y x+y 0^ y^ x — y ^ , , 
.-;-^ = zrr5> 73-^'=-Ti^J fiiid^candy. 



y 05 a^ + y' y^ x^ y 

Divide the and by the ist, then 

X y __a?-r-y aj' + y* 
y a? x + y' y * 

Put (c==y«5 
thei^ « + - =5 -^ (21* + 1) : 
.'. «* + I = o, and z + i — z^ — z. 
The latter gives « or - = i + V^« 

-+ 



%±'J% 



y (4 ± ^ Vi^) 



=: I ± \/i& - 



I ± V^ I ± v^ * 



.'. — = a ; hence y = - , and a? = ~ . 
^y ^4 4 



46. 

y 



QiMuurAmoin kquatiokb. 21 



« — 4y + 8 = o, J 

From the ist, a!» + a«y*+y**aoy'-(y*+agf); 

.-. (x«+3»y*+y»)+y(»+y*)+^»*^+aoy» = -^y*, 

a!+y*+|=±|y; 

.-. a+y*-4y = o, and aj+y* + Sy»o. 

Sal}trscting the anii 

y*-8=o, and y* + 9y-8»o; 

.-. y = 4, and ««8; y(y* + i) + 8(y-i)«o; 

.-. y*+i=o, y+8/-8 = o; 

.-. y-J, y = 8(5T»V6), 
a!--4, aj = 8(i9T8V6). 

47. a^+y' + ^^ + aS^-^Sl findxandy. 

a^ + aj»y*=468 j' 

Dividing (a) by (i), we hare 

-^L- = 36; .-. ay = ± 6 VJB+y ; 
sB + y ** 

.-. from (i) {(« + y)' - la (as + y)*} (« +y) = 13. 

or (a? + y)» - la (a? + y)* = 13, 

whence a;+y=i, or (13)* and .•. a!y = 6(i3)*, or -6; 

.-. a!-y=±5. or ±{-ii(i3)¥; 
.-. a = 3, or - a, Of ^ [13* ± {-11(13)*}^. 

y a, or 3. or i [13* T {- 1 1 (13)*}*]. 



22 SIMULTANEOUS EQUATIONS. 

48. a^ + y4=i + 2^+3^y' l,finda:andy. ' 

From the ist, a?* — y' = i +a?y ; 

Divide the 2nd bj this equation, 

then aj + y= 1 + 05, and a^ — a?y4-y* = o; 
.•. y = I, oj* — a? = a ; .'. aj = i&, or — i. 

Also 2y*+is=o, and a* — aw+«^== — ^: 

4 4 

; ^( T^ , ( k find a? and y. 
y" (a-a?) = Jaj (a? - n) J ' ^ 

By subtraction, n (Ja? + oy) = ajy (a? + y). 
Bymultiphcation,&c. -^=:g_^|-_jj; 

< 

. ay^ ay-n(a? + y) + n* 
oft a?y— (Ja; + ay) +a5* 

For hx + ay substitute its equal — {x + y) ; then 



i^y 



{xyY — ^-^ (» + y) + oJajy = ahocy — o5n (a? +y) + ahn^ ; 



n 



•*• (^)' {^ - (i» + y)l == w'«* {w - (^ +y)} ; 
.*. X +y = n, and xy^±n (ab)^. 

Whence a? = - {w ± (w» T 4n Voft)*}, 

y = ^{^T(^'T4w^^)*}- 
AgainJ + a| = f(x + |) = ±(a5)*(n.|); 



SIMULTANEOUS EQUATIONS. 23 

(aj+y)*-(a;-y)'=a6 j 

From (i), 

(aj+y)-(a;-y) 
32 



(« + y)*-(a,-y)*' 

H^o.4- + D'-(r-l)'),or@%»-0.«<ii 

••• a? = 5, and y = 4. 

57. (ay* + a?)* + a^ = y(aj + 9)*+3y 

From(x). VZ^^^^^3^^^ ^^^ 

Squaring, y' + g + g>^F+^ _ « + i8 + 6 V^T? . 



.3 



y i» 



24 SIHULTAlfEOUS EQUATIONS. 



Dividing (3) by this equation 



* 



y^y; .-.05=93'. 

From (a), y'(y+ i)*-4y» + y; 



I , . I — , 1 



Whence y = g {3 ± V105}, or g {3 ± V-87}, 

a; = 5 {19 £ V105}, or ?{_i3±\Cll7}. 

I 

Similarly 3^ = ± {a*<*-^-^.J^<«^^-«^>}^. 

69. a? » =y3j y " '^'^j fijii ^ and y. 

Let tlx+i/g--t; 

8 3« 3 

Hence ar = y, and ar + a;* + - = - +-; 

■^ 4 3 4 



(smuLTAiasous equations. 26 

62. Qcyz = ^31, xyw = 420, xtsw « 660, yzw « 1540 ; 

find a?, y, le;, w. 
Multiply all together, then 

{xyzwY = 231 X 420 X 660 X 1540 

= (iixai)x(6x7xio)x(6xiixio)x(ax7xiixio) 
= ii^x io'x7*x6*; 

.•. xyzw = 6x7x10x11. 

But yzw = 1540=: 2x7x10x11; 

.-. x = ^. Similarlj y = 7, «=ii, 1(7 = 20. 



63. «' + ay+y* = 37 j 

a^ + aj35 + «* = 28 I ; find a?, y, «. 

y'+y«+«" = i9 j 

Then a;(y-i5)+y»-«» = 9; .". (a:+y + «)(y-«) = 9. 
Again a^-y» + «(a?-y) = 9; .'• (a5 + y+«)(a?-y) = 9; 

.-. y-« = a;-y; .'. y = I(a: + «). 

2 

Hence ar" + - (aj + «) + i£±fL :« 37; 

. 7x' + 4^ + g" ^ 4X37 ^ 37 
a^+xz + z* 28 7 ' 

.-. 7+4* + e» = 22(i+e + ^), if«=-; 

7 ^ 

.-. 3oe" + 9^=12. 

Hence ^=- or — ^; 
^ 5 

.•.a.(x + l4.i) = 28; 

••• «=±4> «=±a, y=±3- 

Also a:» f I - 4 + — ") = 38 ; 
, V 5 35/ 

/. a;=±— V3, « = +-V3, y = ±-V3- 
o 3 3 



26 SIMULTANEOUS EQUATIONS. 

64. X + y + e =ii (i) i 

a;'+y' + «»=49 {2) I; find a?, y, 2?. 

y;5 = 3aj(«-y) (3) J 

Subtract (2) from the square of (i) and divide by z, 

131 — 40 
then ay + xz+yz^ iXZssgg, 

Hence, and from (3), 

36 — a;« _ _ ^xz , 
a; + « "'^""3» + «' 

Substituting these values of y successively in (i), we get 

^af + 7XZ + z^ == II {3X + z) (4), 

a^+xz + z* = ii (a? + «) -36 (5). 

Multiply (5) by 3 and subtract from (4), 

hence 4xz-'2z' = — iz%z + io8] 

. ^-.?!_Zliii±54 

•'^" 2Z 

2Z 2 

Substituting in {x + zY-ii {x + z)-xz- 36, 

/ 3g»-iig+54 \ / 3^'-33g + 54 A- g'-iig-i8 . 

V ^« A ^« / ^ '. 

. 3(3g'-iig + 54) _ g'-iig-i8 
2«* 2;»-ii« + i8' 

Subtracting both sides from 4, we get 

g' + 33^- 16:^ ^ 3^' - 33^ + 90 , 
2z^ «"- iia4-i8 ' 

. (g-6)(g-^7) _3 (g-5)(g-6) 

Whence — 6 = 0; /. i5 = 6; 
54— 6 X 5 



• • 



SIMULTANEOUS EQUATIONS. 27 

65. a + y + «=i3l 

a^+jf^+z*= 91 >; find a?, y, and «. 

Subtract the 2nd firom the square of the i8t, and divide by a, 
then xy + xz-\-ye=s — 2 — z^ss^g. 

For xz write y*, 

then y(a? + y + «) = 39; 

/. a; + « = io, and xz^sg; 
.*• a; = 9, or i, and e^i^ or 9. 

66. x-^-if + z = 13 ] 

af+y* + z* — 6i |; find a;, y, and jt. 

x{y + z)-2yz J 

Subtract the and from the square of the ist, and divide by 2, 
then xy -{-xz +yz = -2 = 54. 

From the 3rd, osy+xz + yz^^z ; 

/. y2= 18. 
Multiply the ist by Xj subtract the 3rd and substitute for yz, 

then a?* =s 13a; — 36. , 

Whence a? = 4, or 9; 
.*. y + i5 = 9, taking the former value of a?, 
tfz=i8; 
•'• y = 3> ^^ ^y *^^d £ = 6, or 3. 

Ex.43. 

16. Let aj + y = the hypoihenuse,. 

and X'-i/ = one side. 

Then by the question, 4ajy — 2 (i), 

and 4{(ic + y)' + (aj-y)'l = 5l(a + y) + (a;-y)}....(2), 



28 SIMULTANEOUS EQUATIONS. 

from (i), (4i»y) = V^ the remaining side, 
from (a), 4{^ + y^)^5^ (3), 

from (i) and (3), 4 («" + ;^) = 5^, 

whence 40^ — 5055 + 1 = o, or 40^ (a: — i) — (ic* — i) = o ; 

/. a; — I = o, x=i; 

' /. ic+y = S, the hypothenuse, 

a? — y = - , one side, 
(4a?y)' = V^j the other side. 

18 Let a?, y, « be the numbers. 

Then x + y + z — Q,6 (i), 

oc^-y*^y'-z' (^), 

aj»+y' + «» = 300 (3), 

From (2), a?" — ay" + 0* = o ; 

••• 3y" = 300 ; ••• y = io» 

% a^ + «* = 200, 

a? + ;5 = 16; 

a? — ;5 = Viioo — 56 = lii ; 
/. a; = - (16 + 12) = 14, « = 2. 

19. Let X, ff, z, w he the numbers. 

Then x + tf + z + w = 44 (i)> 

xr/-{-zw=25o (a), 

a?^+yw? = 234 (3), 

xw + yz = 225 • (4)- 



INEQUALITIES. 29 

From (iz) and (3), {x + w){i/ + z)= 484. 
From (i), y + « =44- (« + «?); 
/. (cc + tt?)' — 44(a? + tt?) + 484 = o; 

From {2) and (4), {x + «) (y + «?) = 475. 

Then, as before, a; 4- « = 25. 
From (3) and (4) {x + y) (« 4- w?) = 459» 

hence aj+y = 27; 
.-. (a: + «?) + (a? + «) + (a5 + y) =224-254-27 = 74; 

.-. 2a; = 74-44 = 3o; 
/. aj=i5, y=i2, «=io, t(? = 7. 

Inequalities. 

Ex.44. 

1. w^ 4- 1 > n* 4- n, if n^ 4- 1 > n{n 4- 1), 

or if (n* — w 4- 1) > w, or if w* 4- 1 > 2w, 

which is the case since n' — 2n 4- 1 or (n — i)' is essentially 
positive whatever be the value of n ; 

.*. w' 4- 1 > 2n ; 

4. — ^ = aJ3 4-aJ"y4-a?y"4-y^>y*4-y3 4-^4-y^ 

and Koc^ + Qc^ + oc^ + a^, 

^•^- "r--|'<4^» ^^* >42^^; 

a?^ — w* , a^ — V* 

• X'-'V> ^, and < ^-. 

^ ^ ' 4y3 

9. Show that (a 4- J 4- c)^ > 27060, but < 9 (a^ 4- 5^ 4- c^). 

Now db < {j-^j 9 cd< ( ) ; 



.'. aoca< ( . I . 



1 



30 INEQUALITIES. 

But ^±*.^±^<-^li — L_ii — li ; 

... aj«;<(^±A±£±i)\ 

a + b + c 



As the value of d is arbitrary, assume c?= 



Then dbcd< 



3 



(^7<*. 



.-. {a + l + cy>!Z7abc. 
Again oJ < (^-j-J ; 

4 

.-. a^ + h^ + 3db{a + b) or (a + J)^ <a3 + 53+3 (^^ jp . 

4 

a' + 5^ /a + i\^ 
.-. > 



m- 



Al«,5i±*>(2±*)', 

a^ + 5^ + o» + A» /a + Jjfc + ^Y 
4 ''V 4 / 

3 

... a3 + 53 + c'>3A3>3^^±i±fy; 
.-. (a + J + c)3<9(a3 + 5* + c»). 

11. a5*+y'>aa!y; .'. jb' + a^» > aas'y, and a;^+^>!W!y'; 

or a^ + y3 > a;»y + a;y'. 



INEQUALITIES. 81 

Similarly a5' + «*>a:'«+a»^ 

and y5 + «^ > y*« +y«* ; 
/. by addition • 

Ex. 46. 

10. Since a : & = c : rf; .'. a — ft : a=c — <2 : o (i), 

and a*-i* : a^'^c'^'-cP' : c* (a). 

a—h C'-d 



Now from (i), 



a c ' 



and from (2), 



a c 



but by hypothesis a>c; and .*. a*>d*; 

/. a — &>c — <?, and a^ — h'^xf^d^ 
and .'. a + rf>6 + c, and a* + (?*> J* + c*. 

11. (i) a-c-J4"d^=a-c-i+ — 

_ a(a — c) ""ft(a — c) __ (a — ft)(a — c) 
a a 

w (M)-(W)-(^k)-^9 

_ Jc + a* — oc — oft 
"" abc 

_ a{a^c) ^h{a'-c) _ (a — ft)(a~c) 
"" abc abc ' 

' ' a c? 6 c abc 

12. Since vaa?; .'.2?-= — -y. 

aj 3 

•'• y = 7a; is the equation required. 



32 ARITHMETICAL PROGRESSION. 

18. Let the value of diamonds s= r x square of their weight, 

3 

and...... rubies ^sx (weight)*, 

then (i), ra* = «wJ», 

2. 

and ra* + «wiJ3 = c, 

3 

abo ra* + 8h7=zc; 



.•• « = 



3 

(w + 1) i^ 



Again, mra^ -\-ra^^mc\ 

mc 



.'. r = 



{m+j)d 



29 



value of the diamond = ra?* = 7 — . .,, ^ £. 

(w + 1) a» 



3. 



and value of the ruby = «a?« = £. 



Arithmetical Progression. 

Ex.46. 

42. Now |,^ ^^+(^-i)^^ ^i3zL7!^, 

r 

Let n= i; then -7= ^. ^ = 2 the ratio of the first terms, 



n = 3; then -T-rT7> = -^-7 =--, 

^' a' + d' 1 + 9 5 



the ratio of the second terms. 

49. Li the 1st series, the j>*** term = iJi + (j> — i) 3 = 3p — i, 
and f" =:3 + (j-i)4 = 4jr-i. 

If these terms be identical, 3;p = 4j ; .*. ^ = ~ . 



ASTTHMETIGAL PROaBEflSION* 88 

Hence, every 4th term of the ist series is identical with eveiy 
3rd term of the 2nd series ; there are, therefore, 25 terms iden- 
tical, or common to the two series. 

Ex. 47. 

1. 2^ — i* = (i + i)^ — i^ = 3x i*+3x I + I, 

33-.2^ = (2 + i)*-a* = 3X2* + 3x a+ I, 

&c. = &c., 
(n + i)* - n* = 311* + 3n + I. 

Adding the vertical columns we have 

(n + i)*-«^ = 3(i' + a* + 3*+... + n») + 3(i + 2 + 3+... + n)+n; 

.-. i»+a*+3*+ + n»=r- + ^+^=-gn(n+i)(an+i). 

And if »= 15, we have 

i* + a* + 3* + &c. to 15 terms = -7.15. 16.31, 

or the snm == 1240. 

8. Given series - 2* + 6 - (4" + la) + (6"+ 18) - &c.; 

.-. the sum = a' — 4* + 6' — &c. to ar terms 

+ 6(1 — i* + 3 — 4 + &C. to 2r terms) 

= 4{(i-4) + (9- 16) + (25 - 36) + &c. to 2r terms} 
+ 6 {(i - rj) + (3-4) +&C. to 2r terms} 

= ""4 {3 + 7 + 11 + 15 + &C. tor terms} 
— 6{i + i + i+ &c. to r terms} 

= -(8r* + 4r)-6r 

= -2r(4r + 5). 



34 OBOMETBIOAL PBOORE88IOM. 



Geometrical Pbogression. 



Ex« i8. 



1. Let « as the 5*^ term, a the i"* term, r = the common ratio, 
then « = ar*"' = 5 x 2^ = 5 x i5 = 8o. 

2 
6. Here aa« — 21, r = — , »» = 7; 

3 

/. the 7*** term = ar*"* = - 21 x(--)=- — • 
^ \ 3/ «43 

11, Here = -, r = -^, n = 8; 

3 ^ 



.-. >8f = - . 



. (-i)"- 



_ 6561 — 356 ia5i 

3' _3_i ~ -384x5 ~ "384" 



15. Here o = |, r = -(^ ; 

°'-^ — V'^ 






5 (Vs + Vi) 



-|l-(-V»n-f;W5-V«). 



19. Here a=-, r = — 2. 

3 4' 



i-^- 3 \ 4/ 3 7 «i 



aEOMETBICAL PBOaBESSION. 85 



28. ^-i=.^^fi+?)=./z^. 



32. Find the value of •oajr. 

lO 

otherwise, S = '027 ; 

/. iooo;8= 27777 

1008^ %'^j'j'j 

• • AJ — — — __ — — - ^ 

37. Let a be the first term, and r the common ratio of the 
series, then 

the p^ term = ar^^-P. *....(i), 

and the j**^ term = ar«^ = ^ (2), 

the n^ term = ar""S 

j&om (i) and (a) we have 

r*^ = — , whence r = (75)*^; 



.•• ar*-" = af^\ r*"' = p(^« = (lU-. 1^ 






40. (a + i + c + rf)»=:(a + J)* + (c + rf)» + iZ(a + i)(c4-rf). 
If r be the common ratio, then 

(a + J) (c + rf) = a (i + r) or' (i + r) = (ar + ar^y = (i + c)" ; 

.-. {a + b + c + dy=: {a + by -^ {0 + dy + 2{b + cy. 

41. Let r be the common ratio, 

then a» + J» + c*>or <(a-i + c)% 

according as i+r*+»^>or<(i— r-f r*)*, 

....♦ I '^r^+r^> or <(i + r' + r*-!Zr + iZr*- ar^, 

iir(i- 2r + r*)> or <-iir'. 



36 GEOKETBICAL FB00BE8SI0N. 

If r be positiTey then (i — r)" > — r, 

negatiTe, (i + r)*> + r. 

Hence d'-^h^ + c* ia >(a-6 + c)». 

44. Let a, ar, ar*^ See, to an tenns be the series, then 

r**— I 

ar+ar^-\'ar^'\'&c. ion tenns^ar.^ = Sj (i), 

f^ — I 

r**— I 

and a + ar* + a^ + &c. ton tenns = a.-T = 8^ (2). 

r*— I 

Biyiding (i) by (a), we have 
which, Bubstitnted in (i), gives 



a = 



The i^*"* and g^ terms of the series are respectively 

ar'^ and ar^^^ 

and by the question, the^^ and ^ terms are the first and last 
terms respectively of an a.b. series of {m + 2) terms ; 

/. the common difference =— ; : 

m + i 

/• the r"* mean = the (r + i)*** term = ^ ^ 

r [\ m + I J m+ 1 ) 






OEOMETBIGAL PB0GBB8810K. 87 

Ex. 49. 

1. Let iS=i-3aj + 5a"-7a5»-f &c.; 

"" I +a I +0?' 

(I + x)* 

8. Let iS=i.i + ii.3 + 4.5+8.7 + ... + (an- i).a*^; 

.*. 28==^ 2.1 + 4.3 + 8.5 + ...+(2n-3)2"+(3n-i)a*; 

.-. - iS= I +4 + 8 + 16 + &c. + 2* - (un - 1) a* 

= i-(2n+i)2* + 4.^37 
= -2*(3n-3)-3; 

.-. 5= 2* (an -3) + 3. 

9. Let iff = 1.20? + 2.305* + 3.40^ + 4.5a;* + to infinity, 

then x8^ + i.^as* + 2.335* + 3.405* + . 

.•. (i— a?) iff = i.2aj + 2.2a? + 3.235* + 4.2aj*+.. 

= 2a?(i + 2aj + 335* + 40:* 4. ). 

Now, let5= i+_2a?+3a5" + 4a:5^ ^ 

tlienajjB= aj + 2a5* + 3aj9+ ... 



9 



f 



.-. (i-aj)-B=i + a? + aj* + aj»+ =- — ~; 



• /ff=-^^— 

Similarly, for the sum of n terms. 



I — a? 

I 



■ 



88 HABMOKICAL PBOGSESSIOir. 



Habmonical Psoqression. 

Ex. 60. 

7. Let a be the i0t term and d the common di£ference of an 
A.B. Beries of wliidi the nfl^ and vf^ terms are -^t- and -^ 
respectiTelyi then 

the m^ term « a + (m — i) ^== tj7» 
the n*"* term = a + (n — i) c?== t^, 

whence f ni — n) a « ■ i^^y J .*• » « ■ ,^^7.. ■ . . 
^ ' JOT MN{m^n) 

The (w + n)*''term = a+(«H-n— i)6=a+(f»- i)6 + nJ 

Jf ■*■ JVaf (f» - n) "■ (w - n) -^if ' 
/. the (m + n)^ term of the Harm. Progression is 

9. Since a, &, c are in Harm. Progression, 



i = 



a-\-c 






HABMONIGAL PSOaBUSIOH* 89 

-{m";'-^(^r(^')v..}, 

since all the temus of the series aie poflitiTe, n being a pomtive 
integer; 

17. Let r be the common ratio of Qeom. Progression. 

Then oT = (or)" = (ar«)' « &c. ; 



2^CS 



Hence a;, y, « are in Harm. Progression. 

Piles of Balls and Shells. 

Ex. 61. 

9. Nnmber of balls 

= g» (n + i) (an 4- 1 + 3<i).- ^ w («t + 1) (2111+ 1 + 3d). 

Here n = 25 + 34 = 59> "^^ «»*34; 

£?=: 100 - 59 = 41 = 75 - 34; 

.•. required number =5 ^ (59x60 x 442- 34x35x192) « 104700. 

14. in(n+i)(n+2) : ^» (n+ i) (2n+ i) «6 : 11; 
.•. n + 2 : iW+ 1 = 6: II, whence n= 16; 

/r g»(n+i)(n+ 2) =^.16.17. 18 = 816, 

and -^n(w + i)(2n + i)=-g. 16.17.23 = 1496. 



40 permutatioks and combinations. 

Permutations and Combinations. 

Ex. 62. 

4. The whole number of letters is ii; and a recurs 5 
times, and e, 4 times; 

/. required nmnber 

I. a. ^.4. ^.6.7 .8.9. 10. II ^ ^ 
— o 'T o 1 2 = 7 .a .0.10.11 

I . a . 3 . 4 X I . rj . 3 . 4 
sa 13860. 

10. j>3-l>a==w(n-i){(n-iz)-il=w(n-i)(n-3)=:;^, 

= w(n-i)(rj-ii)(n-4)=-^, 
&c. = &c. 



••• (ft -i^a) iP4 -Pi) • • • CP«-i -i>«-a) 

^ Pi^Pi-'Pn ^ P^Pi'PA'Pi'^'Pn _ P%Pi*''Pn . 

14. (I) g. = 36. 35 .34 .33. 3^ , 3^6^^,. 

X. ^ . 3 .4.5 
(il) Remove the given patrol from the 36, then 

C> 35-34-33-3g ^^ 60. 
■• I. a. 3-4 ^ •* 

20. If GJ^ denote the number of combinations of m things 
taken r at a time, 

Then the total number 

c«.= G^+ a2+ 0^+.: + og:-"+ c«r 

i.ii i.i^.3 

= (l+ l)»*-I=:!Z^-I. 



PERMUTATIONS AND COMBINATIONS. 41 

Similarly, (7« = 2" — i. 
Hence a'"- i = 65 (a*— i); 
.-. a" + i = 65, 
a* = 64 = a*; 

23. Combinations of consonants, 4 together 

19. 18. ly .16 
1,2. 3. 4 ' 

combinations of vowels, 2 together = 5-ll. j 

.*. number of different sets, each having 4 consonants and z 

TowelB= ^9-i8-iy>i6 ^5^. 
i.:2.3.4 1.2 

.*. number of words =-^^ ^—^-^ — x =^ x I.2.3.4.5.6 

= 27907200. 

26 G ^'^ = w(n-i)(w-2)...(n-r + 2) 

• i.2.3...(r — i) ' 

and 0/)==*iZ:I±ixC7r'>; 






^ — — — — —— ^_-^.^__— — — — — — — — — — — - 



But v.^, 

^ i.2.3...r 



_ n(n— i) (n~ 2)...(n — r + 2) n+ i 
~ 1. 2.3...(r— i) r 






28. 



^ ,rt 2n(2n-i)...(2w — r + i) 
^j»* ' = r-T — z > 



I. 2...r 



Q (r~i>^ an(2n-i)...(2n-r.+ 2) . ^ frti>_^ ^^(^^-i)'"(^^-^) 
*" l.2...(r-l) ' "* l.2...(r + i) 

6 



42 PESMUTATIONS AKD COMBINATION. 

Now GJ^ > C^^'\ and also > CJ^'\ 

an — r + i ,2n — r 
> I, and — ; — < I ; 

/• r < w + - , and >n ; 

/, r = n. 

30. The number of combinations of the p things taken 
I, 2, 3...^ together =^, 

the nmnber of combinations of the q things taken i, 2, 3-*«; 
together = q ; 

therefore the number of combinations of the i? + J things 
taken i, 2, 3-..(;? + g) together =^j +i> + ? 

= 0? + 1) (j + i) - I. 

Hence the whole number of combinations of p -{-q-k-r 
things, p being of one sort, q of another sort, and r of a mird, 
taken i, a, 3"»(i> + J 4- »•) together 

= Cpj+i^ + j)^^- (i?j+i? + 2') + »*= (/>J+p + ^+i) (r + i)-i 

= (i? + i)(? + i)(r + i)-i, 
and so on for any additional sets of like things ; 

therefore the required number = (/> + 1) (j + i) (»• + i) ... — i, 

where ^ + j + r + ...«= «. 



Binomial and Multinomial Theorems. 



Ex. 63. 

4, (3.-.-^)-..g{.-i«»}-.-{,^4(f) 

^ 4^ /aaa?y 4.5.6 / aaa?Y 4.5.6.7 f%ax^^ ^ ) 
i.aV3/ i-^-3\3/ i.a.3.4V3/ J 

= o"H ^ H o h ^g ^ — +&C. 

81 243 7^9 ^187 ^5^^ 



BINOMIAL AND MULTINOMIAL THEOREMS. 43 

-- + I- + 2 ,- - + i -+r — I 



5 a 3 a?'^ 5 r» r a^' j 

5 25 ij*5 

I • i3i . 3 . &c. r . 5*^ 

9^. (i+a?-H»')^ = (i+i»)^+5(i+aj)^aj^+io(i+a5)^aJ*+io(i+a?)^a^ 

= i+5aj+iOiB*+ioaj5+5aj^+a?^+5a?*+2oa?'+30«* 
+i2oa;^+5a^+ioaj^+3oa?*+3oaJ^+ioa?'+io»* 
+:zoa;'+ ioa^+5aj^+5aJ'^+aj" 

« i+5aj+i5a?»+3oaj5+45a*+5ia^+45a^+3oa?» 

+i5iB»+5i»'+a?". 

10,. (i + a; + a* + &c.V = (73^)* = (i -.»)"' 

= i + iia? + 3a^ + 4aj5 + &c. 

14. (i + 3aj - ofy = {i + (3 - a?) xy. 

The terms involving os^ are 

i.r»^^ ' ' 1.2.3^^ ^ ' I. a. 3. 4^^ 

^ Hence coefficient of 

a:* = 10 — 10 X ^7 + 5 X 81 = 415 — 2J0 =» + 145. 

17. In the expression 

n{n-i){n'' i) &c. (n-r+i) ^^^^ , . 

for r write 5, and for x put (J + c), and we have 

n(«-i)(«-a)(n-3)(n-4) ^^j „ . j, 



u 



BINOMIAL AND MULTINOKIAL THBOBElfS. 



Now the term in (6 + 0)^ inYolying 6*c» is lo&V; therefore 
the required coefficient 



n(n-i)(n-2)(n-~3)(n - 4) 



lax 10 



X 10 



_ ii(n-- i) (n- g) (n- 3) (n-4) 

12 

22. (i + aa?+ 3»» + &c.)*= (i -«)"* 

= x+4^ + 4.4±i^^^^^4(4+i)(4 + ^).<^-(4-l-r^i) 

i& I . a . 3 . &c. r ' 

.*. the required coefficient 

_ 4-5'6.&c. (r-\'^) ^ 4.5.6.&c. rjr-^- 1) (r + 2) (r + 3) 
i.a.3.&c.r i.a.3.4.5.6.&c. r 

= ^(r+i)(r + 2)(r + 3). 

23. Here n= 10, ^ = o, y= i», *•= 3> « = o, < = 4, w = i ; 

•'• the required coefficient 

(10 
= {2,.\3.\4 ' ^'- 3^- 5^- <5i = 51030000CX). 

28j. The middle term will obviously involve a?", and to find 
the coefficient of a?" we have 

y for the equations of condition : 



i* 


? 





la 


I 


10 


2 


8 


3 


6 


4 


4 


5 


2 


6 






o 

I 

3 

4 

5 
6 



.'. the coefficient of the middle term is 

■^11^215 + 16:16} 

= I + 13a + 3970 + 18480 + 34650 + 1663a + 934 
=73789; 

••. the middle term = 737890;". 



BINOMIAL AND MULTINOMIAL THEOREMS. 45 

30. (0 (« + i + <^Y = «• + 9<^ (i + c) + 36a» (J+c)»+ &c. + (i+c)^. 

Whence we see that a' gives one term ; 9a' (5 + c) gives 
two terms, 36a^(J 4- c)* gives <Are6 terms, and so on; so that 
the whole No. of terms wUl be given by the A.B, series ' 

i + 2+3 + &c. + io = (n- 10)— =;55. 

{%) No. of terms in the expansion of any multinomial 

(o.+ a.+ a,+ ... + a,)- is r (r + i)(r + a) ... (r + n-i) ^ 

Here n = 4, r=4; 
.•. the No. of terms = ^lAi — 12 = /»^. 

31j. The middle term being the (n + i)*^ term 

_ 2n(2n-i) (y^+i) « I 

— — ■ — __— — — ^^— ^^— — ju^ ^ - 

1.% n X 

= ''3>5> (^^ - ^) , ,^>, (vide Ex. 53, :^8,) ; 

I. a. 3 n ' ^ '^'^ *' 

.*. (a? + - ) = a?** + 2w . aj**"\ - + an . ^^Hl . (c**^*. -I + &c. 
\ xj X 2 of 

X . ji .3******'^ 

(ijw— i) , I . I.I 

I . a \ a^v 

\ I '3' 5 (an-i) 

-f p_ . -* • 

I . a • 3«occ. ••• n 

*' \l+x/ Vi+ao;/ V ifW 

i+aa? I. a Vi + aa?/ 
n(n+i)(n+a)..,(n+r-i) [ ^ )\ &c. 



46 BINOMIAL AND MULTINOMIAL THEOKEMS. 

32. The aeries 

_ I II \2 J I I V2 I V^ / I ,^ 



-(-r- ©"*-©•• 



35. The Talne 



I ^ i-^-3 J 

, f / ^ . n (n — i) (n — rj) « ) 
-5 Jn-n(n- i) + — ^^ ^-^ -^ — &c.|- 

= a(i-i)"-Jn(i-i)--« = o. 



T» ,. n (n — i) 14 

38. By question — ^ <^ = — ^ , 

•^ ^ I. a 9 ' 

I. a. 3. 4 243' 

. (n-a)(n-3) ^ 7x9 ^ ^ . 
3-4 ^43x14 54' 

9 

whence n = -. 

3 



39. (i + a?)* = I + WiOJ + n^ + ^33?^ + &c. +*n^iaj*^' 4- 71^*"+ &c. 
and (1+ a?)' = I +i?iaJ +i>aaj' +i>3aj' + &c. +Pr^iar^ +Pfaf+ &c. 

,'. (i + 05)*"**= the product of these two series, also 

(i + a?)"^ -i + {n +i?)iaJ + {n + p)^ix^ + &c. + (n +i?)X+ &c., 

and these values of (i + a?)*^ are identical; therefore the coeffi- 
cients of the same powers of x are equal; 

/. {n +J?)r = W^ 4 W^., .JP, + W^j .i?a 4 &C. + W,^r-i 4/>,. 



BINOMIAL AND MULTINOMIAL THEOBEMS. 47 

43. Let a, J, c, d be the coefficients of the (r + i)*^, (r + a)***, 
(r + 3)*^, (r4-4)**' terms respectively of aa expanded 
binomial, whose index is n. 

Then tr-"("~')-^"-^+') ^_ n(n-i)...(»-r) 

I -a »• ' I. a (r+ i) ' 

^_ n(n-i)... («-r-i) ^_ n(»- i)...(«-r -a) 



I. a, 



•(»- + a) ' i.a (r+3) 



Hence ^=:^; ... * + i=!L+I 
ar+i' a r+i 



^+1 ...J « . _ w+i 



Similarly, ^+ I =^, and ^+i=:!i±l; 

... (r+i)g+l) = n+i=(r + 2)(| + i) = (r + 3)(^+i); 

( J + c) a 



.'. r + I = (r + a) 



(a+i)&' 






By addition, and dividing by y + iJ, we get 

b i^+J'^c + rfr 
.•. a& {ac+hc + ad + bd)== {b + c) (zac + bc + ad); 

.-. {bc + ad){b'-c) = {b + c)(!iac)-ndbo--7,b*d 

^%{flo^''bH). 

45. Let « = a' + &», p = %ah, and P= (a + bf, then 

(^ +p) = (a + J)«, and (5+p)« -P\ 

.-. P.P*.P*.&c.=(«+i?)^.(/? + p)^.(«+p)f &c. 

.^ p. P*. pf. P^ &c. in inf. == ^ {i + « f^U^i^Il^ J^V 

i.a.3 W J 



48 INDETERMINATE COEFFICIENTS. 



Indeterminate Coefficients. 
Ex. 54. 

then 

Let a? = a, then i = u4 (2a) (^a*) ; /. ^ = — , 

aj=:-a, i=5(-2a)(2a»); /.£ = --L; 

V /a? + a a; — a\ 

a?» + a» Q^'-d?^ 






%a^ %a^ ' 



A ^a?+Z> = --^; hence (7=o, i> = -^,;. 



• • of - a* 4a3 (a?- a) 4as (a; + a) 2a» (a?* + a») * 

8. 0^ - 14a:" + i6aj + 20 = ^aj* (a? - 5) - 4 (aj* - 4a; - 5) 

= 2 (a? — 5) (aj* — aa? — !j). 



liet ,^ ^ . . Q — r-rz = 1 + 



jac4- c 



Qi^ — 7ar* +8aj + io a? — 5 a:"— 2aj — !j' 

Then 3aj»- 8a; + 16 = J. (a?* - aa? - 2) + (^ + (7) (a? - 5). 

This equation is true for every value of x ; let then x^ $^ 
5(3X5-8) + i6=:^(:i5-io-a); 

• • XX. — • 

13 



IKDETEKMINATE OOEFFiaSNTB. 49 

_ laag'+aag — 310 
_ {inx + 62) {x - 5) 

13 

whence 5 = , 0= : 

13 13 

3g' — 8a! + 16 51 608+31 

" 3aj» — 14a:* + i6aj+ 30 "" 36 (x— 5) ~ 13 (x* — a« — a) * 

12. Let- -^H^^^ ^^=_i_+-:B_ + _^. 

(a?— I) (a:— a) (« — 3) » — i a? — a a?-3 

Then 3a?- a=-4(a?-2) (a?-3) + JB(aj-i) (aj-3) 

+ <7(a?-i)(aj-a). 

Let aj =s I, then i « -4 {— i) (- a) ; /. -4 = - , 
a: = a, ... 4 = JB(i)<-i); .-. £=-4, 
a; = 3, ... 7=C7(a)(i); /. (7«2. 

-3 ^li_?J ^\ 3 3 3* 3 / 

Hence _3£ll*— — = i + ix--Z-!e»- -2^x»-... 
^®''** (a; - 1) (x - a) (» - 3) 3 9 54 3H 

7 



X 



50 REYEBSIOK OF 8EBIES. 



Beyebsiok of Series. 

Ex. 55. 

4. Let y — a«« = 5aj + caj* + doc^. 

Assume x^Az^Bz^^Oz^ + Ds^-^-.... 

Then 5aj = 5^+ JJ8«'+ hCsi^ + hDss^ +... 
caj» = + cA^z* -^ 2cABz^ + c (J?» + ^(7)«^+ ... 

.-. z^hAz-\- {hB + cA") z^+ {bC+ %cAB + dA^) z^ + .... 
Hence hA — i; .\ A — j-^ 

•■• ^= ^ ' 

bD =>-(<!& + oAC + idA'£) ; 






&c. ■= &C. ; 



.-. « = ^(y-.)-i(y-.). + 2^{y-a). 



-^^'(y -<■)-+ 4c 



11. a?* — 3a? + y = o. 

When y = o, then a? (a?" — 3) = o ; 
.*• 0? ss: o, and x = V3« 



BEYEBSION OF SEBIE8. 61 

We shall have a series conesponding to each of these Tallies 

of X. 

(i) Let a! = ^y + jy+Cy + iV + ^<+... 
Then y =» y 

3 3 3* 3* 

(a) Let «»a + Jy + <y* + afy' + «3r*+... 

Then «» = a» + 3«?y (J + cy + <i5y*+ «3f» + ...) 

+ 30^ {6 + y (c + rfy + ...)}• + y» {> + y (c + ...)1* 

= a» + 30% + 3o*cy» + yi*dyi + yt*ey* + 

+ 3aJV + 6ai<^ + (6aW + 3ac»)y*+ 
+ ft»y* +3J»<!y*+... 

-3« = -3<»-3^- 3<iy* - 3<%^ - 3«S* -••• 

+ y =» + y» 

... 0= (a»- 30) + (3a»5 - 3& + i)y + (30*0 + 3o5» - 3c) y» + ...; 
... a = V3, & = -|. ''""tI' rf*-;35»&c.-&c. 
. g.«^._i„^V3«._J_vs-35V2«4-... 

..«-v3 gy ^^y^ 162^ 51843^ 



• •• 



• •• 



n 



52 SUMMATION OF SERIES. 



Summation of Sebies. 

4 

Ex. 66. 

2. Asstiine 

« 

Change n into n + i ; 
.'. i» + 4*+7'+... + (3n-a)*+(3n + i)» = -4 + J5(n+i) 

+ (7 (n + i)» + 2> (n + i)3 + -E?(n + i)f + ... 
By subtraction 
9n' + 6w + 1 = JB+ (7(an + 1) + 2) (3n* + 3n + 1) 

+ ^(4n« + 6n* + 4n + i) + ... 

Equate the coefficients of the same powers of n, tiien 
JETso, and the coefficient of every term after JS is = o ; then 

3-0 = 9, 3-0 + 2(7=5, 2)+<7+JB=i; 
.-. D = 3, C7=-S, 5=-l. 

Heuce i» + 4» + 7«+ ... + (3n-2)« = ui-- -3^ +3n^ 



of w; 

ktihen«=i; .'. -4 = o; 



y inte^gral 



.-. i» + 4» 4- 7' + ... + (3n - a)* = - (6n" - 3« - i). 

5. Assume 

i.a» + a.3* + 3.4'+... + n(n + i)» 

Then i.a'+ii.3' + 3.4'+...4-n(n + i)*+(« + i)(« + a)' 

= -4 + J5 (n + 1) 4- (w + i)» 
+ -D(n + i)3 + JS?(n + i)* 
+ i<'(n+i)*+... 



BUXHATION OF 8EBIE8. 08 

+ -P(5n*+ion^ +...) + ... 

Hence F= o, and the coefficient of eyeiy higher power of 
n=so, 

4£r=i, 6Jr+3D«5, 4S+32) + a<7=.8, -&+/)+ (7+5»4; 

.'. i.»*+a.3* + 3.4»+... + n(n + i)» 

=^ + Si+22:+^ + ?!. when «-i, A~o 

0404 



Ex. 67. 

1. Let fif=i + i + i + &c. + - + -4T; 

a 3 n n+i 

» 3 4 w + i/ 

.'. 1= + +&C. + — 7 — ; — ;+ — ; — : 

I. a a. 3 3.4 n(n+i) n + i 

therefore the required snm 



I n 

I — 



and if n = 00 , the stun » i. 

4. Assume -+-^ + i- + &c. + — -^ +-— ^ = iS; 

3 8 13 5»»-^ 5^ + 3 

.". •5+— + ^ + &C. + \ =/g-i; 

8 13 18 5n+3 3' 



64 SUMMATION OP.SEItIlfi& 

therefore tlie'zeqtured sum 

I I n 



15 5(5^ + 3) 3(5^ + 3)* 



and if w = 00 . the sum = — . 

15 



8. The w*^ term = '*'*"^ 



{2n - i) (2/1 + i) {2,n + 3) {w + 5) 

_i f n + i 

T 6 \{2,n - i) (izn + i) (2w + 3) 

!L±i I. 

{an + i) (i»7J + 3) {2n + 5)) * 



" 6I1.3.5 3.5.7 5.7.9 ••' (2ri-i)(ii«+i)(an+3) 



n 



3-5-7 5-7-9 *" (an-i)(an+i)(2w+3) 

n + i 1 

{5&» + 1) (an + 3) (2n + 5)]" 

_ I if 'i i^ I I 1 

"^90 611:3:5 3.5.7"*" 5-7-9 *" (an-i)(5&n+i){!j»+3)j 



n + i 



6 {2n + i) (5&n + 3) {2n + 5) 

w + i 



90 6 [12 4(i2n+i) (an+3)J 6(an+i) (an+3) (i»n+5) 

^J_ . JL I f4!L±4 . ^l±5l 

90 y2 24. ifln + i) {2n + 3) l^n + 5 2n + 5J 

I I _ n(n4- 3) 



40 8 (an + i) (an + 5) lo (an + i) (an + 5) * 
Hence fi«, = — ; putting n = 00 . 



Ex. 68. 

1. Assume i + -H h... + - 1 ; 1 \ — = A 

a3 n n+in+a 



SUMMATION OF 8EBIE8. 



u 



Multiply both sides bj a^ ~ i ; then 



«* a5» , aj* , , af*^ oi^ a!^ ' 
134 ^ «+i n+a 



— I 



X 

% 



3 



4 5 






(a^-i)^; 



— I — 



aaT' 



a 2^ »«*^ aa* 

a 1.3 a. 4 3.5 '•• n(n + a) 



= («--i)fir- 



a^» a?**' 



' w+i n+a 



Let a; — I ; then we get 



+ T^+-r~: + ...+ 



1-3 ^-4 3-5 



n (n + a) 



^£{3 J _^^, I- w(3n + 5) 

aja n + i n + aj 4(n + i) (n + a)* 



When n becomes infinitei the sum 

** n 



^{-3('-3 ^ 



4. Assume i + -H h — + ...+ H 

a 34 n ^n + i 

Multiply both sides by j> + go? ; then 
^ a 3 4 n n + i 



»/9. 



n— I » » + iJ 



''{p + qx)Si 



... „+£±i2a,+^±i£.^ + :.. + 2£+(!l±lLz 



I. a 



»-3 



n{n + i) 



qx 



.«+i 



= b + 2-)^-S^ 



56 



SUMMATION OF SESIES. 



Up + gx^Of a? = -; we may have p =s — i, a — 2; then 

3 I. JLi-L-5- 14.. , ^ + ^ L 



I — 



»« • 



n + I a' 
When « s= 00 , the siun = i. 



-•+1 



- 35 ac* a^' a^ a;' 

6. Assume 1 + -+ — + ... +-r— + — ; — + — ; — 

a 3 » n + i n + a 

Multiplj both sides hyp + qx + ra^; then 



= 5. 



i^ 



+^+^+^ + ...+£^ 



n + a 



+^+^+^+...+^+5^ 



ras* 



n4- 1 n + 2 

raf*^ ronT^* rx**^ 



2 n n + i n + 2J 



= (jp + ja? + rxi^) S, 



Let J? + jaj + nc* = o, and a? = - ; then 



4p + aj+ r = o. 

Also ap + 32 + 6r «= 19, 

and 6p + Sj + I2r = a8 ; 

Whence a + (-6) i + --^^^-. i+ ^^ 



• 77 T • •• 



19 



2 1.2.34 2.3.48 

an (« + 1) — 7w (n + a) + 6 (n + 1) (n + a) i 
n(n + i)(n + 3) '5^ 

n + i'W |n+i"^a(n + a)j*a-*»' 

1 a8 I ag I 
-I . rH '^^—•—7+ ••• to n tetnna 



i.a.3'4 a. 3. 4*8 3.4.5'i5 

^i « + 4 



(n + 1) (n + a) • a"*' • 
When n = 00 , tiie sum = i. 



SUMMATION OF SERIES. 57 

Ex. 59. 

1. Let I — ^ — qaf be the scale ofreUuian. 
Assume /ff = i + aa? + 3»" + 40?' + ... + naf"\ 

and — 2'aj*)S= — ja^-ajcc*- ... -(n-2)ja:^"' 

— (» - i) jiB^— njaf*' ; 

+ (4 - 3P - aj) a?* + ..i + {n'-'{n'-' i)^ - (n - a) j} oj*"' 

- {«p + (» - i) 2} a?" - wjof*'. 

Hence ap+ 2 =3l. .^ = 2 ^--i- 

/. (i - aa? + a?*) 5= I - (n + i) a^ + naf^ ; 

^ I ~ (n + i) a;^ 4- na^^' 
•*• ^~ (I - aj)» • 



5. liCt - = — « ; and i — o« — qz^ be the 5cafe of relation. 
3 

Assume /S^= i + 7« + i7«' + 55«^ + *? infinity, 

then ^pz8^ —pz — 7p«* — 17^2* — ....i. 
-j«*)S= - j«'- 7g'23- 

.-. (i -|)« - j«*) 5=1 + (7 -jp) « + (17 - 7p - q) «» 

+ (55"-i7P-72)«^+ 

Hence 7»+ 5'=i7l 
i7P + 7?==55) 

. ^^ 1+5^ ^ ^ ^ + 5^ - ^ L.=:-i L. 

' i-a«-3«* (i-3«)(i+«) i-3« i+« i+» j.f 

3 



68 INTCBFOLATIOH OF 8EBIES. 

Ex. 6o: 

1. The orders of differences are 

4, lo, ao, 35, 56j Ac. or rf, =4, 

6, 10, I5i 21, &0. or d^^Sj 

4, 5, 6, &c. or ^3 = 4, 

I, I, &c, or ^4=1, 

o, &c. or c?j = o; 

^, ., , , . II. 10 ^ . II. 10. 9 . II. 10.9.8 

.*. the lath term == 1+11.4+ .6h ^.4H ? — 

^ I. a i'^-3 ^•^•3*4 

= I + 44+ 330 + 660 + 330= 1365. 

12. By giving to n the values i, 2, 3, &c. successively, we find 
the series to be i". i + ^". 4 + 3'. 7 + 4*. 10 + 5'. 13 + &c., 
whence <?i=i5, <?s = 3^, e?3=i8, d[j = o; 

.*. the sum == 25 + -^^ — -^. 15 + "^ — ^. 32 + -^^ — -^^ — . 18 

^ 1.2 ^ I-2.3 "^ 1. 2. .3. 4 

== «5 + 4500 + 78600 + 227700 

Ikterpolatiok of Series. 

Ex.61. 

1. If a, hj Cf &c. represent the ist, 2nd, 3rd, &c. terms 
respectively of the series, we have 

a - 6J + 15c — 20iZ+ 156 — 6f+g = o, 

1. ^t_ thx 20f? + 6(J+/) - 15c- (a + flr) 

whence e. the ^5*** term « — ^^ — '^^ — ^ ^ *^ 

' ^ 15 

_ 1000 + 900 — 450 — 220 

= 1^ = 82. 

B. If we suppose the 4^ order of differences to vanish, then 

a — 46 + 6C — 4£? + e = o; 

. jj . a + e + 6c \ V19 + V23 + 6V21 ' , ^ 

.. a= ]; J = -2^—^: — ^—-'^ 1 V20 = 4*6904101. 

4 4 



CHANCES OB PBOBABILTTIES. 59 

5. Here dj =: 3^ d^^ 3, d^ » i, d^^ o. 

Then y=o + («-!)<?. + fcifclilrf. 



I. a 



(g - i) (a? - g) (a; - .3) J 
I . a . 1 *' 



Let a:=ii, then w=i+ ^ -■% + -^='^, 
•♦ * 4 3* lao 138' 

'' " % 8 16 ia8 

"'^*' 2^-^+4 3a;^ia8 ia8' 

««ai y=, + i5 + I5_^^66| 

*' " 4 3a ii»8 ia8' 

,'. the terms are, 

195 a8o 385 663 - 
' ia8' ia8' ia8' ^ ia8' ^' 



Oranges ob Pbobabilities. 



Ex. 62. 



1. The chance of throwing an ace the first time = 7 ; and 

the chance of nof throwing an ace the second time = 7 ; therefore 
the chance of throwing an ace in the first throw only is 

4, For the balls to form pairs, one of the first 7 tetters 

must be drawn from the first bag ; the chance of this is — . 

Then the chance of drawing the same letter from the second 

bag = i; 

7 I I 
/. the Required probability* = -^ x - = — . 



60 



CHANCES OK PROBABILITIES. 



6. The chance of diawing a white ball each time =-; 
theiefoie the chance of doing so ii times gucceBsiyely 

«i.i.i»&c. to II footors = "77 =^ :rr:fi • 

9. The combinations of lo things, 3 at a time 

10.9.8 _ 



The chance of drawing 



1.^.3 



= ^. 



% gs. and I sov. = -—- , 



lao. 



its value = — x 186 shs. 



2 gs. and I sh. = 



_ 5 . 



liJO 



s= X 2,15 ... 



2 X ^ X ** I 

I ff. isov. andish. s= — rrr-^; =Tr::^^*^o... 



lao 



2 sov. and i g. = 
2 sov. and i sh. = 



2 sh. and i g. = 
2 sh. and i sov. = 



120 

1^0 

1^0 

- 3?_ 

120 

q &h. = 

I 

<i sov. = 

•^ 1^0 



ir^o 

== X (166 ... 

1^0 ^ 
= — X6I5 ... 



X460 ... 



lao 



= — x66o ... 

1^0 



= X 20 

120 ^ 



I2&0 



x6o 



Hence, value of expectation = — x 385^5. = 32*1 == £1. I28.i*2d. 

11. The chance of third ball being white = - . 

22 
Combinations of 3 things, a at a time — ^^— == 3. 

Of these, the chance that a particular couple is drawn &= - ; 

3 
.*. the required chance = the compound chance 

I II 
236 



CHANCES OB PBOBABILITIES. 61 

13. The snm of the favorable and im&Yorable events equal 
the No. of combinations of ^5 things, lo at a time 

_ 25.a4... i6 

"* I .2( ... lO 

The combinations of 5 black balls, 3 at a time = 5*^*5 

ao white balls, 7 «= — i—zJiUi; 

1.2... 7 



.\ the sum of the favorable events = ^'^'^ x — 1— 2iml. 

I. a. 3 i.a...7 ' 

.*. the probability required 

5.4.3 no. 19... 14 , 25.24... 16 60 
"1.2.3 I. a... 7 * I. a. ..10 "^53* 

14. The No. of combinations of 52 things, 13 at a time 

52 . 51 . 50 ... 40 
1.2.3... 13 

The No. of combinations of 48 things, la at a time 

__ 48. 47. 46. ..37 

I. a. 3.. .12 

With each of the latter set, only one of the 4 aces may be 
combined; .*. chance first required 

^ 48.47 ...37 ^ . 52.51 ...40 

I . 2... 12 I . 2 ... 13 

- 39'38>37'(J3X4) ^ 9^39 ^ 703 =XnearIv 
52.51.50.49 20825 1602 16 ^* 

Also, chance of dealing one ace to each person 

16. Chance of dipping into the ist or 2nd urn = - ; 

/. the chance of drawing white firbm ist urn = - x 5 . 

^ 29' 

and the chance of drawing white from 2nd urn =: - x ^ ; 

* 3 

.•. the required chance =-(5-|-3)= — ^ 
^ «V9 5/ 45 



62 CHANCES OB FBOBABILITIES. 

18* The ey^i chance of throwing an ace in x times 

logi'z ^ 

19. Let a, i be the numbers of white and black balls re- 
spectivelj ; n the number of bags. 

The probability of drawing a white ball from each bag = -^ . 



black 



The proV. that the n balls drawn will all be white = ( ^ X 



The probability that the n balls drawn will be 



(n — i) white, and i black = n 



(a + Jr' 



(n-.) ,and2 ^n{n^i)^^2^ 



(n-^r) ,aiidr ^njn^ih^n-^+i) ^TV^ 



The most probable number corresponds with that expression 
in the above series which is greatest ; 

xf n(yi-i)...(n -r+i) dT'V , ^, 

i.i*...r i.2...(r+i) "^ » 

a w — r ^ w J — a 

ft r-fi' a + 5 



SCALES OF NOTATION. 63 

I .«...r i.ia...(r— i) 



n — r + i h nb + h 

a 



.-> i; /. r< , , . 



Herew=iia, a = 7, i = 2i; .'• r> — , and < — • 

Since r is an integer, it must = 2 ; 
/. the most probable number of white balls drawn = w — r = 10. 

20. The number of ways in which this can be done is 
expressed by the coefficient of x^^ in the expansion of 

{x + x* + a^ + os^ + x^ +a^)S 

which coefficient, found by the Multinomial Theorem, is 

6 + 24+12,-}- i2, + i2f\-i+i!Z + 2,4+4 + 6'{'i!Z=^i2,5; 

but there are 6* = 12^96 possible throws, hence the required 

chance =-^. 
1296 



Scales of Notation. 



Ex. 63. 



1. 3/53^ 

3 ) 1793 » 

3)m o 

5)66. I 9/53^ 

3 >>gg o 9 J 597' 8 

SJJI' ^ 9)J^ 3 

a I 7 3 

.'. the No. is ijiioioj^a. .*. the No. is 7338. 



64 SCAUSS OF NOTATIOK. 

3. 34402 



9. 



11. 



103^, since 19,, = 1034 


5 




. 


1203 






5 






13233 

5 
212231 






Hence (34402)4 = (212231)4. 


tt 


4; 95088918 
9074 

4548 
3754 

9«49 
9074 

<95i 

9«58 

9«58 
9«58 


\(^t4tee 


25400544 U" 2 


3fle757o^'i(^62te 


24 




30 


121 J 140 




102, J2,ej 


121 




2,04 


1221 j 1505 




i04<>357 


1221 


• 


^404 


12222 ) 24444 




10586^6531^1 


24444 




^53^ I 



.% the sq, root = 41 lij, .\ the sq. root = 62 te, 

14« Let r be the radix of the scale, 

- then i^r* + 3*^ + 5 « 4954» 

orr4 + 3^^4942 
2 2 ' 

whence r = 7. 



SCALES OF NOTATION. 65 

16. Here we have to express 17 19 by means of the terms 
of the series i, 2^ %^y «^, &c.; and to transform 17 19 into the 
binary scale, we have 

2^429 1 whence the number 1719 is equivalent 

2/214 1 to iioioiioiii in the binary scale, 

2 ) 107 o which is expressed by i + 2 + 2' + 2^ 

^prTL J + 2^ 4- 2' -h 2' + 2" ; and therefore it 

4^' will be necessary to select the weights 

^ Z^ ^ I lb., 2lbs., 2nbs., 2nbs., 2^ lbs., 2nbs., 

2 223 o ^2? lbs., and 2" lbs. 

2/6. I 

^)jt o 

I I 

18. Let ao, a^, a^, 03, &c. be the digits, so that 

N— Oq + loa, + lo'tta 4- io»a3 + &c. 

of which the general term is io"*a« ; then with the same digits 
differently arranged to make N\ the general term contaimng 
a^ will be io*a„; and therefore the general term o{ N*^ N^ 
will be 

(10* — 10") a«= 10* (10*^ — i) a„ if w > n, 
or = 10"* (10""" -^i) a^ it n>m; 
since this term is divisible by 10 — i ; 

.•. every term in N '^ N' is divisible by 9. 



Logarithms. 
Ex. 70. 

1. log 5*4 = 3 log 3 + log « - log 10 = 1*431363 

+ •301030-1=732939. 

log i7'5 = log 7 + ^ log 15 - ^log 3 - log 10 

= '845098 + 2*352182- '954242 - I 

= 1-243038. 

9 



66 LOGABITHMS. 

3. 256 = a« = {%^)^ ; 

.•• the required log 256 = — s= 5*3. 

3 



Exponential Equations. 

Ex. 71. 

!• X log 20 = log 100 ; 

. logioo 2 

1*3010300 



logao 1*3010300 ^^' 



7. /J.,) ^- log 1000 ^ 3 . 

log 4'5 •653^1^5' 



•'•"^?5m875 = ^'3x^: 



13. fi^y=^; ^.,^^log^9-log^i^^,gg 

V14/ iii' log 17 -log 14 ^^*4» 

.*. logy = 1*66^4 X log 14 - log 63 ; 
.*. y = i:a754. 



Intebest and Annuities. 



Ex. 72. 



1. Amonnt if = P/'i + -) , 

and P=66o, r = *o5, w=:5, ^ = 4; 
/. if = 660 (i + •0125)'^; 
.*. log if = log 660 + 24 log i'oia5 

= 2-94902^39 = log 889-25 ; 
.*• the amount = 889*125 = £889. 5*. 



INTESE8T AND AKKUITIES. 67 

8. Heie3P«»Pri +-j , and j«»a, r—'oj; 

.'. ««— r— ^-2 — = 36-894. 
a log 1*015 

12. Let a;, y, « be the present values of the sums wliich 
Ay JB, O lespectiyelj haye to leoeiYey then 

a? + y + « = 35«> (Oi 

and X X (i'04)» «y x (i*04)' = « x (i*04)" ; 

X X 



(i:o4)*' (ro4)^' 



. J, ^ 3500 X (1*04)^ ^ 4^58*282 
I + (ro4)» + (ro4)< 3-3415159 ' 

whence x^&iT,*]^ 'jb. \\d. nearly; 



z 



(ro4)» (1-04)' 

{1-04)* (104)* 



15. Let P. =s his property at the end of the a^ year; therefore 
the interest in the (a;+ i)^ year = rP^ ; and his expenditure in 
the X +1)*"* year «= (a? + 1) wrP^^ 

.•. the property left = P, (i + r) — (a + 1) wrP„ 

whence, putting %i—\ for a? we have 

(i + r - a^r) P^x = 0; 

Similarly his expenditure in the ^ year is inrP^^^ which is 
also the property he had left at the end of that year. 



68 INTBBEST AND AKNUITI£S. 



9 



. _ iV (i +r)* _ ago X 'o; x (I'oy) 

^ ' (ro7)».-i 

147 1 -aa 
"8^ = ^77107. 

or A =£17. 14a. iXjrf. nearly. 



r(i + r)*' ^ ' ' 

, loff-4 — loffP— logr . . 

whence n=s ° , — .^ . . — ^^ 

log (i + r) 

^ log 85 - log 946-63194 - log -05 

log 1-05 

^1% years. 



22. Let X the required annuity ; 
the present value of the ist annuity =| -i-^i-g— _| x 50, 

and ^ f (1-0335)'^- 1 I 



. f (ro3a5)'<-i ) ^f (i-03a5)'^-i ) ^ 
• • l-o3a5 X (ro3a5)«J VosaS^ (i*o325)"J ' 

= 34*o<535 

= %4- i«- 37^- 



THEOBY OP EQUATIONS. 69 



Theory of Equations. 



Ex.1. 

+ V~T\ / I - V^' 



5. x{x^7){x+y)(x-'^-^ — -)(a?- ^""g J ^o, 

or a? (»* - 49) (»" - ^ + 1 j = o , 
or gx^ — 6x* — 43905^ 4- 29405* — 980? = o. 



9. The coefficient of the 4th term is equal to the, product 
of every three roots with their signs changed 

= (-2).(-iMi) + (-^).(-i)(3) + (-^)(-i)(4) + (-2)(+i)(3) 
+ (-2)(i)(4) + (-2)(3)(4) + (-i)(i)(3) + (-i)(i)(4) 
+ (-i)(3)(4) + i.3-4 

= 2 + 6 + 8-6-8-24-3-4- Iii + 12 --29; 

.*. th^ 4th term = 2905*. 

11. Since 7, 6, 4, 2 are divisors of 336, which = 7. 6.4. ^ 
and 7+6 + 4+2 = 19, the coefficient of the 2nd term ; 
.'. 7, 6, 4, 2 are the roots of the equation. 

Ex. 2. 

1. The equation is then divisible by a? — 4; to find the 
quotient. 



Division / .1 

^ +4 



I — 19 + 132 — 302 + 56 
+ 4- 60+288-56 



I- 15+ 7^- 14+ o 
/. a^ ~ 15a;* + 72a? — 14 = o is the required equation. 



70 THEORY OF EQUATIONS. 

3. The equation is divisible hj 

(a?- j»)(a?-6), or a"-8a + ia; 

Diyision i 

+ 8 

- la 

1 + 6 — 31 — 120+ 0+ o 
/. a?* + 6x* — 31a? — iJ»o = o is the required equation. 



1-2-67 + 2100 + 588- 


- 1440 


+ 8+48- 


■ 248 — 960 




-la- 


• 72 + 372 + 1440 



Ex.8. 

3. The coefficients being rational, 3, + /i/2 and — ^2 must 
be roots, therefore the equation is divisible hj 

(x -^3) {x'-' fji) {x + V^) = a?' — 3«' — 2ic + 6. 
Division 



I 


I — 10 + 29 — 10 — 62 + 60 


+ 3 


+ 3-21+30 


+ 2 


+ 2-14 + 20 


-6 


— 6 + 42-60 



I— 7 + 10+ 0+ 0+0 

.•. X* — 7aj + 10 =^0 = (aj — 2) (a: — 5) ; 
.'. a? = 2, and x=^5. 

4. Let the roots be a, a, i. 

Then a + a + b= 7, and a(a + &) +aft= 16; 

.'. a* + 2a5, or a" + 2a (7 — 2a) « 16; 

•*• 3a* — 14^+16, or (3a — 8)(a — 2)=o. 

Hence the roots are 2, 2, 3. 

11. Let the roots be a — rf, a, and a + rf, 
then their sum 3a=9; /. a^^^ 
also their product a (a* — <Z') = 15 ; 

.'. 9 — <Z* = 5, whence <Z=±2; 
.*. the roots are i, 3, 5. 



THEOBT OP EQUATIONS. 71 

13. Let a — 3«Z, a — rf, a + rf, a + ^d he the roots, 
then the sum 4a » 10; the product of eyery % and n, 

6a* — IOC?" = 35. 

Hence a = ^, c?=-; .'. the roots are i, a, 3, 4. 

15. Let -, a, or be the roots, 

then a^ = 217, a = 3, 

and ^ + a + ar==^ + 3 + 3r= 13, 

or 3>'— ior=! — 3, 

whence r = ^, or - ; 
^ 3 

.*. the roots are i, 3, 9. 

20. Then a+- + 5+i = ^; 

a 00 

•••(•-:)('-J)-?— f- 

03 6 a 

.*. the roots are 3, - , a, - • 

^3 ^ 

Transformations. 
Ex.4. 
2. Le.,.*.,ft«>(|)'-2(|)%p(|)-25„; 



72 TBANSPOEMATIONS. 

11. Let a?— 2 ^tfy or x^y + 2, and substitute. Then 

(y + 2j)^ = y< + ioy* + 4oy9 + 8oy*+ 8oy+ 31a 

^{y+^)\- 2t/^+i6f + 48f+ 64^+ 32 

- 15 (y + ^y = - 153^^ - 99y" - i8oy - i!Zo 

^i2{y + 2y=^ '■'i2y^- 4%- 48 

-76(y + ^)= - 76y-i^2 

-80 = - 80 



.-. the reqd. equation \ay^ + i2y^ + 4iy^ + 26y^ — i6oy — 336 = o. 

11. Another method. 

I + ij — 15 — i!i — 76 — 80 
+ 2+ 8-14- ^2- 256 



1+ 4- 7-2^6- 

+ 21 + lii + 10 - 


128- 
3^ 


-336 


1+ 6+ s-^^- 
+ 2+16 + 42 


160 




1+ 8 + 21 + 26 
+ a + ao 

1 4- 10 + 41 
+ 2 


- 



I + 12 

.'. y^ + lay^ -t 41^3 + 26^' — i6oy — 336 = o is the transformed 

equation. 

15. ' 2 + o — 13 + 10 — 19 

+ 2+ 2- II- I 

2 -f 2 — li — I — 20 
+ 2+4-7 

2 + 4- 7 - 8 
+ 2+6 

2 + 6 - I 
+ 2 



2 + 8 

.'. 2y^ + 8y3 — y' — 8y — 20 = o is the transformed equation. 



TBAKSFOBXATIONB. 



78 



Ex. 6. 



1. Transform the equation into one whoee roots are thoee 

of the original, diminished each bj - or 2; thus 

•J 



+ 1^-4 



1-2-5 

+ 2 



I -ho 



. •'•y*-5y — 4 = o is the required 

equation. 



11. The roots of the transformed equation are to be those 
of the original, increased each by — or ^ ; thus 



3 + 15 + 25- 
^ 3 


■3 
"5 
9 


3+XO + */- 

- 5-^ 
^ 3 


15a 
9 


3+ 5+ o 

-4 





3 + 



•••3y*-^» 0' a7»'-i5»"0 
is the reqiiired equation. 



14. To take away the third term, die general formula is 

- w (n — i) mk^ + {n— i)pk + y = o. 

2 



Here « = 3, wi = i, /? = 6, j = 9 ; .'. i = 3, or i. 

10 



74 



TKANSFOBMATIONS. 



•'• ^ + 3y* - ao = O. 



•*• y*"-3y*— i6 = o. 



(1) I - 6 + 9 - ao 


(2) 


I — 6+9 — ao 


+3-9+ o 




+1-5+ 4 


I— 3+0— ao 




I-5+4-K5 


+ 3+0 




+ 1-4 


I + O + O 




I-4 + 


+ 3 




+ 1 


1+3 


1-3 



17. Here - n (n - i) wA:* + (n - i);?^ + j = o^^ves 

(3*)"-^(3*) + i = o; •••* = ^- - 

The two roots of the equation in Je being equal, there is onlj 
one transformed equation, which by the process in (14) is found 
to be ajy^ + 6yjf - 59 = o. 

19. Let m=^if and a, b, c I, be the n roots. 

Then -n (n— i) A:* + (w - i) pA: 4- y = o, 

2i 



Eliminating A;, 



nA;+p=o. 



n — I 



•'• JP* : jp* — izy = w : i; 
i.e. (a + S+c + +?)' : (a» + &* + c»+ +2*) = ^ : i. 



SYMMETRICAL FUNCTIONS, 75 



Symmetrical Functions. 



Ex, 6. 



1. Compaxing a:* — a? — i = o, with a^ +px^ + qx + r — oi 
we have ^ = o, j = — i, r = - i. 

Then 8i +jp = o ; .'• fl^i = o. 

82+j>8i + 2q-oi .\ 8^=^ + 2. 

8s+j!>8, + q8, + 3r = o; /. ^3 = + 3- 

8^+jp8^ + q8^ + r8^==o; /. 8^=^ + 2. 

Si+jp8^+q8^ + r8^^o; /• 8^-=^ + 5. 

86+p8^ + q8^ + r8s^o; /. /8i = + 5. 

2. To find flLa; put a? = -, and in the transformed equa- 
tion find 8^. 

Ex, 7. 

1. Let the transformed equation be 

y-^ + 5 = o, 
then if a, i^ c be the roots of the proposed equation, 

^ = (a + J) (a + c) + (a 4- J) (5 + c) + (a + c) (6 + c) 

= a* + J" + c* + 3 (aJ + ac + he) 

= {a + b + cy+{ab + ac + ic) 

=i {(ib+ac+hc)f since the and term of the equation 
is wanting, 

= 40. 

Similarly B= {a + h) {a + c) (6 + c) 

=^{a + b + c)(ab + ac + bc)--r 

= -^ = -39^ 
Therefore the transformed equation is 

aj3 ». 40a? — 39 = o. 



76 BTMMETBICAL FUNCTtONS. 

Ex. 8. 

2. Let afi — iV+ Qx-^B^ o be the transfonned equation, 
then (i) P^ah + ac + hc^q, 

Qs:tab.ac-\-ab.hc-\'ae.hc^a{abc) +h{dbc) •\-c{ahc) 

« aic {a + b + c) =|w, 
Rssab*ae.bc = a^bV « r» ; 
therefore the transformed equation is 

a?* — joj* + prx — r" ss o. 

4. P=a' + i« + a* + c« + J* + c*«=a(a« + 6* + c*) 

a > 

therefore the transformed equation is 

aj» - ii (;?» - aj) a?" + (p* - ^y'j + 52* - is^pr) x 

- (i^ V •" ^P^'* + 4?^^ - 22^ — r*) = o. 
7. P«a3 + J5 + c3«53=i?3-.3^2r + 3y^ 

= j3 + 3r» - 3pyr, 

substituting these values in the equation 

aj5 - iV + 0a? - r = o, 
we have for the transformed equation 

aj3 — (jp3 _ 3^2 + 3^) ar» + {q^ + 3*-" — 3p3^r) aj — r^ s= o. 



or 



STMMETfilCAL FUNCTIONS. 77 

Ex. 9. 

1. Dividing the equation by a?*, we have 

or — loo; 4- JZO h —, »= o, 

X x^ 

whence a? + - = 4, or 6: 

X 

/. iB*— 4xc = — I, or a?* — 6a? = — I, 
whence a? = i^ + V3> whence a? = 3 ± 2 V^ 5 
therefore the roots are 3 ± !^ V^* st ± \/3« 

6. I is evidently a root. Therefore, dividing the equation by 
05 — I, we have 

jc4 — ^ajs + i^ ^a? + i =0, 

whence f a; 4- -j ^ ^a; 4. - j 4. 10 =: o, 

whence aj + - = 4, or ^ ; 

X % 

therefore the roots are i, ij, - , 2 ± V3« 

10. Here - i is a root ; therefore dividing by (a: 4- 1) we get 
a^ — 935** 4- laa?^ — 2003* 4- 1%^ — 93^ 4- 1 = o ; 

Assume a? 4- - = ^ , 

X ^ 

then y*-4y* 4-2-9^*4- 184- iay-iio = o, 
or y*- 13^*4- lay = 0; 

••• y(y-i)(y-3)(y+4) = o- 



79 8TMMBT1UCAL FUHCnONS. 

Patting a5+-«o, i, 3, —4. sucoeasiyelj ; we obtain all 
the roots of tlie original, 

14. a^+i«o, a*-i=o; 

.-. a?+- = ±va, a?--«o; .-. a?=±i. 

3? a; 

Similarlj « — = ±V— a; aj+- = o; .'. a? = ±v— i; 

14,. 3?^ = - 1 = cos (aX 4- 1) IT ± V^ sin (aX + 1) w in its most 
general form. 

By Demoivre^s Theorem, 

(aX + l) W , ^/ . (%\ + l) TT 

aj = cos^ i— + V— ism-^ ^— . 

9 9 

Hence, by giving to X the values o, i, a, 3, 4 successively, 
the roots are found to be — i, cos (n + 20) ± V^ sin (n + ao), 
where w is i, 3, 5, 7 successively. 

Ex. 10. 

1. X=aj*-iw;* — 15^ + 36 = 0; 

.•. Xi = 30?* — 4a;— 15 = (a? — 3) (30? + 5) = o, the ist derived. 
Hence a? — 3 is the Greatest Common Measure of X and X, ; 

•*• (^ ■" 3)* is a factor of X 
By division, it appears that X= (a? — 3)* (a: + 4) ; 

•'. the roots are 3, 3, —4. 



T»TI 



»;■• 



SYMMETRICAL FUNCTIONS. 79 

.5. X= aa?* — ijjaj»+ 19a?* — 6a? + 9 = 6, 

X, = 8a?3 - 360?* + 38a; - 6 = o. 

The G. CM. of X and Xj is a? — 3; 

.-. X=(aj-3)»(aar»+i) = o; 

/. the roots are 3, 3, + f — J . 

•I 

12. The Greatest Common Measure of X and X, is 

a?^ — yaj'H- i3aj*4-3a? — i8 = o = i? suppose, 
then 4aj3 — aiaj*+ii6a; + 3 = o = i?i the ist derived of i>. 

The a. C. M. of D and i?, is a? — 3 ; 

.•. (a? — 3)* is a factor of D. 

Hence J9=s(a;-3)» (aj'-a?-a) = (aj-3)*(a?-ii) (a?+i). 
Similarly (a? — 3)^ (a? — 2)* (a? + 1)* is a factor of X; 
.-. X=(a;-3)^(a;-.a)»(a;4-i)"(a; + 4)j 
.•. the roots are 3, 3, 3, o,^ a, — i, — i, —4. 

13. LetX=^a;"+.aif'*+ai?"'» + ...+Xaj + X = o, 
and r=:^aa?" + jB (a + /8) aT' + C'(a + a/8) aj*-*+ ... 

4-X{a + (w - 1) . /8} a? + i (a + w/8) 
= a (-4aj*4- jRc*"' + ... +Xa? + i) 

+ /8 {jRr»"' + ii(7a?»"»+ ... + (w- 1) Kx + ni}. 

Now X, = w4aj*"' + (n - i) jBa?""* + (w - a) Ca?*"^ + ...'+ X= o, 

since X= o has two equal roots, 

= n (Ax"^' +JBaj*-»+ Cb*-3+ ... +X+ ^) 

.-. -B^'' + iiC7aj**"*+... + (w--i)Zaj + wX==wX-a?Xx. 
Hence r=aX+/8(nX-ajX,) =0, 



80 STMMETBICAL FUNCTIONS. 

16. X«aj* + {a5* — f^ — <«o, has two equal roots ; 

Proceed as for finding the a. c. m. of X and X\ 

Then 5J:=(a:' + |{)-S:,-(3ra:*+|j-aj + 5e-^). 

Hence, a value of x which makes X=s o, and X* = o, must 
also make 

6 AQT 

b o 

therefore one of the equal roots is a root of 

5*" S*" 15 



Ex. 11. 

2. Let « = -; then i — iiy + 36y»-36y» = o; 

Since the roots of the original are in Harm. Prog., 
the roots of the transformed are in Arith. Prog. 

Let a — A, a, a + h be the roots of the transformed. 
Then 30=1 and a(a' — A*)=->; 

/. a = -, A = 7; •'• 2> -> - are the values of y; 
3- 6 6' 3' a ^' 

•*. 2, 3) 6 are the roots of the original equation. 

9. Let a, /3, 7, B be the roots of the proposed equation. 
If a/8 = 30, then 78 = = 40, 

a + /8 + 7 + 8 = -i, . 
a(/3 + 7 + S)+/8(7 + 8)+78 = -6a; 

.•. (a + /9)(7+S) = -6!^-30 — 40=:-i3ii. 



SYMMETRICAL FUNCTIONS. 81 

Hence 

a + /8 - (7 + S) = Vi + 4 X 13^ = '^fs^ - ^3 1 
.•.a + )8= 11; 7 + S=-ijj; 

a-)8 = \/ii' — 4x30= i; 7 - S = Viri' - 4 x 40 = ± 4 V- 1 ; 

Ex. 12. 

I. Since ( — ^J +(— -) is positive, Cardan's Method 
is applicable. 

Leta?=sy + «; theny^ + «^4- (3y« — 9)(y + «) — 14 = 0. 
Also let 3y« — 9 = 0, then y^ 4- «^ = 14 ; 

.-. y^ -«3 - \/i4»-4X 33 = ± !i Va^. 

Hence y»=7 fV^, «* = 7TV53; 
.-. a: = (11-6904158)* + (^^-309584^)* = 3-591. 

II. Assnine v + - = », then y^ » « 3 s-s o. 

^3 3 ^ i^7 

Since ( — -j - (~) is negative, Cardan's method is not 

applicable. 

Let y = w cos ^, 

then m' cos^ r cos ^ -^ = o, 

3 «7 

«zi 8a ^ iai3 ■ 

or cos^ z cos ^ = o. 

3m' a7»i* 

Now cos^ ^ — 2 cos ^ cos ^^ = o ; 

4 4. 

... ii=3, ... ^ = £V85, 
3m» 4' 3 



J I ^ 1213 /, IJJ13V8JJ 

and - cos ^^ = — =? . /, cos ct0 = — ^ ^ ■ . 
4 ^ ij7w3' "^ 13448 



11 



82 BIQUA01LA.TIC EQUATIONS. 

Henoe, hy the tablesi 



a ./^r- 



/. y = - VSa COB ^ = 5*9io5a78. 



Also y = ^ V^ ooa (?:!!±^ , or "i V85 cos (^^il^lSf) 

«_2 V88008(48<'i5'i7"-6o3), 
3 

or — VSa cos (71** 44' 4a"*396) 
3 

= - 4-0194923, or - 1-8910355 ; 

/. aj« 8-5771945, -1-35282507, -77563- 



Biquadratic Equations. 



Ex. 13. 



Then 



1. By Des Cartes' Method. 

Let (a5"4-ea?+/)(ic"—ea?+5') =0^-30?* — 420? — 40 = 0. 

Equating the coefficients of the same powers of a?, we get 

^+/-e» = -3, e5r-6/=-42, j& = -40. 

a^X2/-(6«-3-^)(6«-3+^):-64-6e-+9-l^=-4X4o; 

.-. e* — 6e* + 1696* — 1764 = o. 
Let e" =y, then y^ «. gy* 4. i5^y _ 1^5^ — q. 

By trial of square numbers, 9 is found to be it root of this 
cubic ; 

.-. « = 3» /=5(9-3+H)==io, 5' = -4- 

Hence a?* + 3a? + 10 = o ; .-. a? = -- {— 3 ± V— 31}, 
and aj" + 3a5— 4 = o; .'. aj = 4, and — i. 



BIQUABBATIC EQUATIONS. 83 

2. ajf— ^535" — 6oaj — 36=0. 

By Euler's Method. ^ 

Let x^y + z + u. 
Then a^=y' + «* + M* + i*(y« + yM + «w); 

.-. os^'-iza^ (y* + «* + w*) + (y* + «* + w»)* 

= 4 (y»a» + y V -H«f V) + 8y^?w (y + « + <*), 

or , aj* - iiiB* (y* + «» + w") - 8a?y««* + (y» + «* + 1**)« 

- 4 (y'«* +y V + «v) = o. 

Comparing tlis equation with the proposed 
- as* — iJ5ic' — fioa? — 36 = o, we have 

• ' • , i^^ 60 I s 

y* + «' + w' = -^, y«w = -^=^, 

(y* + «' + u'^y - 4 (y"«" + y V + « V) = - 36 ; 
.-. y V + y V + « V = 9 + - (— y =^ . 
Hence the values oit/*, z\ u' are the roots of the equation 

a 16 4 

Let f = -; theii ^5 — 50^"+ 7(594 — 3600 = 0, the jroots of 
4 / 

which are 9, 16, 25 ; .*. the values of ^ are - , — , — . 

4 • 4 4 

Hence y=:±^, « = +2, w = ±^. ^^ 

Now 0? =y 4- ;5 + w, with the condition that ysw = + — ; 



r:j 



0? =5 - 3 + jj ^ 5 «. _ j^ 
05 = 4- - - a - - == - ^. 



84 LIMITS OF BOOTS. 

By Ferrari's Method. i^ 

Take away the second term by putting aj=y + 1, the re- 
sult is 

y* - %* — i^y + 21 = o. 

Then y^ = 6y' + i6y-ai, 

add a^* + i* to both sides; 

/. (y* + Jfe)'==(ji^ + 6)y' + i5y4-fc" — ai a perfect square; 

i6* 

/. {2k + 6)(J(^'- ai)= — = 64; 

.•. ^ + 3i" — 21A; — 95 = 0; of which 5 is a root. 

Hence y* + 5 = ±(4y + ^); 
•'• y*-4y+3 = o; .-. y=i, andy=3, 

y*+4y + 7 = o; .•. y = -a±v^. 
The roots of the original are .'. 4, 2, — i ± V— 3. 

Limits of Roots. 

Ex. 14. 

1. Transform the equation into one whose roots shall be 
less by h than the roots of the given equation, 

then (y+A)3-4(y + *)'-4(y + *) + «o = o, 

or {h? - 4A»-4A 4-iio) +(3**- 8A - 4)^ +(3A-4)y*+y^=o-..(i). 

Substitute the numbers i, 2, ^, 4, &c. successively for A 
in (i^, and the first number whicn makes all the coefficients 
positive will be found to be 4; 

/. 4 is the No. next greater than the greatest positive root. 

4. ^ = iK* — aa?^ — 3JB* — 15a? —3 = 0, 

f'x = 4rc3 - 60^ — 6a? — 15 = o, 

ya? = lizaj^ = liza; — 6 = 0, 

f"'x= 24X— 1JJ = 0, 

/^^a?= ' 24 



RATIONAL BOOTS. 85 

And substituting the numbers i, a, 3, 4, &c. successively 
for X in these expressions, we find that 4 is the first numb^ 
that makes them all positive ; 

/• 4 is the superior limit to the roots* 

10. There are but ttoo changes of sign, viz. firom + to — , 
and from -^ to +1 therefore there cannot be 'more than 2 positive 
roots. 

Again, there is but .one .permanence of the same sign, viz. 
— , — y therefore the equation cannot have more than i negative 
root. 



Eational Boots. 

Ex. 15. 

1. Here the limits of the roots are 7 and — i, and the 
divisors of the last term are therefore 6, 4, 3, 2; 

.'. a = 6, 4, 3, 2y 
?a = -^=-4» -6, -8, -12, 

J, + 22= 18, 16, 14, 10, 

Qi + 22 
?a=^^^-^— = 3> 4y X, 5» 

?a+(-9)=-<5, -5, X, -4, 
?3=^-^ = -I| X, X, -2; 

therefore 6 is the only commensurable root, since it does not 
satisfy the equation 305" — i8a? + 22 = o ; .*. dividing the given 
equation by X'-'6, we have 

a^-3a: + 4 = o, 
whence « = i {3 ± (- 7)*}. 

4. aj3-.5ic»-i8aj + 72==o=/(a:) suppose. 

Changing the signs of the alternate terms, we have 

a^ + 5^5* — iSx — 72 = o. 



86 KATIONAL BOOTS. 

Hence tiie roots of f{x) = o lie between 7 and — 5. 

But /(I) -50, /(-I) = 84; 

/. the only admissible divisors of 7 a are those which when 
diminished by i, divide 50 without remainder, and are 

fi> 3» ^> -4; 

also the only admissible divisors increased by i which divide 
84, are 

/. 6, 3, 2, - a, — 3, — 4 are the only divisors which need be 
tried; 

/. a= 6, 3, a, - a, - 3, - 4, 

J, = 12, 24, 36, - 36, - 24, - 18, 

?i + (- 18) = - 6, 6, 18, - 54, - 42, - 26, 

' ffa=-i, «» 9f /*7» ^4i 9> 

?a + (- 5) == - 6, -3, 4» ^i^, 9> 4» 

?3=-i, -I, 2, -II, - 3, - i; 

/. the roots are 6, 3, — 4. 

9. 8aj» — 26ic* 4- no; + 10 = o. 

Let y = 8aj, then y^ — 26y* + 88y + 640 = o ^fijf), 

the roots of which lie between 27 and — 7. 

But /(i) = 703 and /(- i) = 535 ; 

.*. the only admissible divisors of 640, which, when diminished 
by I, divide 703, are 20, 2. 

And the only admissible divisors increased by i which 
divide 535 are 4, - 2 ; 

/• the only divisors which need be tried are 20, 4, ± 2 ; 

/. a= 20, 4, 2,-2, 
J, = 32, 160, 320, - 320, 
J, + 88 = 120, ^48, 408, — 232, 
ja= 6, 62, 204, 116, 
ya- 26 = -20, 36, 178, ,90, 

?3 = --i> 9)89, -45; 

.*. 20 is the only commensurable root, for it is not a root of 
the equation 3^* — 52^ + 88 == o ; 



Sturm's theobex. 



87 



.*. -o- = ^ is the only commensurable root of the given 
equation. Divide the given equation bj 2:17 — 5, and we have 

4^ — 3a? — a = o ; whence « « g (3 ± V41). 

12* Assmne a? = ^. and we have 





Let 


BMBltiL 


DivlionL 


Pn«Mriaik 


y= I... 


7656 


i,»,3,4,6,8,ii,ia,aa,&c 


3 8 


=s 0... 


907a 


I, a,3,4,6, 7, 8, 9, la, 14, 16, i8,ai, »4,&c. 


4 9 


=— I... 


9350 


i,a,5,io,i7,ao,&c. 


5 10 



.*• the only nnmbers to be tried aie 4, 9, both of which are 

roots of /(y) == o ; 

.'. a;==^ = - and ^ the only conmiensorable roots. 
6 3 * 

Now divide the given equation by (3a: — a) {%x — 3) « o, 
and we have 

aj*+ iia:+7 3so, 

whence «= -{— n ± (93) }• 



Sturm's Theorem. 



1. Here X= 0^ + 20^ 



Ex. 16. 

3« + «> 

3,^335* + 6ic*—9a? + 6^0?+ I 

yc^ + 43?* — 3» 



or 305" — 90;+ 9 
3g« + 4a?-3 

.'. Xa = i3aj— 12. 



xbyi, 



88 



stubm's theobbm. 



13 



i3aj-ia^39«*+5w?-39iw3a:+88 

88a?— 39 



1 144a;- 507 
1144a;— 1056 



Otherwise, 

since X, = X^Q, — X, 



i» 



if a? = — , then X. = o, 
.13 

and Xi is positive; 
/. X3 is negative. 



+549 

/. X = aj5 + iWJ* - 3a? + a, 
X, = 3a;» + 4a;-3, 
Xa=i3a?-i», 

Since there is one change of signs in the ist terms of X, X„ 
Xg, X3 ; tiierefore there is one pair of imi^inaiy roots in X = o. 

The limits of the real roots of X^ o are 4 and — 4. 

X Xx X, X3 
If a? s= 00 , the signs are + H- + — , one change 

aJ = o, + - 

aj = — J, + - 

a?=-i», + + 

aJ = -3> + + 

aj = — 4, — + — — , two changes. 

Hence, the real root lies between — 3 and — 4. 



6. 



Here X = 
X,= 
X,= 
X,= 



Ql^ — 2'JX 

«"-9> 
xA- %y 

+. 



-3<5, 



Hence all the roots of X=* o are real and lie between the 
limits +6 and —5. 



Sturm's theorem. 89 

JL ^i ^g ^3 

Ifo^ssoo, the signs are 4^ + + +1 .••onerootisposi- 
ojsso, — — + + ltive,andtworoots 

0? = — CO, - + — + j are negative. 



x = 6, + + + +) 

x = 5, - + + +3 

- - -^ n 

+ - o +j 



••.•••••••••••• 



0? = — !^, 

0? = — 3, + o - + 

a?=:-4, + + - +1 

a = -5> - + - +j 

Hence the positions of the roots are indicated hj 

{6, 5) '7 {-I, -a}; {"4, -5}. 

14. Here the frmctions are 

X =0^ — aj* — 4a*+4a?+ i, 
JCi = 4i»3 - 3aj« — 8a? 4- 4, 
^a=-7»* + 8aj + 4, 
JC3 = 4aj — 5, 

Since there are two changes in the signs of the ist terms of 
Xand its four aiudliarj functions; /• X^o has two pairs of 
imaginary roots. 

15. The functions are 

X = a?^ — 30* — !Z4x^+ 95a:* —463? — loi, 
X, = 5aj*- izsc^ - jax^ +. 190a? — 46, 
X, = ^7603* - 1^090" + 35oaj 4- 2,663^ 
X3 = SSySQ^Spcf" - 4io54aj + 359401^5. 

Since in X3, four times the product of the extremes is > the 
square of the mean, therefore every possible vMue of x will 
make X3 positive. 

Now X = X^C.— X^; therefore the sign of X^ is always 
negative ; nence whatever be the value of x, the signs of X3, 
X4, X| must be + ± -. 

12 



90 sraRM's thsobeh. 

Therefore Xa o has one pair of imaginary roots, and the 
position of the real roots will be indicated by the chaoiges of 
sign of the functions given aboye. 

If a; » Qo , the signs are+ + + +1^^ .. 

_ ' ^ _ _ J. . I • • ^J^® root IS +, 

"^ ' . I two roots are —. 

aj«-co, - + - +j 

The limits of the roots of X= o, are + 5 and — 6. 
If a: = 5, the signs are + + 



+ +) 



a? = 4> - + 

a? = o, — - + +1 

a? = -i, + - + +J 

a = -3> + - - + 

aj = -5, + + 

0? = — 6, — + 



::} 



Hence {5, 4}, {o, - 1], {- 5, - 6}. 

19. Here the functions are 

X =«' — aa?^-3aj5 + 4a?'-5a; + 6, 

Xx = 7ic** - looj* - 9a^ + 8aj - 5, 

Xa=aa^ + 6iD3— ioar*+ I5aj-iji, 

X3 = 6ijaj*— jox^ + 1230^ — 1630? + 10, 

X4 = — 6ri9i»' + 12,660:^ — 28300; + ^933, 

X^ = - i8o7i6o5ic» - 47635809^ + 10503345^5, 

Xe = - 4^353^ + 65080, 

X^ = — ; 

.*. if a; = 4- Qo , signs + + + H give i variation, 

0? = — 00, — H hH h— ... 6 , 

the proposed equation has 5 real roots, and one pair of imagi- 
nary roots. 



I 






sxubm's theorem. 91 

The limits of the roots of -X'= o are + jj and — 5. If 

Vara. Van. 



a? = o, signs H 4-4-+H — 

a?=i, + +4-4— 

X=Q,, 4-4-4-4 



3 x=o, signs4- 4-4-4-4-- 3 

3 aj=-i, H 4-4-4-4-- 3 

I ic=-a, -4--4-4-4-4-- 4 

•c ^^ 3' •- ^ — .^ -i- -i- -l- — . ^ 

a?=-4> -4--4-4-4-4"- 4 

a?=— 5> — I H4 h— 6 

hence it appears that the roots lie thus 

{2,i}y {a, i}, {-5» -4}, {-5) -4l, {-a, -i}. 

20. Here 

X=af+ Z4x^ 4- 12533* — 37635* — 175^60?' 4- 5592X — 4080, 
Xj = 30:^ 4- 6033* 4- 250x^ — 564ar* — 17260? 4- 2796, 
Xa = 533* 4- 68aj* 4- 5%oc^ - 908a? 4- 1018*4, 
X3 = 529053 4- 1635603* - 296470? 4- 13934> 
X4 = — 1315944103*4-2576611603 — 14426791. 

If X4 = — <za^ 4- Jo3 — c, it is obvious that (— 2a) (— 2c) is > h\ 
hence the roots of X^ o are imaginary, and every possible 
value of 03, will make X^ negative. 

Now X^^X^Q^ — Xe; a value of x which makes -^ = o, 
will make 03^ negative and hence X^, the absolute function is 
positive. 

Hence the signs of X4, X^, X5 will always be — ± 4- ; 
.•. X= o, has two pairs of imaginary roots ; 
the limits of its real roots are 4- 2 and — 6, 

^ uLi ■^% ^3 ^^4* 

m 

If 03 =00 , the signs are 4- 4- 4- 4- — 
03 = 2, 4-4-4-4" 



: 



a=i, -4-4-4- 

03 = 0, - 4. + + 

03 = — 2, — "f 4- + 

0/ ~~~ "~ Am .•••........... ■™" ^~ "X* t" ^" 

03 = — 
03 



one root 



-6, + - + + -) 



one root 



therefore the positions of the two real roots are expressed by 
{a. A {-5. -6}- 



92 APPROXIMATION. 



Appboximation. 

Ex. 17. 

2. To find a real root of «?• — 5aj-3=s3; by Newton's 
Method. 

Let X=a?' — 50? — 3, and -T «= 30?* — 5. 

Ifaj= a, thenX=-5| /. the real root of X=o Uea 
x^ 3^ Xb + 9> between 2 and 2*5, but nearer to 

x^rs, Z« uj tl^® la^r- 

Let a? = a*5 + 8. 

Then S'=-f5,) =-^« — L«--oo909; 

.*. X » ^'4909^ a first approximation. 
Again, let a; ss a'49 4- S'* 

Then S':= - f 1) « + :5H25i = + .ooo85; . 
.*. X a ^49086, a second approximation. 

6. A root of 6x^ — 141a: + ^^63 = o, is found to lie between 
— 5 and — 6. 

Let 05 = —y, then 6y^ — 141^ = z6^. 

5 ••• y ••• ^ 



750 ... 6y* ••• 1^96 
— 705 ... — I4iy ... —846 

45 ... +2,63 ... 450 

Then 450-45 : 263-45 = 6-5 : Sthe ist correction ; 

/. S = — = '53... and y=s 5*53 is a ist approximation. 
Again, 5-59 ... y ... 5*60 



io48*o6i ... 6y^ ... io53'696 
— 788*19 ... — I4iy ... — 789*6 

259*871 ... + 263 ... 264*096 



▲FPBOZIMATION. 98 

As befoie, 4*225 : 3*129 » *oi : i* the and ooneetion ; 
/. S' s —2 — z = •0074.. • and y a 5*5974 the and approzunation. 

Again, 5'597 *** y **> 5'598 

1052*0 ... 6y> ... 1052*56 
789-17 ... — I4iy ... — 789*318 

262*83 ... + 263 ... 263*14 
.•. •31 : '17 =5 -ooi : -0005 the 3rd conection. 

Hence y = 5*5975 "^^ .*.«=- 5*5975. 

. i 

11. A root of («*+ 3a; + ir + (20" + 1)* » 2*617, lies be- 
tween o and I. 

By Double Position 

o*«* tin •■• I 



I ... (a:*+3aj+i)* ... 1*7099 

1 ... (2»*+l)* ... 1*2457 

2 ..► 2*617 ... 2*9556 

Then *9556 : •617 = i : S a ist correction ; 

/. 5 =s ^ — '- ss *64... and a; » '64 is a first approximation. 

Again, '6^ ... x ... '64 



1*48683 ... (ic* -h 3a? + i)* ... 1*49324 
i'i2397 ... (2£B*+i)s ...1*12713 

2*6108 ... 2*617 ... 2*62037 

Then '00957 : '0062 = 'oi : 8* a 2nd correction ; 

•0062 
.*. S' « ^ = '00647, ^^^ •*• ^ = '^3^47 neaxlj. 
957 



94 



APPBOZIMAnON. 



14. One root ol ihe equation is found to lie between 3 and 4, 



4 

A 

7 

J. 
10 

13 

16 
JL 



193 
3 

i9'6 

3 
19-9 

3 

ao*» 

3. 

aM"5 



~a 

19 

49 
39 
88 
48 



10 

57 
67 

147 
214 
264 



478 
42*537 



136 

579 

14179 

5-88 


520*537 
44*301 

564-838 
46*092 


147-67 

5*97 

153*64 
6-o6 


610*930 
8*037 

6i8*96|7 
8-09 


i59'7o 
1-03 


627-0I6 
8*2 



i6o-7l3 

I 



-a 
201 

642 



841 
156-1611 

997-1611 
169-4514 



1 166-6 1 25 

3Q'9484 
1197-560 9 

31*353 



ia28'9i4 

^544 

i^3i'45|8 
a'54 



161 '7 

I 



6353 

■6 



6|3l5l*9 i|2l3l5ro 



1234-00 

-SP 

1234-5I0 

•5 



962^-35484874 
597 



365 
299-1483 3 

65*85167 
59*87805 



5*973<5» 
4*92583 



1*04779 
•98760 



6019 

4940 

1079 

988 

91 
85 

5 

5 



a root is 3*35484874. 



17. Now 3 is the iSist figure of the greatest root. 

Let a; = 3 + ;^, 

and we have the transfonned equation 

5af^ - 7aj'» - 60? - I = o, 
the first figure in the root of which is 2,. 

Put therefore a/ = a + -77 , 

X 

and we have for a new transformed equation 
the first figure in the root of which is 2,6. 



APPROXIMATION. 95 



Again let a?'' = a6 + -777 , 

and effect a third transformation, which will be 

6o3a?'"3 - 653a?"'' - 5!zx"' -1=0, 
the first figure in the root of which is i ; 

which furnishes the converging fractions 

7 185 195 &c 
3' a' 53' 55'^' 



18. The limiting equation is the first derived of the given 
equation, and is 



.*. {x — J) (a? — c) + (a? — a) (x^cj^ {x—a) (x-rl) — m'— »V^'=o , 
i.e. 30?* — ^^ {a + b + c) x + ah + ac+lc=^m^ + n* + p*. 

Now, since {a + h + c)* is > 3 (aJ + oc + Sc) 
the roots of the limiting equation are reaL 

Hence the 3 roots of the proposed are all real. 



(96) 



GEOMETRY. 




MISCELLANEOUS THEOBEliS AND PROBLEMS. 

1. Let ABG be the triangle ; and AD the line bisecting 
the vertical angle £^0 and meeting the base in D. 

Then, in the triangles BAD, CAD, 
we have BA » CA^ AD common to 
both the triangles, and the angle 
BAD = angle CAD. 

Therefore (Euc. I. 4), BD^DC, 

zsAlBDA^lCDA. 

Therefore (Def. x.) AD is perpen- 
dicular to BG, and it has also been 
shewn to bisect\B(7. 

2. Beferring to the figure in the preceding example, since 
in the triangles BAD, GAD, we have BA=CA, BD^DG, 
and AD common to both triangles. Therefore (Euc. i. 8), 
z BAD = z CAD, or AD bisects me vertical angle ; and 

z BDA = z CD A, or AD is perpendicular to BG. 

4« Let BAG be the triangle, AD the line bisecting the 
angle BAG, and the base BG. Produce AD to E, and make 
DE^AD, and join BE. 

Now, in the triangles ADG, BDE, 
we have AD = DE (by construction) ; 
BD = BC (by hypothesis) ; and 

z BDE^ z ADG (Euc. l 15). 

Therefore (Euc. i. 4) BE==AG, 

and angle BED = angle i?uil G; 

but angle DAG^ angle JS42) (by hy- 
pothesis) ; 

therefore z JSEZ) = z BAD; 

and .'. BE^BA (Euc. i. 5) ; 




XISCELLANEOnS THBOBEHS AND PB0BLBII8. 



97 




but BE^AC. 

llieiefoie BA^AG, 

or the triangle BA C is iaoeoeleB, 

6. Let AjBh^ the two given points, and CD the given 
line. Draw AF perpendicular 
to CD J and produce it to Ey 
making FE eqnal to AF^ and 
join BE catting CD in G. Join 
also Aa. Then AG and BG 
make eqnal angles with CD. 

For since AF is eqnal to 
FEj and FG is common to 
the two triangles AGF, EGF, 
and the included angles AFG^ 
EFO are equal ; 

therefore the base AG is equal to the base EG^ and the 
angle AGF to the angle EGF; 

but the angle EGF is equal to the vertical angle BGD ; 

therefore the angle AGF is equal to the angle BGD; that 
is, the straight lines AGj BG make equal angles with the 
straight line CD., 

9. Let AB be the given straight line. 

Upon AD describe an . eauilateral 
triangle ABCy bisect the angles at A 
and B h^ the straight lines AF, BFy 
meeting in F\ throu^ F draw FD 

Parallel to ACy and F^ parallel to BC. 
I'hen AB is trisected in the points D, 
E. For since ^ is parallel to FD, and 
FA meets them, therefore the alternate 
angles FAC, AFD are equal; but the 
angle FAD is equal to the angle FAC, hence the angle DAF 
is equal to the angle AFD; and therefore DA is equal to DF. 
But the angle FDE is equal to the angle GAB, and FED to 
CBA (i. 29), and therefore the remaining angle D^^is equal 
to the remaining angle ACB. Hence the three sides of the 
triangle DFE are equal to one another, and DJF^has been shewn 
to be equal to DA ; 




therefore AD, DE, EB are equal to one another. 



13 



88 



MISCELLANEOUS THEOREMS AND PBOBLJ^MS. 




13. Let 2>, J^y jP be the middle points of the sides AB, BC^ 
^0 of the triangle uICjS. Draw the per- 
pendiculars EOj FO^ meeting in O. The 
perpendicular at D also passes through 

a. 

Join OD, a A, OB, 00. Since 
AF^FO and FQ is common to the 
triangles AFG, CFG, and the angles at 
F are right angles, therefore AG i& 
eqnal to GC. Similarly it may be 
shewn that GO is equal to GB, and therefore AG is equal to 
GB\ but AD is equal to DB, and DG is common to the 
triangles ABG, BDG; therefore the angles at J) are equal, 
and therefore right angles; or the perpendicular at D passes 
through G. 

17. From B the right angle of the triangle ABO let BE 
be drawn perpendicular to the hypo- 
thenuse, and BB bisecting the angle 
ABO : the angle EBB is half the 
difference of the angles BA O, BOA. 

Now z EBB = ABB -- ABE 

^ABD^O. 

Also zEBB^EBO-DBO^^A-BBO; 

.-. EBB = ^{A^O), amce ABB^BBO. 

25. Let E, F, G, Hhe four points at equal distances from 
the angles of the square ABOB. Join 
EF, FG, GH, HE', EFGH is also a 
square. 

Since AH^EB, md AE^BF, 
and the angles at^^l and B right angles; 
therefore HE^EF, and angle AEH^ 
angle BFE. Similarly it may be shewn 
that EG and GF are each of them 
equal to HE and EF; therefore the 
figure HEFG is ecjuilateral. It is also 
rectangular ; for smce the exterior angle FEA is equal to the 
interior augles EBF, EFB, parts of which AEH and EFB are 
equal ; therefore the remaining angle FEH^ the remaining 
angle FBE; and therefore is a right angle. In the same way 
it may be shewn that the angles at F, &, H are right angles ; 





MTBCELLAWEOft THBOBBMS AND PB0BLB1I8. 



99 




and therefore EFGH being eqnilateral and rectangnlar is a 
square. 

29. From the angle ii of the triangle ABC let ilD be 
drawn to the middle point of the 
opposite side ; the squares of AB^ BO 
are together double the squares of 
AD, DB. 

From A draw AE perpendicular 
to-Ba Then (Eucl. IL la) 

AB'^AD' + DB'^^BDxDE, 

and (EucL ii. 13) 

A0^^AD^'\'DC^-20DxDE 
^AD'^-BB' -%BDy.de, 
whence AB^ + ul(7* = ^AD' + xBZ)*. 

32. Let ABG be any triangle; and let JSP, GQ be the per* 
pendicnlars on AG, AB. 

Then (Eucl. 11. 7) 
AQ'-BQ'^AB'-iABxBQ, 

and (Eucl.i.47y and Axiom i) 
AG^-BG'^Ai^'-'BCf. 

Therefore 
A G'^BG'+AB^-2AB.BQ. 

Similarly ^ 

AB'=BG^+AG*'-%AG.GR 

Therefore BG*=^AB.BQ + AG.CP. 

35. Let ABG be the triande having angle ABO 
j4GB= twice the angle BAG. Draw BD ^ 

bisecting the angle ABG, and BE perpen* 
dicolar to AG; then (Eucl. i. 6) 

AD^BD^BG; 

and therefore DE-FG. 

Now (Eucl. II. 12) 

AB" = BD^ + .4i)» + 7.AD.DE 

=^BG^ + AD.{AD + 2DE} 

= BG^ + AB.BG. 




« angle 




100 



MISCELLAKE0U8 THEOREMS Ad PBOBLEMS. 




36. Let ABO be a triangle, and AP^ BQ, CB perpendicu- 
lars from the aneles A^ J?, C on the 
opposite sides. The circle described 
onACsiB diameter will pass through 
the points JS, P, because the angles 
ABUt APC are right angles; hence 
angle APB ^Ansle AGjB=^9^ right 
angle — angle CAB; and BPB = a 
right a3igh-APB = GAB. Simi- 
larly the circle described on AB as 
diameter will pass throu^ the points 
<?, P; and ^APQ = ZABQ = a right 
z- angle CAB. 

Therefore z CQP^ a right z - APQ = CAB, 

ox zBPB^zCPQ, 

and therefore the complements of these angles are equal, 

or /.APB:=^jlAPQ. 

Similarly it may be shewn that 

zBQP^^BQB, mdzCBP=^CBQ. 



37. Let ABC be the given triangle, and P the given point. 
Bisect BC in I) ; join AD, PJ); 
and from A draw AE parallel to 
PB; join PE; PE bisects the tri- 
angle ABC 

Since AE is parallel to PB, the 
triangle API) is equal to the triangle ^ 
EPJ); to each of these equals add 
the triangle PBC; .\ CPJE= CAD = half the triangle ABO. 




40. Draw C^parallel to AP; and therefore equal to it; 
therefore also BE = PD, and 
BP^DE. Since CE is paral- 
lel to AB, 

PB: CE{::DP:DE:: 

BEiBP):: CEiPQ, 
otPB:AP::AP:PQ; 

whence AP' = PQ . PB. 




MISCELLANEOUS THEOREMS AND PROBLEMS. 



101 



42. Let AB, CD be the chords intersecting in E] from 
centre 0, draw OF^ OG perpendicular to 
ABy CJ)y and therefore oisecting them: 
then 

difference of AE, EB^ %EF, 
and difference of GE, ED = %EG: but 
AB"^ CD' = 4Bir* - 4(7fl^» 

= 400^ '^40J^ 
= 4£!Z?^^4jg^G^ 

= {7,EF)^^ (zEG)^ 

= {AE-EB)'^{CE'-EDy. 




46. Let AB be any diameter of the circle, and AD^ BF 
tangents at A, B; also DF a tan- 
gent at any point C: and let be 
the centre; join DO, FO; the angle 
DOFis a right angle. 

Join CO. Then since CF^FB, ^ 
CO s= OB; and the angles at C and 
jB being right angles, are equal, there- 
fore aCFO^aOFB, and CFB is 
bisected by FO. 

Similarly it may be shewn that 
the angle ADC is bisected by DO. And since the angles GFBj 
CD A are equal to two right anffles, therefore CDO and CFO are 
equal to one right angle, and therefore (Eucl. i. 3^) 

the angle D OF is a right angle. 




48. Let A be the centre of the 
central circle, and J3, C the centres 
of two others. Then, the triangle 
ABC being e(]^uilateral, the angle 
BAC is one-third of two right an- 
gles; and (Eucl. i. 15, Cor. %) there 
may be six such angles around the 
point A\ and therefore, six circles 
can be placed round the given circle. 




102 



MISCELLANEOUS THEOBEMS AND PBOBLElfS. 




50. Let AB be the common tangent to the two circles. 
Draw CD parallel to AB, touching 
the inner circle, and let EF be anj 
other tangent to the inner circle. 
From O9 the centre of the onter 
circle, draw OH perpendicular to 
CD J and OL perpendicular to UF^ 
cutting the circle in Q. Then OH 
is less than OG (Eucl. ill. 7), 
therefore OHis less than OL; and 
therefore BG is greater than EF. 
Similarly, if any other line were taken, it might be shown that 
BC is greater than that line ; therefore, &c. 

53. Let the circles touch each other in the point B^ to 
which let a tangent BA be drawn, and 
from anj point ^ in it as centre, with 
any radius, let a drcle EFO be described. 
Draw the lines AED, AFH, AGI\ then 
will the parts DE, HF, IQ be equal. 

For since AB touches the circle (Eucl. 
III. 36), 

DAxABwABiAE, 

for the same reason AB : AHw AF : AB\ 

therefore, ex a&quo, DA : AH :: AF : AE : 

but AF^ AE, therefore DA = AH; and 
therefore DE = HF. In the same way it 
may be proved that IQ^HF^DK 

66. Let A,B\^ the centres of the given circles; join AH\ 
then AB will pass through the 
point of contact. On AB describe 
a segment containing an angle 
e^ual to the supplement of the 
given angle, in wnich arc take any 

Soint G; join GA, GB, and pro- 
uce them to meet the circumfer- 
ences in D and E; at which points 
let the tangents DE, EFhe drawn ; 
then the angle F at the intersec- 
tion of the tangents will be equal 
to the given angle. 

For FD, FE being tangents 
at D and E, each of the angles 
ADF, BEFia a right angle; there- 





MISOELLANEOUS THEOBEMS AND PROBLEMS. 



103 




fore the angles DFE and i>(7i? together are equal to two right 
angles, or DFE is the supplement of DOE^ and is therefore 
equal to the given angle. 

57. Yxom'A and Bj the extremities of any diameter AB^ let 
fall the perpendiculars AC, BD on any 
chord FH] and let the perpendicular 
BD cut the circumference of the circle 
inJ5?; thtuED^Aa 

Join AE\ then (Eucl. III. 31) the 
angle AEB is a right angle; therefore 
AE is parallel to CD; and AG is 
parallel to ED) therefore the figure 
AGED is a parallelogram; and there- 
fore AG^ED. 

60. Let AS J CD cut one another at right angles in E) the 
sum of the squares of AE, EB, GE, 
ED will be equal to the square of the 
diameter. 

Draw the diameter AF» Join FB, 
BG, FD, DA ; then ABFh&mg a right 
angle is equal to AED, and there&re 
BF is parallel to GD, and (Ex. 59) 
BG= FD. And since the angles at E 
are right angles, the squares of GE^ 
EB are equid to the square of GB; 
i.e. to tiie square of 1)F; but the 
squares of AE, ED are equal to the 

square of AD; therefore the squares of GE, EB, AE, ED are 
equal to the squares AD^ DF; i. e. to the square of AF, ADF 
being a right angle. 

62. Let MN be drawn parallel to AB the diameter of the 
circle AMN; and from any point P in 
AB let PM, PNhe drawn. Take the 
centre of the circle, and join MO, NO; 
and let fall the perpendiculars MF, DG 
on AB. Then since MN is parallel to 
AB, the angles AOM, BON are equal, 
mdOF^^OG. 

Now (Eucl. II. I a), 

MP^ = MO' + 0P» -f aOi^x OP, 

and (Eucl. ii. 13) JVP» = NO" +OP''-%OGx OP, 





104 



mSCELLANEOUS THE0BEM8 AND PROBLEMS. 



whence the squares of JKP, .KP are equal to twice the squares of 
MO^ OP^ or twice the squares oiAO, OF; Le. to the squares of 
AP, PB. 



63. Let ADB be a circle described 
on the radius AB of the circle ACE. 
Draw any line AO meetlne the circle 
ABD in 2>; .^ is equal to DC. 

Join DB. Then the anele ADB 
being in a semicircle is a right angle; 
and therefore BD being drawn from the 
centre B of the circle ACE bisects AC. 
(Eud. III. 3.) 




^ 65. Let 5-4(7 be a circle; AB any chord in it; and G any 
point in its circumference. Let 
AC J jSC7meet a diameter per* 

?mdicular to AB in D ana E. 
ake the centre of the circle; 
draw die diameter BOO. Join 
GC, CO. Since the angle 
OCB is equal to the angle 
OBCy and BGC to FAD, and 
that CBG and BGC together 
are equal to a right angle; 
therefore OCE and FAD to- 
gether are equal to a right angle, and therefore to FAD and 
ADF together; hence UCE is equal to ADO; llierefore the 
triangles COD, COE are equiangular, 

and DO : OC :: OC : OE; 

therefore OD.OE=OC^OA\ 




67. Let ABC be a triangle whose 
acute vertex is A. On BC describe the 
circle BDC, cutting the side AB of the 
triangle in D; join CD; and from A 
draw AP a tangent to the circle at P. 
Then (Eucl. iii. 31) the angle CDB is a 
right angle; and (Eucl. il. 13) BC^ is 
less than AB* + AC^\)y zAB.AD; i. e. 
(Eucl. III. 36) by zAP". 




MISCELLANEOUS THEOREMS AND PROBLEMS. 



105 



68« Let CFH^ CQE be the circles toaching each other in 
the point (7, A and B their centres, 
and D the point without them, such 
that the angle ADC is ^equal to the 
angfe GDB. Let jD(7 cut the circles 
in G and H. Join AG^ Bff, and 
draw the tangents BE^ DF. Then 

BE^^BO.BG, mSiBF^^BG.BH^ 

also the triangles BGAy BCBheiag 
similar, we have ' 

BG : BG^AG : BG, 

or BG I BG^AG : BG. 

And for the same reason we have 

AGiBG^BH 

or AG \ BG^BG 

therefore BG.AG 

therefore BG 

otBG.bg 

or BE^ 




BE, 
BH; 
BG'^BG.AG : BG.Bff; 
BG^BG : BH, 

BH.BG, 
BF*, 



BG^^BG' 
BC'^BG' 
and BE: BG^BG : BF; 
therefore BE.BF=BG\ 



70. From the point E without the circle let EA, EGB be 
drawn touching the circle in A and 
C and let EB meet the diameter AB, 
drawn from -4, in the point B. Join 
AG; the angle AEu is double of 
GAB. 

Through G draw the diameter 
GOF; then the angle FOB is a right 
angle, and therefore equal to EAB, 
and EBA is common to the triangles 
EBA, GOB, therefore the angle 
GOB is equal to AEB. But GOB 
is (Eucl. III. ao) double of GAB; 
therefore the angle AEG is double of 
GAB. 

14 




106 



MflSGBLLAKEOUB THEOSEMS AND FBOBLEMCL 



point 



78. Let AB be the given Btraiglit line, C the 
in which the cirde is to touch it, D 
the point through which it most pass. 
Draw CO perpendicular to AB. Join 
CD, and at the point 2> make the angle 
ODO « aiu?le DGO ; the intersection of 
Htkd lines vO and DO is the centre of 
the circle required. 

Since the angle 

DCO^CDO, CO^DO, 

and therefore a circle described from the centre O, at €ke dis- 
tance ODf will pass through C7, and touch the line AB in (7, 
because OC is perpendicular to AB. 




75. Let A be the given point in the drcumference of the 
circle whose centre is 0; B the given 
point without. Join BA and produce 
it to D. Join OD; and through A 
draw OAE] and draw BE parallel to 
ODj cutting OAEmE. Eis the cen- 
tre of the circle required. 

Since (Eucl. l. 29) the angle ODA 
is equal to ABE, and OAD to BAE, 
therefore the triangles ODA, ABE are 

similMT, and OD being equal to OA, AEvnR be equal to EB; 
therefore a circle described with the centre E^ and radius 
EAj will pass through B, and touch the circle ADE in the 
point Af since the line joining the centres passes through A. 




78. Let AB be the given line in which the centre is to 
be ; BC the line which the circle is to touch. 

Li BC take any point C, and 
draw CD at rieht angles to it ; and 
make CD equid to the given radius. 
Through D draw DO parallel to 
OB; is the centre of the circle 
required. 

Through draw OE parallel to 
DC; therefore CO is a parallelo- 
gram; whence OE^DC, i.e. to the 
given radius. With the centre O, 





MISCELLANEOUS THEOBEMS AND PBOBLEMS. 107 

and radius OE^ describe a circle; it will touch CB in E^ 
because CO beine a parallelogram^ and EOD a right angle, 
CEO is also a right angle. 



80. Let ABO be a triangle having the obtuse angle 
ABG. Describe a circle about 
it, and produce BA to D^ mak- 
ing -4 J? =a AB. From D draw 
DIE parallel ix> AG^ meeting 
the circle in E; join BE, cut- 
ting AO in J^; !fei^ will be a 
mean proportional between AF 
and FC. 

For (Eucl. VL t) 
BFiFEiiBAiAB, 

and since 5-4 =-42); .-. BF^FE. 
Now (Eucl. in. 35) AF.FO^^BF.FE^JSF*; 

.-. AFiFB^FBiFO. 



84. Let ABO be an equilateral triangle inscribed in a 
circle, about which an- 
other, DEF^ is circum- 
scribedy touching the cir- 
cle in tiie points Ay B^ 0. 

Since BA touches 
the circle, the angle 
BAB ^AGB (Eucl. iii. 
3^); but AOB^ABO; 
.-. BAB ^ ABG, and 
they are alternate angles, 
therefore BF is parallel 
to BO. In the same 
manner it maj be shewn 
that AB is parallel to 
FEi therefore ABOF is 

a paridlelogram ; and the triangle ABO is equal to AFO. In 
the same manner AlBO may be shown to be equal to each of 
the triangles ABBy BOE; and therefore it is one-fourth of the 
circumscribing triangle. 




108 



XnCELLAITEOVB THBORBIfS AMD PB0BLB1C8. 




87. Let ABO be an iaoaedefl triangle. Vntw DE paial- 
lAiaBOiyimBE. Then 

BE*»BO.DE+CE\ 
Draw EQ perpendicular to BC. 
Then (End. ii. 13) 
BE* »CE*^ BC + %BC. BO 
-MCE*+BO{iBa-BCI). 
Draw .^UTperpendicnhur to BO; 

then (End. vi. a) KG : 00= AE: EG, 

or B0-2G0:2G0'^AE:E0} 
... B0-2G0:B0=AE:A0=J)E:BC; 
.: BO^aGO^DE, 
or 5(7- a (5(7 -5G^, i.e. 2BG-B0=J)E; 
.'. BE*='OE*+BO.DE. 

89. Let J[J3(r be a triangle having a right angle 0. 
Draw AJD bisecting the angle A 
and meeting BO in D ; then 

DE^DO. 

AO* \ AD* '^ BO '. Q.BD. 

Draw 2)^ perpendicular to 
AB. 

Then (End. vi. 3) AO: OD=ABi BD, 

and (EucL vi. 4) AB'.BO=BD: BE, 

or AB'.BD'^BO'.BE', , 

therefore AOi OJ) = BO: BE, 

eni AO* : OB* = BO* : BE* ; 

therefore ^ (7» : AB* = PC7» : BE* + BO*, 

also BE* = BD'-DE* = BB*-DO* = (BD + DO) {BD-DC) 

= BO{BD-DC)',, 
whence it follows that 

AO* : AB* =B0* : BC(BB-BO+BO) ^^BO : »£i). 




HISCELLANEOUS THEOBEMS ASH PBOBLEUS. 



109 




91. Let ihe tangent 8FT meet the radii GA, OP prodnoed 
in 8 and T. Draw PJf at right angles 
to AG fiXi^ PN id BO. Then 

triangle SCTi triangle AGS 

^GS.CTx GA.GB, 

also 08: CP=GP: GM, 

and GT: OP'^GP: ON; 

therefore 08. 8T: OP* = OP* : ON. GM=> OP* : GM.MP; 
therefore triangle BGT\ triangle ACB^ CP* : CM. MP 

= triangle -4 0!B : triangle CMP. 

94. From airfr point D within the equilateral triangle 
ABG, let perpenoicnlarB DJS, DFj 
DGhe drawn to the sides ; they are 
together equal to BH a perpendicu- 
lar drawn from B on the opposite 
side A C. 

Join DA, DB, DO. Since tri- 
angles upon ike same and equal bases 
are to one another as their altitudes ; 
therefore 

ABCiADC^BHiDE, 

also ABG:BDO=:BH:DF, 

and ABC : ADB = BH; DG; 

whence ABC : ADC+BDC+ADB=^BH : DE+DF+DG, 
in which proportion the first term being equal to the second ; 

therefore DE+DF+DG^BH. 




95. Let ABC be the inscribed triangle, 
tangents at B and 0; and from 
A draw AD, ^JE? parallel \jq BF, 
CQ respectively, meeting the 
base J5G in D and E. 

The angle 
ABF = i BAD -/.EGA 

also the angle 

AGa = /.ABL^/.OAE\ 



Draw BF, GG 




110 



MiaOKUAWEOUS THEORBMS AND PBOBLEMS. 




therefore ihe triangles BAD, ACE are similar, and 

zAOE^zADB; .: AD^AE. 
Hence BD : 0E= ^BAJD : A EGA =^AB' : AC*, 
AB, A being homologous sides of similar triangles. 

98. Suppose A and D to fall 
both wilJtm the perpendiculars BE, 
CF; then 

AC'^AB'+OB' + nAB.FD; 

and BD*='AD*+AB* + iAD.AEi 

therefore 

AO'-BD^i CD* - AB^ == CD* - AB* 

+ 2AD(1>F-AE) : CB*-AB', 

ix AC* - BD* i CB* - AB* ^ J)F* - AE* 

+ %AD {DF- AE) : J)F* - AE* 

= DF+AE+ iAD : DF+ AE 

^BO+AD: BO-AD. 

101. Let D be ai^ point in the peroendicnlar to AB, 
through G', draw the tangents DE, DFtoma circles, and joiit 
DA, DB. 




Since AC^CBx B^r-B-r : AG-CB; 
therefore AG*- GB*'=B*-f*, 
and AG*-J^ = BG*-r». 
Therefore DE*='AD*-AE*=GD* + AG*-B^ 

= GD* + BG*-r*^BD*-r'::^lf^. 
Therefore DE= DF. 



MISCSLIiAKBOUS THBOXBHS AHD PBOBLBMl. 



Ill 



104 Let the angle BAG be bisected by {he line ADi and 
through any point D in this line draw O^DE meeting AO a 
perpendicular to AD ia (7, and the sides in ^and E\ then will 

OEx GF^ED : FD. 




For through D draw jS (7 parallel to ^(7, and therefore perpen- 
dicular to AD\ then the angles BDAy CD A being right angles 
are equal, and BAD = CADy and ./li) is common to the triangles 
BDA, GDAi therefore BD^DC. But DO being parallel to 
AG, 

GE : ED :: GA : DG :: G^^ : JRD ;: GF : ^ 

since the triangles GFAy BFD are similar; 

therefore GE : G^J^s: ED : JRD. 

105. Let ABGD be a quadrilateral, whose sides are bisected 
in Ey Fj Gy H. Let the points 
of bisection be joined, and 
draw the diagonals AG<, BD. 

Since AB^ AD are bisected 
in ^ and J7 (Eucl. vi. %), EH 
is parallel to BD\ and for the 
same reason FG is parallel to 
BDy and therefore to EH. Li 
the same waj it maj be shown 
that EE is parallel to HG, 
and therefore the figure EFGH is a parallelogram. 

Again (Eucl. vi. 19), the triangle EBF is to the triangle 
ABG in the duplicate ratio of EB to AB, i,e. in the ratio of 




118 



inaCSLLANEOUS THBOBEMB AND PROBLEMS. 



1 : 4, therefore EBF is equal to one-fourth of ABGy. for the 
flame reaaon HDO is one-fourth of DAG^ whence EBF and 
HDG are together eooal to one-fonrth of the quadrilateral. For 
{he same reaaon HAK and (7i^C together are equal to one-fourdi 
of the quadrilateral; therefore the four triangles together are 
equal to half the quadrilateral; and consequentlj HEFQ is 
equal to half of jISOD. 

107. Let AGB be an iflosceles triangle, having the angle A 
equal to the angle B. 

Take the point D in AGj such that 
AB^zDG. Produce AG to E until 
AE=i 2,AGf and on DE as diameter de- 
scribe a cude, in which take anj point P; 
then AP^7.GP. 

Join OP, GPj AP. From the construc- 
tion it is evident 

A0^%D0,9XiSiD0^%G0, 

and conseqnentlj AOxOG=^ DO" = OP*, 

or A0\ OP^OPx OG; 

therefore the triangles AOPj POG are equiangular; 

therefore AP i PO^PG i GO, 

or APi PG^PO x.GO, 

or AP \ PG^DO \ GO^T. 

therefore -4P= %PG. 




i; 



109. Let AEBy GED be two circles touching each other in 
E, and a straight line AG in A 

and 0; draw tne diameters AB, ^ 

GD\ ^(7 is a mean proportional 
between AB and GD. 

Join ADy BG; these lines (Ex. 
io8) pass through the point of 
contact E. And since GA touches 
the circle in -4, from which point 
AE is drawn^ the angle GAD is 
equal to the angle ABE in the 

aftemate segment ; also the angle A GD being a right angle is 
equal to the angle GAB-y therefore the triangles AGD, ABG are 
equiangular, and 

BA : AG^AG I GD. 




KISCELLANEOUa THEOBEMg AND PROBLEMS. 



iia 



113. At C and D the extremities of the chord CZ), let 
^perpendiculars to it be drawn 
meeting a diameter AB in E 
and F\ E and F are equallj 
distant from the centre 0. 

Draw OQ perpendicular to 
CDy and therefore bisecting it^ 
then Off is parallel to DF) 

whence 

ODi OF^HGiHO^HGiHE, 

since the triangles HOOy BEG are equiangular; 

therefore (Eucl. v. i8, 15) DG : OF^ GG : OE, 
but GD^GG\ therefore OF^ OE. 




115. Let Ay B be the given points, and GD the given 
straight line. Join BAy and pro- 
duce it to meet GD in E. Take 
EF a mean proportional between 
EA and ER Join FAy FB, and 
describe a circle about the triangle 
AFB'y it will be the circle re- 
quired. 

Since ^jp' is a mean propor- 
tional between EA and EBy EF 
touches the circle (Eud. ill. 37) 
which passes through A and B. 




117. Jj&t ABGD be the given trapezium, and A the angle 
from which the bisecting line is 
to be drawn. Draw the diago- 
nals AGy BB; and bisect jBD 
which is opposite to the angle A 
in the point E. Join AE, GE; 
and through E draw FEG pa- 
rallel to AG. Join AG'y AG 
bisects the trapezium* 

Since BE=i EB, the triangles 
AEBy AEB are equal; as also DEG, BEG; therefore the figure. 
AEGB is ^ual to the figure AEGB. Also (Eucl. i. 38) the 
triangles AEGy GEG are equal ; take awaj therefore the com-^ 

15 




114 macELURBOUB THBonemi jam pboblems. 

mon part EHO, and AES= QHG. To the figure AECD add 
AEH, and take away its eqoal QHG; and to AEOB add GHG, 
snd take away its raoal AEH) and the triangle AGD is eqnil 
to the trapesium A O GB, oi die given txapeDom is bisected by 
AG. 



118. Let ABO be the nven triangle, and P the g^ven point 
in the side BG. Divide BC in 
the points D, E, F, &c. into the * 

reqaired number of eqnal parts. 
Join AD, AE, AF, AP; and draw 
i)(?,^ff,i?7paraUelto^P. Join 
PG, PH, PI; they will divide the 
triangle into the required number 
of equal parts. 

For the triangles ABD, ABE, i 
AEF, AFC, being as their baaea 

will be equal. And since DG is parallel to AP, the triangles 
DQA, BGPaie equal, therefore DBA = GPB. And since the 
triangle ADP=AGP, and AEP=AEP, therefore ADE=HPO. 
Also APE^AHP, and APF~=AIP, therefore AEF=AHPI, 
and AFG^PIC; therefore the parts PBG, GPH. HPIA, IPO 
are equal to ABD, ADE, AEF, AFO, and are consequently 
equal to each other. The same may be proved whatever be the 
number of parts. 



119. Let ABODE be the given rectilineal figure, and D 
the angle which is to be the 
vertex of the triangle, the base 
being in the side AF. Join 
CM, and draw BO parallel to 
it; join OG, DG, DF. Draw 
Cff parallel to JDfl. Joini>fl: 
Draw JX" parallel to 7>j; meet- 
ing ^i^'in K. Join DK; HDK 
wfll be equal to ABCDEF. 

Since BQ is parallel to 
OA, the triangles BA G, BOG 

are equal; the figure therefore is equal to GODEF. And since 
GD 18 parallel to CH, the triangles GOD, GHD are equal. 
Again, sinoe EK is parallel to DF, the triangles DEF, DKF ate 
equal; whence the whole figure ABODEF ia equal to die tri- 
angles DHG, DGF. DKF, Siat is, to the triangle DSK. 




MISCELLANEOUS THEOREMS AND PROBLEMS. 



115 




121. Let ABO be the given isosceles triangle. On AO 
describe an equilateral triangle ADGy 
and from D toaw Z>j& perpendicular to 
A C; it will also bisect A C and pass 
through J?. On DE describe a semi- 
circle; and from B draw J5i^ perpen- 
dicular to DE, meeting the circle in 
JF; with centre E, and radius EF, de- 
scribe a circle meeting jEJZ) in G; draw 
OHy <?/ parallel to I>A, DC respect- 
ively; the triangle GHIis equilateral, 
and equal to AiC. Since G^His pa- 
rallel to ADy and GI to DO, the triangles GHI, ADO are 
similar; but ADO is equilateral; and therefore also GffI is 
equilateral. 

Also (EttcL VI. 8. Cor.) ED : EG^EG : EB, 
and (Eud. vi. a) ED : EG = EA : EH; 

.\ EG : EB^EA : EH; 

and therefore (Eucl. vi. 15) the triangles EGH, EBA are equal. 
But GHE^ GIE, and BAE^ BCE; 

therefore also GHI^BAG. 



124. l^AO be the given base. On it describe a segment 
containing an angle equal to the given angle; bisect the arc 
ABO in JB, and Join AB, BO. The triangle ABC is a max- 



imum* 




Through B draw BE parallel to A 0, and therefore a tangent 
to the circle at B. Take any point D in the circumference of 
the circle ; join AD, and produce it to meet BE in E; join OD, 
CE. Now the triangle AEO is equal to the triangle ABO on 
the same base and between the same parallels ; and the triangle 
ADO has the same vertical angle as the triangle ABO; but 
AEO is greater than ADO; therefore the isosceles triangle ABO 
is greater than ADO* 




116 MISCELLANEOUS THEOREMS AND PBOBLEMS* 

126. Let it be required to divide the circle AKBR into anj 
nmnbery (sttppose four) of equal 
parts. Ihyide the diameter AB 
into the required number of equal 
parts in the points 0, jD, E, On 
A G^ AD, AE describe the semi^ 
circles AOG, APD, AQE; and 
on BC, BDy BE, on the opposite 
side of the diameter, descn oe the 
semicircles BIG, BED, BGE. 
The circle AKBR will be divided 
into four equal parts bj the 
equally compounaed semi-cir- 
cumferences AG GIB, APDBB, AQEGB. 

For AB : AD = semi-circumference ABB : semi-circ. APD, 
and AB : BD = semi-circumference ABB: semi-circ. BED. 

Hence AB: AD^- BD^ARBx APD^BED, 
but AD + BD^ AB; .\ APD + BED = ARB, 

i.e. the compounded boundary APDEB is equal to the semi- 
circumference ARB of the original circle. 

In the same manner it may be shewn that each of the com- 
pounded boundaries BGEQA, BIGOA is equal to the semi-cir- 
cumference ARB] and the same is true whatever be the number 
of divisions. 

Again, the circle on AB : circle on AD = A& : AD^ ; 

also, the circle on AB : circle on AE = AB* : AE* ; 

hence, circle on AD ; circle on AE^ AD* : AE*^ ; 

therefore, circle on AE^ circle on AD : circle on AE 

^AE^-AD'iAI^; 
therefore, 

circle onABia space APDEQA = AB* : AE* - AD* 

= AB* : {AE+ AD) {AE^ AD). 

Bisect the line DE in F; 

then AE+ AD = 2AF, and AE-^ AD « DE; therefore 
circle on AB : % spaoe APDEQA = AB* : %AF x DE, 



XISCELLANEOUS THEOBBMS AND PKOBtBllfl. 



117 



or, circle on AB : apace APDEQA - AB^ : AFx DE\ 
also, ciide on AB : space BGEDHB^AB* : BF\ BE; 
hence, circle on AB : space APBHBOEQ 

* • 

« AB" : ^IFx J0J5+ 5!Fx D^ 

^AB^\ABy.DE 

^ABiI)E% 
bnt JtfJ? is equal to one-fourth AB\ therefore the space 
APbHBOEQ is one-fourth of the whole circle on AB. 

Similarly, it may be shown that each of the other spaces is 
equal to one-fourth of the circle on AB. The same is evidentlj 
trae whatever be the number of equal parts. 



129. Let BC be the given base. 
Draw BA at right angles to -5(7; pro- 
duce AB to Dy and make BD equal to 
the difference between the hjpothenuse 
and the other side. Join D(f, Make the 
angle DC A equal to the angle BDC. 
AJBGis the triangle required. 

The angle ADC being e^ual to the 
angle AGD^ the side AC ^a equal to 
AD J and the difference between A G and 
AB is equal to BD] i.e. to the given 
difference. 




131. Let -4 C be the given hypothenuse. Oxx AC describe 
a segment of a circle, containing an 
angle equal to half a ri^ht angle, in 
which place the straight bne AD equal 
to the sum of the sides. Join DC, and 
make the angle i? 05 equal to the angle 
ADC. Then ABC is the triangle re- 
quired. 

The angle ABC is equal to the two 
interior angles BDC^ DCBy i.e. to 
twice the angle ADCy and is therefore 
a right angle; also AB-k-BC is equal to AD^ i.e. to the sum 
of tne sides. 




118 



ICnCELLANBOUB THBOBBMB AKD PBOBLBlfS. 



184 Let j4(7be ike given lij* 
pothenuse; divide it in the point D 
in extreme and mean ratio. On 
AC deacribe the aemidrde ABOf 
and from D draw DB at ri^ht an- 

flea to ^ (7 to meet it. Jom ABj 
Wf ABG ia the triangle required. 
B7 oonatmction, 



we have AC\ AD^AD : DO^ 

BC\BC\ 
AD. 



and (Eucl, vi. 8. Cor.) AGiBG 

therefore BC 



^ 




Again, gince A G : AB^AB : AD; 
therefore AG: AB^ AB : BC; 
therefore the sides of the triangle are in continued proportion. 



139. LftAB be the given base; on it describe a segment 
ACB containing an angle equal to 
the given vertical angle, ana com- 

?lete the circle. Draw the diameter 
f'E perpendicular to AB^ and there- 
fore bisecting the arc AEB in the 
point E. Divide AB in the point B 
m the ratio of the sides; jom EDf 
and produce it to meet the circle in 
a Join -40, 5C; .4j5(7 is the tri- 
angle required. 

Bj construction the angle AGB 
is^ equal to the given vertical angle; 
and (End. vi. 33) the anrie ACB 
is equal to the angle BCB, or the angle A GB is btseeted bj 
CD ; therefore (Eucl. vi. i) ACiGB :: AD : DB, i. e. in the 
given ratio. 




( 119 ) 



MENSURATION. 



* . ■ 



Abeas op Plane Fiqures. 
Ex. 1. 

6. The height = 4 x sin i6* 45 = 4 x "287539 « 1*150555 ch. 
.-. iirea= 1-150555 x 5-5 = 5-32801 sq, ch. = ar. 21 p. 7^ yd. 

8. 2f rf. =s f ^ ; and £33 i5«. lOjd. *= £33-84375 ; 

/. area of garden = 33-84375 x 95 « 3249 sq. yd. ; 
#•. side of square = V3249 = ^y yd. 

13, 5 gaiDeas«ia5oc7.; 

/. area of rhombus = i^=: 180 sq. yd. ; 

.-.height .^^^^ = 35 feet; 

45 

••. required angle = sin"' (^ = sin"' (^ = 53^ 7' 48". 

Ex. 2. 

8. The remaining angle of the triangle is 83® 30'. 
Jjet a? :=s the side opposite to this angle ; then 
x t 72 = Bin 83* 30' : sin 38** 40' ; 
.^ ^^^^ Bm83^3o\ 



•^ , • 



I sin 82^ ^o' . , 



120 



ABEAS OF PLANE FI<^UR£S. 



acres; 



n. Areaoffield = ^'^^^^'^3 = ri5ia7 

a X lO ^ ' 

.\ rate per acre «-^;^^ = ^5*578i 

= £5 ii«. 6ld. 

13. If a; be the required angle, we haVe 

- X (275)". sin 0? = 6 acres = 6x4840yd.; 

V. sin a? = y ° . , = 768 = sin 50* 10' a8" ; 

,-. a; = 50" 10' 38". 

Ex.3. 

2. Let ABCD be the trape- ' 
zoid. Draw AEDF per- 
pendicular to BG* Let 



« > 4 



penaicuii 
BE=x\ 



then 




6400 lOOOO . Q,OOX 

/. 400 = -^ h ; 

9 9 3-. 

200 7200 j^ 

.'. x^- or aj=i2ft.: 

3 9 \ 

.-. AE^ d = (20» -««)*= 16 ft. ; 

/. area = 8 x (72 + 38f ) = 8 x i lof = 885^ ft. 

3. Referring to the figure in the last example, we have 
BE^AE cot ^ 12' = 36 X 7ai2227 == 25'964. 
Similarlj FC — AE cot 46* 15' = 36 x '957292 = 34'46a5 ; 
.'. AD=BC- {BE + FO) «= 137 - '60-426^ = 76*5735 ; 

.Mi*ea=:i X 36 X (137 + 76*5735) = 18 >< ^^i'5735 = 3844'333- 



ABBA8 OF PLAVB nOUBIS. 181 

9. Ab iZTf 30' 18 the sappleoMiit of 57^ 30', we have 
area = i {690 X 467 + 4^8 X 359} rin 57* 30^ 

« 437941 X '84339 = aoo677*o6 yda. 
== 41*4622^. 



11. .•• 2>-B= Vaao* - 100* « V38400 = I95*96, 
and BF^ Va65« - jd' -* V653115 » 255-588 ; 

/. area =:^ x 378 x (195-96 + 255-588) = 189 x 45i'548 

2 

= 85345-57ayd. = 17-6333 acres* 17 ac. 2 r. 21 p. 

12. Let ABOD be the tcapeziiuBy AC being the diagonali then 

AC^ V335» + 426* = V293701 = 54i-94«, 

area oiLABC^ ^ x 335 ^ 4«6 = 7i355yd. 
and area o£^A2)0 

as V624'97i X 83-029 X 237*971 X 303-971 = 61266*484 ; 
.•. whole area = 61266*484 + 71355 ^ 132621*484 yd. s 27-401 a. 



Ex.4. 



9. The formnlse are^ 



^ i8o* ^ i8o* 



n. n 

and here » = 5 and a =-3 ; 

•-. r = 3 cot 36^ = 3 X 1-376382 = 2-06457, 

and r =» ^0086036^— 3 x 1701302 = 2-55195. 



16 



122 ABEAS OF PLAHE FiaUSES. 

13. The angle sabtended at the centre by each side of the 
nonagon is ^ — s 40P. 

And since the ladluB of the ciicmnscribing ciide is iz^ the 
area of each of the 9 triangles into which the polygon is divided 
by lines drawn from the centre of the circle to its angular 
points is 

-xia*sin40f* = 7!ix '642788 = 46'%8oy ; 
.•. area of polygon = 9 x 45*a8o7 = 4i6'5a6. 

14. Side of polygon = 2r tan = ao x tan : 

.\ area of each of the 25 triangles = 10' tan 7^ i^^; 
/. area of polygon = 2500 x tan 7* i:&' = 315*823. 



Ex.6. 



4. Gircomference = zirr = 1760 yd. ; 

880 



i» 



area = ttt* = i L = 246498*623 yd. 



= 50*93 ac. 



6. See fig. on page 123. 



Let -42) = 18, Gi&=:io; then u4C=Vio» + 9* = Vi8i, 
tan ACE = -2- = '9 = tan42® nearly. 

Hence, sector : ttt* = 2 x 42** : 360*^ ; 

/. sector = — X -TT X 18 = i^2'6. 

30 ^ 



AREAS OF PLANE FiaUBEB. 

7. Ab 2s. 6d. is one-eighth of « pound ; 

/. Io-^g — Softsaiea of thealeoTe, 

or wr*=si6oft.; .•. rssf — j ; 

.•. length of arc = wr = V i6oir « ^^^2*6^6 «= 2%*^ ft 

13. Bj the fonnnla we have 

length of arc : 2irr :: n : 3(So, 

and here 11=^29^, ^"^9; 
.-.length of arc« 5^.5^.^.533. 



1S8 



14. Let ABD he the arc, C 
the centre. 

Let radius = r and CE^x. a 
Then 



4 



(1). 




and (r-a!)» + 4il* = 5l! (a); 

4 4 



.*. (r— a;)* = ^ and r— «= 
^ ^ 4 



13 



(3), 



fix>m (3) and (i), f» — a4r + 144 = r* — ^5*5 ^ 

whence r = — ^ : 

3* 

2 3^ 289 '^ ^' 

.-. z 4 CB== 112*29, 
and length of arc : 27rx — ^ :: 112*29 : 360; 

.-. length of arc = 8<^7^ / "^'^9 ^ -088. 

^ 360 X 16 ^^"^ 



1S4 ABBA8 OF FLAKE FIGUBES. 

Ex. 6. 

8. 10 : awr :: 39^ : 360; 

. ^„ 10x360 , 

^ ^x^ I iox3(5o 18000 ^^ 

/.area of sector »-x ^ — x 10 = ^yz'KtUitL 

2 ygir 79W ' ^ 



9. By the formula in the teact, we hare 

ao : a57r :: n : 360; 

4 X ^60 a88 ^^ • » f# « 

57r IT 

10. Area of sector s: half the area of triangle « i^ioyds., 

and, length of ate : '2vr :: 60 : 360 ; 

.•. arc a= — ; 
3 

.'. area of sector = -jr = laio, 
whence r = (- — j = 48*072 yds. 



Ex. 7. 

• • • ♦ 

1. n^ being the number of degrees at the centre,.^ 

cos^» = i^ = |=»=-a857i43=cos 73^23' 54"; 

.-. n = 146^ 47' 48" = 146^7967 ; 

.*. area of sector : 42* x tt :: 1467967 : 360; 

.•. area of sector = 497r x 14*6^^7 = 2259765 in. ; 

ajea of triangle = - x 42' sin 146** 47' 48" 

= 882 X -5476 = 482*983 in. ; 
/. area of segment » 2259765 — 482*983 = 1776782 in. 



ABEAS OF FLAKE FiaUBBS. 



125 



2* If n be the number of degrees at the centre, we hare 
Bm-n= — s=-=s*6ss Bin4i®48 qy ; 

/. » = 83** 37' 14" = 83**6ao6 ; 
•*• area of sector : i447r :: 83*6^06 : 360 ; 

.•. area of sector = — '— -^ = 105*081 ; 

5 

area of triangle = - x 12' x sin 83*^ 37' 14" = 71*554; 
/. area of segment = 105*08 1 — 71*554 = 33*527. 



10. Let the circles be as in 
the margin. 

It is required to determine the 
area of the figure AOBD. Let 
£>E^x; then 



8»- 



=3 300? — a* ; 



w 



30 15' 




.-. COB ADO = 2? ^ 8 - -^ = cos 74*^ 32' 2"'36 ; 

.*. u4i)5=i49^4'5". 

Therefore AOB^zx (180*^- 149** 4' 5") = 61*^ 51' 50", 
and proceeding as in Ex. i, we find 

area of segment ADB^22*%6/^y 
and area of segment A OB » 66*806, 
.*. area of figure AOBD = 89*07. 



126 



ABEAS OF PLANE FIGUfiES. 



14, Let DGE be a semi-ellipae having its semi-major 
axis OC^i*j\ and semi-minor axis 0D= 12I; and let OCI 




he the elliptic segment, whose height {CK) — lo. Produce DE 
both ways to A and -J?, and make 0-4= 0-8= OC=iy{. On 
AB as diameter describe the circle A OB, and produce GI both 
ways to meet it in i^and H. Join 0-F, OH. 

ojj:=oo-oir=i7^-io=7i; 

/. cos 00^= 7f - i7i = ^= HaSs;^ = cos 64^ 37' ^13" ; 

.-. i^0^= 129^ 14' 46", 
and, as in Ex. i, we find 

area of segment FOH^ %26*%y>i^ 
and, segment 001 : segment FOB :: 25 : 35 ; 

^ ^x *; X 226*8^01 ^ 
/. segment 001 — ^ — = 162*02 1. 



Surfaces and Contents of Solids. 



Ex.8. 

6. i8xi|xig = 3i^ cub. ft. = the solidity of the log ; 

•*• 3^1 "" ^3 ~ ^9 ^^^' ^- " solidity of remaining piece ; 
and therefore we have 



required length = -7 



- ^9 _ 29 X 4 



i^xit 



16 J ft. 
7 



SUBFACB8 AND 00NTEMT8 OF SOUOB. 



127 



4. 



Ex. 9. 

Perimeter of dther eaA « 4 x a^ <<■ ~ ft. 

* 4 



.•• lateral embce = — x -^ = -^-J ™ i7o*4375 1- ^» 

area of the tvo enda = 3 x |x |x V3 = -^ V3"'<S*57648q« ft«; 

.'. whole suxfiue = 170*4375 + 6*5764 =» 177*0139 sq. ft. 

The solid content = ^2862 x z^'^S == 83'oa7 cub. ft. 



Ex. 10. 

1. If the triangle ABO represent the 
base of the pjramidy and JE be the point 
in which the perpendicular on the plane 
ABC from the vertex of the pyramid meets 
the base, it may be easily snewn that die 
point JS will be equidistant from the three 
angles of the triangle, and 




.-. dant height = (900 + ^* = ^i^^*» 30-042; 

.-. whole sarSaoa =i x 3 x 5^ x 30'04a + i x (5^)*. ^ 

= a47'8465 + i3*0987 « 26o*945» sq. ft. 

Solid oontent = i X 30.x J. (5|)*.^ 

^ z 2, 

= 10 X 13*098 s= 130*98 cub. ft. 

7. Here a* = 8 x ^ x cot !ia| = 50 x a*4i4ai4 « 120*7107, 

4 

ft* = 8 X 4 X cot 227 = 32 X 2*414214 = 77*2548, 
and therefore db = 10*986 x 8*789 = 96*556 ; 
therefore volume 

= - (120711 + 96-556 + 77*^55) = 589'044 cub. ft. 
3 



128 SUBFACES Ain> OONTEim OF S0UD8. - 

Slant height 

therefote whole sarfisuse 

6*1202 , . 

« — ^ X (3^ + 40) + iao7io7 + 77*^548 

« 418*293 aq. ft. 

Ex. 11. 

4. Area of either end 

-^ X 3'H^6 = 11-0447 sqi- ft- = 1-2272 ; 

therefore quantily of earth excavated 

= 1-2272 X 15 « 18-408 cub. yd. 
18*408 

7f 

128-856 
4'6o2 

i33'458«. « X6 I3-458*- =X6 13*. 5f(?., 
therefore cost of excavation £6 133. 5\d. 

6. Volume of whole globe = - wr^. 

Volume of unknown portion «= 4 tt (r - 5)^ ; 

3 
therefore volume of known part 

= |^{^-(»--5)'}-|^(i5^), nearly; 

4. 
therefore fraction required = 3 =: IS 

3 



4000 800 



J 



SUBFAGES AND CONTENTS OF SOLIDSv 



129 



\ H / 



9. Let ABC represent the conical glass ; F the centre of 
the sphere, and FD a radius at right 

angles to BG the slant sidd of the cone. ^^ x q \ --. » 
If a circle be described about ^at the 
distance FD it will represent the globe 
in piano. H therefore is the lowest 
point of the sphere ; and Gj H, and F 
are in the same straight line, which, 
if produced will bisect AB in O, 

By question, AG = fz^s in. ; and 
<?(?= 6 m., and .\ AG =^6-5 in,, and AGz AG:: FD : FG, 
or 2'5 : 6*5 :: 2 : FG^S^2 in. ; 

.'. FG = '8 in,, 0H= 2'% in, = the height of the segment 
immersed; 

therefore solidily of segment, or quantity of water that 
will run over 

= {(4 X 3) in. — (r8 X 2)] in. x (a*8)* x ;r = 26*27^ cub. in. 



13. Let BAS represent the earth^s hemisphere ; its 
centre, and A the pole. And let 
DAF, GDFE, BGEH represent the 
three zones. 

By question, 
angle 50(7- ^Oi)= 23%-; 
therefore angle 2)00 = 43^ 

AOG^Ge^l. 




Now AM^ r (i - cos 23^!) ^ height of segment DAE\ 
therefore surface of zone 

DAE= 2irr^ (i - cos 23^^). 



Similarly surface of segment 

GAE= 2irr^ (i - cos SS""^}, 

and therefore surface of zone 

GDFE = 27rr» (cos 23** | - cos 66'' |). 



17 



\ 



130 



BUBFACSS AND CONTEIH^ OF SOLIDS. 



Similarty surfece of zone BCEH:=^ zvr*. cob 66*^^ ; 
therefore the Burfaces of the zones are to each other 

as I - cos »3*j : cos 23** j - cos 66'''^ : cos 66**^, 

as -083 : -518 : '399, nearly, 
as 83 : 518 : 399, nearly. 

16. Let A represent the point above ihe surface of ihe 
earth J3SEZ), such tifat AE=^ EU. Produce 
^0 to meet the circle in E. Let the radius 
sr, and EC ^x. 

Then AB* =^AE. AF^ 3r» ; 
••. 5(7* = 3r«-(r + aj)» = r»-(r-a;)% 
or %f^ - 2rx — a?* =5 2rx — a?* ; 

.*• arajssr*, 

am 

a; s - =s height of segment visible ; 




therefore surface of segment =. 2irr x - = wr" ; 

I 

therefore portion of surface visible = wr* -s- 4irr* = - ; 

4 



also volume of segment BED^ 



-- (or — r) - = ^ — ; 
6 ^ '4 24 ' 



therefore portion of volume visible = — wr* ■^ ^ wr* = — 



24 



3^ 



17. Let j5, represent the two places on the sur£Ace of the 
earth; at O apply the tangent GA\ 
join OB (0 bemg the centre of the cir- 
cle), and produce it to meet the tangent 
in A. Join 00. The angle 00 A is 
a right angle. BA is the height to 
which a man must descend. 

%Qo : 25000 :: angle OOB : 360* ; 

.-. z GOB = 2^ = 2^ = 2^ 52' 48", 

125 25 ^ ^ ' 




SUBFACES AND OOMTEMTB OF 80LID6. 181 



diameter of the earth « -^ — - s 79577^8 mi 

^^^ -^^ cosaO^ coBa*5a'48" -998737 39^^ ^^' 
therefore required height = 3983*895 - 3978*8648 5*031 miles. 

Ex. 18. 

7. a* = area of mouth = 3*1416 x -| - 4*90875, 

6* as area of bottom « 3*1416 x - « 7854, 

4 

/. h « *886 : 

therefore content of gUiss = ^ x (4*9088 + 1*9625 + 7854) 

24 

» -2 X 7*6567 ■» 9*152 cub. in., 
24 

and« one imperial gallon s 277*274 cub. in. ; 

/. 77 74 _ 29*969 = No. of Buch glasses in an imperial 
gallon* 

9. Since the cone is right angled at the vertex its altitude 
will be equal to the radius of its base« and therefore equal to tibe 
radius of the hemisphere. 

Volume of the cone = - wr*, 

3 

volume of the hemisphere ^ - irr^ ; 

therefore volume of cone and hemisphere together = ttt^^ 

and volume of the qrlinder =» 2vr^. 
Volume of additional space » irr^ ss ^ cub. ft., since r = i. 



183 ilBTIFICEBfi' WORK. 



Abtificebs* Wobk. 



Ex. 14. 



2. I 8B 120 sq. ft. s supeifidal oontoit of the gable 

when redacea to standard thickness^ 

and ^Q ^ 3Q ^ 4 ^ 3^;^ gq. ft. — saperficial content of the end 

wall reduced to standard thickness ; 

therefore whole superficial content = goo + 120 = 920 sq. ft. ; 

.•. -2 as ^'070 standard rods <^ brickwork ; 

••• 3*379 X 4 = £i3'5i^ =» ^^3 ^^' 4^- 

5. 21 X 10 X a + 15 X 10 X a =: 720 sq. ft. =: superficial con- 
tent of the whole room. 

6 X 4 as: 94 ft. » content of door, 

2x5x4 = 40 ft. = content of windows ; 

.*. 720 + ^ 5= 752 sq. ft. =5 whole amount of work. 

As 100 sq. ft. : 752 sq. ft. :: £375 : 7-52 x 375 ; 
therefore cost of wainscoting = £28*2 — £28 49. 

6. Let X » thickness of lead in inches. 

Then area of section of pipe = '''{(g + ^J "" (f) } > 

content of one yard = tt [^ a? + a?* J x 36. 

Hence 36W (a^ + ^x\ : 1728 = 14 x 16 : 11325 ; 

...^ + 5^^48xi4X_i6^ 
4 iiS^S^ 

Whence x = '20737 in. 



MEASUREMENT OF SHOT, SHELLS, AND POWDER. 133 



Measurement of Shot, Shells, and Powder. 

Ex. 15. 
3. Weight of 4 in. ball : weight of 9 in. :: 4* : 9* :: i : ^ , 

weight of 4iii. ball : weight of 6 in. :: 4^ : 6^ :: i : -^; 
therefore weight of shell 

* ^4 (7«9~*i6) =|jx 513 = 7^14 Ihs. 

2 X 'Z*i.a.'l6 X - 
10. Quantity of powder = 2 8_^-^i4i6x8i 

= 1272 lbs. 

13. 10 lbs. = 300 cub. in. = volume of cylinder ; 

therefore the length = ^ p = -= lO'di in. 

® 9 X 3-1416 9-4^^48 

18. A leaden ball i inch in diameter weighs — oz. 

7 

.•. — : 4 :: i* : aj*, or a^= ^; 

the diameter of the 4 oz. ball ; 

therefore calibre of gun vo^2,y + — ^-^ = 1*074 in. 

49 



PLANE TRIGONOMETRY. 



Tbiqonometbical Fobmuub. 

Ex. 1. 

1. First reducing the minutes and seconds to the dednud 
parts of a degree, 

6o )iy .'. J&= 15-0713889 

•0713889 9 ^^7459»7 

E 

.\ ^4-— = 167459876 

y 
or F^ iff 74' 59''-876, 

4. circular measure = -^ .w^— — ^ ^ ■ = '99484. 

6. cos-4 = '6; .*. sin ul » Vi -cosM = ^64= '8 = ^ , 

"^^^ Bin^"-8"J' 
chd-4 « ^Bm- = V2 (i - cos A) = ^^^-8 = a/^ . 

. A 

A A A A i^tan — 

f% • A ^ ' A A , A ^A 2, 

9. sm-4 = asm — cos— = atan— cos* — = 



i+tan* — 
2 



i + (7-4\/3) 4(«-V3) ^* 



TRIGONOMETRICAL FORMUUE. 135 

Ex. 2. 



A 
3« Bee A^ ^ 



cos"^ cosfA^^^ 



COS -4 COS — COS A cos -- 

A A 

cosjI cos — +sm^sin — 



% ^ A 

■"""■ = i + tan-4taii— . 



cos -4 cos — ^ 

A. o^^n o,A-. ; sin^^ + cosM _ i+cotV^ 

sin 2-4 asin^cos^ zooiA * 

IL secM.cosec*-4 = secM(i + cotM) 

cos' -4 sin'^ 

la cos» ^ - sin» J5 ' , ^ 

sinMsin'-B " cotMcosec«5-cosecM 

= cotM (i 4-cof 5) - (i +cotM) 
= cot* -4 . cot* J5 - 1. 

13. tana^-tan^=.5HL±^^sin^ 

cos 7, A COS ^ 

_ sin {%A - ^) 
cos 2-4 cos A 

2sin^ 



cos -4 + cos 3^ • 
16. -^^^x ^^^ =tan^x-ggl£- 

I+COSa-4 I + C08ui I+C08^ 

_ sin ^ ^ 



136 TRIGONOMETBICAL FOSMULiB. 



18. sin 5il Bin -4 « - (co04ii — cos 6A) 



= - {i - 2 sin" a-4 - (i - !i sin' 3^)} 



2 

sin' 3^ - sin' aX 



21. tan3^.tan^*?i5Li4is!54 



- {cos (3^ -.4) - cos (3-4 -f A)} 

- {cos (3^ -u4) +COS (3^ + -4)} 

_ COB %A — cos 4^ 
"" COS %A + cos 44 * 

32. taB(4 + J)=«A(4+^=«l(f±f«M^ 
^ ' COS (-4 +5) cos (^ + 5) sin (^—5) 

flin*^ — sin*J _ sin*^— sin*^ 

"" I , . . . ri\ fiin-4 . cos -4 — sin£. cos B * 
- (sin a-4 - sm a J5) 

34. sin(^-.B)sinO+sin(5-(7)sinu4-sin(u4-C7)sin5 
« - {cos (.4 - 5 - CJ) - cos (^ - 5 + (7) + cos (J8 - (7 - ^) 

-cob<J5-(74-^) + cos(C7-u1-jB)-cos((7-.^ + jB)} 

s: o, since the first and sixth, second and third, fonrth and fifth 
terms destroy each other. 

40. sin^ + 8in-B+8in(?-sin(^ + -B+0) 
=. (sin ^ + sin 5) - {sin (^ + jB + 0) - sin C} 

masin-(^+J5)cosi(^-5)-2sini (-4+5) cosi (^+5+2(7) 



. A + B ( A-'B A-^B+izC 
% sm -{COB cos 



2, { 2 a 

. A + B< . A-^-C . B+C 
r^asin — -^— 42 sm — - — sin — - 
21 2 2 

^\ sin -4 + sin 5+ sin C7 



} 



]■■ 



. A + B . A+C . B+C ^ . fjt.T>, />x 
4 Bin sm — - — sm + sm {A + B+ C). 



TBIQONOMETIUCSAL FOBXinLB. 187 

42. t«i^ + t«i5+t«<7=^+^^+^ 

COS C COB A cobB 

_ {coB^coB J^cog(^+J)}8mC^^m(^+J)oo^ 

cos^.cos^.cos C 

mnABmB.mn (7+8111(^1 + J+ (7) 
" cobA.cobB.cob C 

= tan^.tan£.tan C+ -^ p 'r' - 

co%A.QO&B.eo%C 

44. 28m — ^- — ^^-— sin— ^^ « cos-4-cos(-B+ C) 

2sin 8m scos(£— C7}— oos^; 

.\^m^A-^B^C)wn-{B\-C-A)Bm-{A+(}-B)«ixi-{A^B-a) 
= - cos»-4 - COS (JBi- (7) COS (S- (7) +€08-4 (cos (J84- C54KXM 

= 1— C08*-4 — C08"B— cos* C7+2C08il COS j9 COS O. 

B 
Aft sin ii Bm ^ _ sm ^ i — cos^ a 

sin^cosi? smB^BmB' cobB "" sin^cosj? ' 

. A . /A j.\ 

f A -OK -D i A iy\ asm— smi i?) 

cos(^-^^ cos J— cos(il — B) ti \2 / 



I — 



COB^ COSjB oosjS 

.A . ^sinB 

s= 2 8m sm ^ 55; 

a c(Mi-B' 

sin^ cos(^ — J) sin^d 
* ' sin j5 cos £ cos B sinB 



. . ^ sm ^1 sm' B 

= a sm^ : — ^1 =r - 

2 8m-Bycos5 

/, 

. ^A . ,-B Bin At 

— a sm* a sm" — . -^ — =1 • 

a a sm-B\ 

a I sin 5 sin* - -4 — sin -4 sin*--B) 

sin J9 

18 




138 TRIGOKOMETBIOAL FORMULiE. 

Ex.8. . 

- • ^ - /l— COBI^' /l/ /l 4- COS ^o*^ 

1. smry/^^ -^^^^^j^^ ^j 

2 

2. COS la^ — cos (30® — 18*)= cos 30* COB i8* +811130** sin i8* 
= ^x3Vio+aV5+^xi(V5-i)=|{V5-i+(3o + 6V5)*}. 

3. tii37«>3o'='-"'75'^i-Bini5' 

^' ^ Bin 75" cos 15" 

1. 8in3^ = 38in^ — 4sin^.i = 48ln^/^— sin'^j 

= 4 sin -4 (sin* 60" - sin* ^) 

s= 4 sin ^ sin ((So® + -4) sin (60** — A). 

2. see (45** + -4) sec (45*^ — -4) 

I I 



cos(45" + -4) cos (45" — -4) cos* 45** — sin* -4 
I 2 2 



1 . , , I — a sin* -4 0QB2A 

— sm*-4 

2 



=s a sec 2A. 



/ A\ / A\ t^"(45"+T) + i 
3. tan(450+:|) + cot(4^+^) = ^ 24-. 



tan 
sec* 






3 a ^ 



sin {go' + A) cm A 



TBIGONOICBTBIGAL FOBMUUB. 189 

7. chd io8*«»aBm54*»ax- (V5 + 1) 

4 

« a-|- (V5 - 1) + ^}»« a am i8*+ a Bm30* 
« chd 56® + chd (So*. 

Ex. 6. 

1. tan(^4-J+C)=. ^_^^J^^^^^ 

tan^ + 7 — dI — 7» 

I — tan Jg tan C 

" tan^(tanir + tan7?y 

.1 — tan^tan C 

^ tan^ + tanJ-htang-tan^tanJtang 
i-tan jltan^- tanjl tan C7-tan^tah C7' 

but since -4 + 5+ 0=90; 

••• tan (-4+5+ C) « 00 , and therefore 

I— tan^ tan 5+ tan ^ tan 0— tanjBtan O^o^ 

or tan^ tan 5 + tan ^ tan 0+tanjBtan O^i. 

^,A rtK C0t-4c0t5— I 

^ ' . cot 0+ cot (-4 + 5) ' 

/. cot (J[ + -B) cot 0^=1, 

cot-4cot-B— I ^^ 
cot5 + cot-4 

whence cot-4 + cotJ? + cot (7=cot-4 cotBcot CL 

5. co8-4 + Bin 0^— 8inJ? = Bin(5+ (7) +Bin (7— sinjB 

B+C / . B+0 . B-CTK 
s a cos ( sm sm ) 

B+C . C B 

.354008 sin — cos — , 

^ 2 % 2 



140 



maOKOMETBIOAL FOBXUUB. 



Similtrljy ocmB+mnO'-dnA^^QOB 



A+G .a 



am -COB — ; 



COB ^ + gin C— sin^ 
" coB^+sinC?— Bin^' 

coB(45'-^)coBf 

3 — T—^ 

COB— 008(4^ 



B^-a B 

COS COB — 
2 % 

A + G A 

COB COB — 

2 % 




1. Since ^ + 5+<7=i8o; 

.•. 8ina(7«8m{36c^ — a(-4 + -B)}=s — ana(-4+jB) 
(flin 2^ + Bin 2B) + sin 2(7= 2 sin (^ +^ cob (^ -^} 

- 2 sin (wi + J3) cob(^ + J3) 
«2Bin(4 + J9){coB(JL-^-cofl(-4+J3)} 
s2 8in (7.2sin^ sinjB 
» 4 sin^ sin^ sin C 

3.' Since(7=i8o*-(^+-B); /. cos (7=-cos(u1 + jB); 
/. COB* (7+ 2 cos^ cos J? cos G 
= - cos (-4 + J5) {- cob (-4 +5) + 2 COB -4 cob5} 
= - cos (-4 + -B) cos (-4 - jB) 
cos 2 a + cos 2B 



Hence cos?-4 + cos*£ + cos' (7 + 2 cos ^ cos B cos C 

I + C0S2il I + C0S2j9 cos 2-4 + cos 2-B 

s 1 J^ 



i 



TBIGOlSrOMETRICAL FORMULiSr. 141 

/. (sin A + mnJB) + sm C 

. A+B(. A'-B . A + B) 

= ij sin Jcos h cos y 

a I a 2 ) 

C A B 

=5 2 COS— X a COS — COS — 

^ J? a 

= 4 cos — cos —COS — . 
^ a a a 

10. TiA'^A^e, R^B^e, G'^o-e, 

4sin^'sin^sinO' = asina'{cos(^'-5')-cos(^' + ^)} 

+ sin(-4' + 5'- co-sin (^' + 5'+ C). 
Now 4sin«(? = 4sin(u4-^sin(J5-^sm((7-^; 

/. 4 sin« (? = sin (u4 +5- 0- ^ + sin (-4 + (7-5- ^ 

+ sin (5+ 0- ^ - ^ - sin (^ +5+ (7- 3^ 
« sin (a (7 + ^ + sin (a5 + ^ + sin (a-4 + ^ - sin ^6, 

or 3sin(? = sin(a(7+^+sin(a5 + ^ + sin(a^ + ^, 
hence, 
(sin 2A-\- sin a5+ sina(7) cot d = 3 — (cos 2A + cos a^+cosaO); 

.*. 4sin^ sinJ?sin (7 cotd — 4(1 + 008 ul cosj^cos 0)^ 

.>, i + cosud cosfcos (7 
or cott/=5 — ; — 5 — ; — 55—5 — Ty— 

sinji smjSsin C7 

= cot-4 + cot-B + cot (7by Ex. 2^. 

Also cosec*^«i + cot*^ = i + cotM + cot»J5+cot*(7 

+ 2 (cot-4 cot5 + cot^ cot (7+ cot jB cot G) 
= I + cotM + cot«5+ cot»(7+ a, 
(since cot-4 cotJ3 + cotj4 cot (7+cotJ?cot C7=si), 
hence cosec*^ = cosec'-4 + cosec'5 + cosec' G. 



142 TMGONOIIETBICAI. FOBMnL& 

Ex. 7. 

t 

1. Here, Bm.i — cdn^^sinJ?— 8mC7, 
or, Bini(^-5)coB-(^+i5) = 8in-(5-C)coBi(J?+C); 

/. sin -(-4 -J?) sin- O^ein- (5— (7) sin — , 

. ^-5 . ^ . B . B-G . B . G 

or, sin J-Bin — sin— = Bm rsin — sm — , 

that is, cot cot — = cot cot~: 

' a a a a 

whence cot— • cot — , cot— are also in Arith. Pros:. 



3. cos (ul - C) cos J? = cos (A + 0- J5) 

= cos (-4 + (7) cos5 + sin (-4 + C) sin Jff; 

.-. {cos(^-a)-cos(u4 + C)}cosJ? 

as^sinul sin Ccos J^=sin(^+ C}sinj5', 

or acot J = ^!^\. . — T^^QoiA + QotGi 

sinji sm O 

.\ cotAf cot 5, cot C7 are in Arith. Prog, 
and.', tan j1, tan^, tan are in Bixmonic Prog. 



7. 



cos J I ^ , COS.B — COS ul _ I — cos (7 
cos-4 cost?' " cos£ + cos-4"~ i+cos (7* 

. A + B . A-B . ,G 

a sin sm a sm* — 

a a _a 

a cos cos a cos*— 

a a a 

.-. tan~(Jl+J?)tani(^-5) = tan*-. 



TRIGOHOICBIBICAL FOBMUUB. 143 

Ex. 8. 

••. 008-4as-=BCoa6oV* 

and mnAsso; 
.*. A^o; or 6c^. 



6. 8m*a-4— gin*ilaa-; 

4 

/. 48m*-4(i— Bm*-4)-8in*-4 = '; 

4 

4 64 64 16 64 
_8m^ __ _ , 

.\ Bin^ = ±5^^^=8m54% sin (180P ± 54'*) ; 

4 

8inI8^ siQ (180P ± 18**). 

9« tan-4+cot^ = 4; 

^•. tan*-4 — 4taii^ + 4=4— 1=3; 

i± — 
^_^^ t«n45'±tim30» 

I I T tan 45" tan 30" "^^"to^J^f* 
.-. A = 75", or i5». 

11. (cos ud + COS 3^) + COS 2ui =: a COS ^ COS a^ + COS 2ul = o ; 

4 

.% (i) COS a-4 = o, or zA = 90% A = 45% 
and (a) cos -4 = — , or -4 = lijo^ 



144 niGOKOMETBICAL FOBMULJB. 

Ex. 9. 

1. Bin (aj + a) = cos {90*- (» + a)} = cos (i? - a) ; 

/. » — a = 90® — a? — a 5 
.•. 2aj = 90', 

2. V^ (cos 3a? + sin 30?) = I ; 

COS (3a? — 45®) = COB 60® ; 
... ..i^-=3S-. 

6. tana.tana;== {ta]i(a+a?)+tan(a — xc)} 

X (tan (a + aj) —tan (a — «)}, 

Bin tt . sin a? sin {(g + a?) + (a — a;)} sin{(a4-ar) — (« — a;)} 
cosa.cosa^^cos (a + a?) cob (a — a>) *cos{a + aj) cos{a— a?) 

_ sin w sin ga? 

"" cos* (a +aj) cos* (a— a?) ' 

hence sina; — o, and cos* (a + a;)cos* (a — a;)s=4cos*acos*a^ 

or cos(a + aj)cos(a— aj)=s± acofia.coB«; 

•\ cos'aj — sin*a = ± acosacosaj, 

and cos"a?±acosa cosaj + cos*a=i; 
.*. COB a? = ± (i ± cos a), 

9. tan*a? = tan(a? — a): .% tan*a5 = — -^ i; 

^ " tana; 

1 — tan* X _ tan x — tan (a; — a) ^ 
' ' I +tan'aj""tana? + tan (a?— a)' 

sin a 
,*. cos liUC = 



sin (iw? — a)^ 



TBIGONOMETRICAL FOBMUUE. 

2 sma» 2ain (ao?— a) 008 M ==8in (4x— a) —sin a; 
/. sin (40; — a)a38ina; 

A a?=— {a + £dn"'(3 8iiia)}. 
4 



14« 



Ex. 10. 

3. Let -4 = tan''-, then ttn^ — -, 

7 7 

I 

a X- 

jRs=tan~'-, then ttnf--, and tan ai?»— — ^»^, 
3 3 j_i 4 

9 

* />! . D\ tan.4 + tana5 7 4 

tan M + a jB) = — j^ ^ = ' ^ « i ; 

^ ' I — tan^ltansiJ? i 3 ' 

7 4 
/. A + a-B= tan"' i, or tan"' - + a tan"' - = 45**. 



5. Let 6 s= sin"' -7- , then sin d = -7- , cosec' ^ = 5 ; 

v5 v5 

.\ cot ^=5 a; ^=*cor'3, *^®^ cot ^==3, 

X /ii . ^\ cot 5 cot — I 6—1 ^ - 

cot (d+ ^) = — . : , ^ ^^ =— — = I «cot45*; 
^ ^' cot A + cot ^ « + ij ^*^ 



-, I 



•\ 0'\r4> or sin"' -^ + cot"' 3 = 45**. 



8. 



tan^ tj —tan"' l^ = tan 



-lA--^ 



I + «x<a * 



tan"' f, - tan"' L « tan"' '. ,f , 



/ 



tan"' L X - tan"' t^ = tan"' ^*-' ^* ; 



19 



146 TBIOONOMETBICAI. FOBMCUE, 



i + «»^<; 



10. t«n-' -^^?^ - tan- ^H^E* 

I ^ a; sin ^ cos ^ 

a; COS ^ a? — sin <f> 

_ A -I I —a? sin 4> cos 4> 

X COS if} a;— sm^ 
I — a; sin ' COS0 

= tan"' (tan 0) ss 0. 



Ex. 11. 

>, sin ^ I J ^ • sin <& i 

1. tan0» — 3 = -T-; and tanA = — r.=_^. 

costf V3 ^ COS0 yi5' 

gin ^ sin ^ __ sin (ff + ^) i i 
"coB^ cos ^"" cos 6? cos0~"V3 V15* 

sin (5 + A) II 
or i / '^ =-7 — h-7 — : 

2. Let ^ = cot"' (a? — i), then cot5 = a?— i, 

^ == cot*' (a? + 1), cot ^ = a; + 1, 

^ ^^"" cot^ — cot^ "aj+i — (a? — i) "" ij ' 

la *^ sin3o'* I ' 

.•. a?" = 4 + a V3 ; .-. a? = ± (>/3 + i). 



— ^ 



5. r COS V = 



TKIGONOMETKICAL FORMULA. 14Y 

COSM — 6 



I — 6C0Btt' 



I — cost? _ I— e coBt^— (cosu — e) _ (i +g) (i — costi) 

' ' I +COSt? 1—6 C08W + C0SW — 6 ~ (l — 6) (l+COStt) * 

and tan* - = tan* - , or tan - = | ) tan - ,• 

6. acos2^ + isinii.d=: a (cosij^ + tan^sinij^ 

=:-^cos(iid-d) 

COS0 ^ ' 



cos^^aintf 



I -tan* J i_ten*d C08*^-8m*^ 
10. cos 9 = 



i+taii»^ i+tan*d cos* (? + sin* (? ' 

cos* ^ - a sin* ^ cos* ^ + sin* 0] 



, . ,. fcos»d-asin'dco8'^ + sin'^) 

and «»' = i + 3cos'4=i + 3-{ — + v- — r- — r-rr 

lco8*d + a8in*^cos*d + sin»W 

(cos* ^- sin* ^ cos* ^+ sin* d) (cos»^ + 8in»d 



,1 (cos'^ 



?cos^g+sin^g ) _ fcos'g + sin'g ) 
+ 8in*^» J "^1(008*5 + sin* ^>) 



; lience m~ 



, . (cos* + sin* d)» (cos* fl + sin* 0)^ 

and .-. - = r (cos* ^+ sin* ^». 



148 TBIQONOMETRICAL FOBHUUB. 

11. tan*+tan^ = tan^fi+taii«^, 

• * 

I -tan -tanC = I - tan*^; 



•*. tan 

2 



a 



^ t«n|(i + tan*^) tan^ 



-^^ g-^= i^ = itaii/9 = tan*; 

i-tan*" i-tan»^ ^ 

13. i-costf=i-co8"a+sin*o(i-c*Bin*^* 

:= gin" a {i + (i - c'flin* ffn^ iJ 

i-(i-c»8m-«)* 

A i-(i-Csin'g)^=. ^^^'^^^-?^^'^ =c'sm'a(i+coBg), 
^ ' I — cos tf ^ ' 

MultlplTing by sin' a and substituting for the. radical term» 
we get 

sin'a— (cos'a — cos^ sc'sin^a + c'sin^a cos 0; 

a I — a sin* a + c* sin* a 

.\ COS0 = r-r-T , 

I — c* sm* a ' 
Dividendo and Componendo, 

I — cos 5 % sin* a — ac* sin'* a 
i + cosd Z — Zsm^OL * 

Q 

.'. tan*-=tan*a(i -c'sin'a). 

Ex. 12. 

tan(g + ^) ^ a + ft . 
tan (tf - ^) a - 6 ' 

tan [0+^) + tan (g — ^) sin 2^ _ a 
tan (tf + ^) — tan (^ — ^) sin li^ "" J * 

Hence i* sin* 2^ = a' sin* a^. 

But 5' cos* 20 =s a* cos' ij^ — aoc cos a^ + c* ; 

a*+c*-i* 
/. cos a© = . 



2. 



J 



TRIOONOHETSICAL EOBMULiS. 149 

4 atan'd + J=5m(i + tan'd); /. (m — a)tan*d=:6-w, 
&tan*^+a=n (i+tan'^); .% (n-&)taii»^ = a-n; 

• : V — ^ -^ — ' 

n^o a* a — w 
/. (w — &)(n-&)a»=:(w-a)(n-o)y; 
/. [mn- (w +n) i + i'} a' = [mn- (w + w) a + a'} J" ; 
••. mn(a*-i*)T(wi+w)(a-i)aJ = o; 

m+n a+b * 



6. From the equations, (a) and (3), 

^aja jra 

i' cos*o> + a* sin"o> = ^ = — sin* a, 

F* 

whence (a' + i*) - (a* - J*) cosi^a) = — (i-cos2a); 

F' 
.•. from (i), (a* - J') cos ao) = — cosaa + JB' (4). 

Again, from (i) and (a), 



/F» ^ . FB 
a + l^Lf — + JS» + --jXa8ina, 

/F^ ^^ ^ra": 






+ 5* + ^i— ^ cos a4 (5). 



from (4) and (5), 



— jCos'aaH co8aa + 5* 

-- H cos %a + -a* 



150 PBOPEBTIES OF PLANE FIQURES. 

/. cot a» = I — cos aa + JB* I -s- — sin 2a=.cot aa+^^cosec za. 



PfiOFERTIES OF PULKE FlOUBES. 



Ex.18. 

2. cos \%A — B\ =eos J2-4 — f- - AVt = cos \%A — — j 

= sm3il = 3 sinul — 4 sin^^ 

4^ ^ 4 

8. The h3rpothenase must be the diameter of the dreum- 
scribing circle ; 

•r. c :t B ah ah 

/. -B = - ; and r = — = — = 



%^ 8 28 a + ft + c' 

1 T> c , ab (a + &)c+ c*+ 2aJ 

hence jB + r = -H ~t— — = ^ / . , . v 

2 a + A + c a{a + ft + c) 

_ (a+_J) c+ a» + i* + aq& 
. ~ 2 (a + ft + c) 



a (a + 6 + c) a ^ ' 



Ex. 14. 

tan J ^ sinJg cos C _h d' + h'-c' , o*+c'-ft' 
tan (7~ sin (7 cosjB^c aoA * 2ac 

^ a^ + b^-c^ 



5. 



PSOPEBTIES OF PLANE FIOUSES. I5i; 

a* h' sin* -4 sm»5 



c» c» sin' (7 sin* C 

o' - &' _ sin' A - sin' B sin (.4 +^) ain {A - B) , _ 
C sin' (7 sin»(^+£) "7 ^^' ^*» 

_ sin (.4 - B) 
sin(^+£)- 

9. tan'^tan:?=./5Epp:./5ESE^ 

»— c a+b—c 



s a + b + c' 



H- Area = - Jc sin^ = - (a^ - J») sin ^ 



a' 



-^/^a za\ sin 5 sin (7 . . 
-i^'^-^^ sin'^-sin'^ ^"^^ 

— ^ /^a )la^ sin 5 sin (7 . . 

"5(« -^) sin(^+^)sin(^-^) "^^ 

I / « ziN sin ^ sin 5 . , ^ 
"i^** ~^ ^in(^-^) ' «"»** 8inO=sin(^+^. 

15. -4 + 5+ C= iSo'; .-. -(A + B-CTj^^o- C. 

Hence, °° Aoc^tr = ;; <^ mp <?= area. 

4tani(.4 + 5-C) 4cot G • a 

Also, area = ^8{8--a) (s-b) («-c) 

_q&c / s{8-'a) s{8'-'b) g(g — c) 
8 \ be, ' <zc * ab 

2abc ABC 

cos — cos — cos — . 



a + i + c ij iJi » 



17 1 __ a8in^ _ / sin^u!l\ i_ f sin^ sin5 sin O H 

— / SIP -^ sin B sin {7 \i 
"" \ abc ./ 



162 FBOPBBTIEB OF PLANE HGUBBL 

19. aoos^+ioosB+eoosO 

a — ?L—tixi2A + — r-^BmaJB+ — = — yyBmzC 
znnA %uaB zsmC 

TsB{Bin2A+mn2B+mn2C) 

^B[%nji{A+B)ooB{A-B)-^%miCcoBC] 

^2BBnC[coB{A-B)-coB{A+B)] 

a^Unn Cn2miA mnB^ since cos C= — cos (^ + 5) 

w^^smAfAaBmnO. 

20. Let be the centre of the ciide inscribed in the tri- 
angle ABCj 

SimilBrly, a-^+^; and >3»| + |; 

. . « . ABO 

,\ 4Bina8my3 8in')r=4co8-co8-co8 J 

BzBmA+ Bin5+Bin C, by Ex. 6^. 

Ez. 16. 

O G 

6. -^aw /.AEO=^B+-y axAir-AEC=A + -', 

2 ^ 

.'. ir-2AC=A-B; 

sin-G em- C 

Aeam, .<1E= CE-i-^ , and £^= GE ."i, ; 



^^^"^ a VsinJ. smJ5/ \a *y' 



.•. (7if « -— T cos - (7. 

a + 6 2 



PROPEBTIES OF PLAHE FIQUBES. 



153 



7. ta.vi5tanicr=|i£=^^H*xjfc2^i2=51l*=-2zf. 
.♦. a—_p{i—taa-Btsai- C)=pcoa-{B+ C?) -s- cob - 5 cob - C 



=-P 



Sin --4 



COB- Bco& G 

2 % 



81111118X17, h^p 



sin--B 



COS - -4 COS O 



c=p 



Sin- (7 



COS - ^ COS - jB 



8. Let the angles be, 20, 3^9 4^9 ^^^^ 9^== i8o^, and the 
angles are 40**, 60% So**, 

• ^'yi.'/^ •><.• J 2 Sin ^^ — cos — 
"N" g-t-c _ 8mul4-8in (7 . sin^ + sina^ . 2, a , 

b " sinJS "" . ctA "" . qJ[ ' 

sin ^^— sm ^^— 

A a + c 
/. cos — = — t- • 

9. The solution may be deduced from that of 7. 
12. Bisect AB in E; then we have 




JSD = ^JS?-^2) = --Jcos^ 

2 

c BiaB 



} 

_ c f sin .4 cos JB — cos ^ sin B ) c tan .4 — 
'^ 2\siaA cosB + qobA sin-Bj ""a *tan-4 + 



2 Sin C/ 



a] SI 



a sin jg COS ^ 

sin {A + jB) 



-tan5 
tan^' 
20 



154 



pboperths of plane piansES. 



10 b '^ a _ Bm B ^ am A _ Bm {A + (7)— sin^ 

c sui C/ sin C/ 



28m 



a Sin — COS — 

2 7, 



••. COB [-4 +.— J = n COB ^ • 



cot 



B-A B+A 

^— ^ g g cosJg+cosJ. 

a . JB — A' B + A^smB^ auxA 
sm COB 



cos^+co8^ _ sInJ3cos^+flin^cos^ 
nsin (7 n sin j8 Bin C7 

Bin O—cosjgsin^ + BJnJgcos jg 
nsinj8sin(7 ' 

since sin C= sin (^ + ^ 

sin P— sin ^ 



1 + 



sin O 
nsinB 



cosB 



I +nco8^ 
nsinjg 



20. Let ABO be any triangle; 
ACBE the circumscribing circle, of 
which CE is a diameter. Draw CD 
perpendicular to AB. 

Then OE.CD^AG.BC; 
.\ CE.CBfonB^AG.GB; 

••. l=^2,BBmB. 

Similarly, a = aiS sin A^ 

and c^2R sin C. 




PBOPEBTIES OF PLANE FIGUSES. 



155 



24 LetjEP=JEB=r, FQ^FB^r^\ 




1) 



Then r, = FQy EP- EN= r-{r-{- r,) sin - ; 
.•. Ti (i + sin— j=r(i— sin— j, 

. A f A .Ay 

I — sm— (cos sin — ) 

i+sin— (cos hsin — I ^ 

« V 4 4/ 

Similarly, 5==taii»(45-~|), and ^ =tau»(45-- ^) . 
Hence (r,r,)* + (r,r3)* + (r^rj)* 

= r]tan-(ir--4)tan- (ir — J?) +tan- (w— -4) tan-(7r— C^ 
14 4 4 4 

+ tani (tt-P) tani (w-- (7)1 
4 4 ) 

»r; since -(tt — -4)+-(ir — 5) +-(7r— (7) = -w. 
4 4 4 a 



(See Ex. 5^) 



25. If OD^ OE^ OF^ t, ; then 
£kABC^&e.ABOC-'£iBOG 



^ABx OE+^ACx OF-'-BCx OB; 

% % % ' 



156 



PfiOPBRnSB OF PLANS FiaUBESL 



Hence /. 8^r^~r^{p-a) 

==r,(j>-J)=r3(p-c); 

if iS^ be the aiea of the triangle, and 
2pit8 perimeter: 



r,r^ rs P P J? 

_ 3p — (a + & + c) 3P^2p 
P F 

hence — == — H h — . 




= 1, 



* . 8 S S ' S ^^ 
Again, rofxr,r. = - x x ^ x = S*; 

.% /8^a= area of triangle = [rjrjVjr^^. . 

26. Through D and i? the 
middle points of AB^ BO, draw 
DFj OEF respectively parallel 
to AE, AB, and join CF, BF, 
DE\ then, because AF is a 
parallelogram, DF = AE^ and 
^J?" is parallel and equal to AD 
or DB; and therefore jBJ* is 
parallel and equal to DE or 
jSGj and consequently FG is 
equal to £8. Hence DFO is the required triangle, and its 




area 



GF 



2 4 

or A^CS : AX>J'6)'=4 : 3. fiETmers' 2V^.) 



SBOPEBIIES OF PliANE FIOUBES. 157 

97 -NT^™ Aa AG h 

27. Now ^=-^ = -; 

., AF AB e Av ^ 



FO BO a' •• a + c' 

••• AAFG = --, — 7-rr-, T-=^ABCx 



2(a + b)(fl+c} {a + b){a + c)' 

Similar^, ABGE^AABOx ^ 



and ACEF=£kABOx 



ia + b){b + cy 

ah ^ 

(a + e){b + e) ' 



i 

\ {a+b){a'\'c) {a+b){b+c) {a+c) {He)) 



^LABC>< ^"^ 



{a'\'b){a'\-c){b-\-c)* 



28. Let 2a be a side of the equilateral triangle. Then 

I • • ' 

- (4a*) sin 6o* or a* Vs — its area. 

K 2a — <f, aa, 2a + ^ he the sides of the other triangle, 
its area 

= {3a. a (a + df) (a -(?)}* « a V3 (a'-c?)*. 

Hence aV3(a*-<J^')* : aV3 = 3 • 5; 

.•. a* — <Z"=-2-a*; .'. c? = -a. 

^^5 5 

The ratios of the sides are 

4 4 

2a — -a : aa : 2a + -a s= 3 : 5 : 7. 

5 5 

If be the greatest angle, aa + <^ is the side opposite ; 

a* -1- r* — i^* I 

/. cos5 = ^2 — ^ — '—- — = cosIao^ 

2x3x5 2 



158 



PBOPEBHES OF PLANE FIGUBES. 



33. Let A and B be the aieas of the inflcribed and ciiciim- 
scribed polygons of the same niunber'of sides; and A' the ai^a 
of the inscribed polygon of twice the number of sides ; then 

* • 
^ --W W'r» ••T T 

-as««r" sm — COS — , ^=nr" sm- -s- cos — • 
n n ^ n n 

also -4'= anr" sin — cos — 5snr*sin— , 

3in 2n n 



hence AxB^ n*r^ sin* — =^'*, 



or 4' = V^ X B. 



35. Produce CA to ui', making AA' = -BJD, Join A'D. 
l^gfli-ofstring^CD+a'iy + arc CGG' + axQ DEiy. 



The 




Now CD = (^'i>- - 04"*)* = (a» - c»)* 
arc GF=^C7coflr'^=^Ocoar'-^ = ^acos-'|; 



.\BxcCGC' = 2AC 



(^-cos-£). 
Similarly, arc DJaZT = aJ5i> (w - cosr^ ^) ; 
length of string = a |(a* - c")* + c('rr'- cos"' -J L 



TRiaOKOHETBICAL TABLES. 159 



Trigonometbical Tabuxl 

II 1 

Ex.16. 

1. sin laj** 14* ao" = ain (iSd*— 123^ 14' ao") 

= sin 56*^45' 40", 
iind sin 56"* 46' = •836446, 
sin 56^ 45*= -S362S6, 

Difil for i' or 60" = + -000160 ; 

.\ di£ for 40" = ^ X '000160 = '000107 ; 
.% sin I1Z3** 14' izo" a= -836^86 + *oooio7 = •836393. 

2. Here cos 41** 14' = 75^*030, 

cos 41** 13*= 75»^^3^ 
DfflT. for I* or 60" = — '000191; 

.•. diff: for 26" = |- X - 'ooo J91 = - '000083 ; 

.'. cos 41** 13' iz6" = 752223 - '000083 = '752140. 

4. sin A = '346105 sin 20"* 15' = '3461 17 

sin 20* 14' = '345844 sin 20** 14' = •345844 

_ k Jill I • 

261 - 273 

Henee 273 : 261 :: 60" : (n) no. of seconds required ; 

^ 261 ^., 
.'. «=:6ox — =57 ...: 

.'. A = 20** 14' 57"* 

7. log tan 23* 24'= 9'636226, 100" : 25" :: 577 : req. cor- 
corr. for25"- 144 '25 [rection 

log tan if 04 as" = 9*636370 i44-a5 



leo 



SOLUTION OF PLAN£ TBUKGLES. 



9. log COS ^ s 9*123456 

log 008 82^ Q,l' = 9*124248 



•% 1566 : 792 :: 100" : required number of 

seconds = 50" ; 
•% A = 82* 21' 50". 



Solution of Plane Tbiangles. 



Ex. 17. 



1. -4+5«90«; ,\ 5«90^-i8^ I4' = 7i*»46'; 




. ^^- • J. 

« • "j^ — sin^j 

.'. log BC^ log -4-B + log sin ^ -^ 10 

« 2'i3o872 log 432 = 2-635484 

log i35'^ = ^'^30655 log sin iS" 14' = 9*495388 







ai; 


6 




192 






250 




7 


224 



.-. B0= i35'i67; 

-4(7= ^432' - (i35'i67)» = \A567'i67 x 296-833 = 4io'3i 

by logs. 



SOLCnON OF PLAHE TBIAmLES, 161 

8. C = 180' -{A + B) = loo" «a' 45", 

AB = BO. ?J^ J .-. log AB = log BG + log sin C - log sin A, 

Bin .ilx 

AC=BC.-J^; .'. \ogAO = hg BO +log em B -log am A, 

log BG {= g2,6j') = 2*966939 log 5(7 = 2*966939 

log sin (= 100** 22,' 43") = 9*992835 log sin B = 9*873812 

*A.C.log sin-4 (= 3i**i3') = 0*285439 A.C. log sin -4 = 0*285439 

/. log ^J?= 3*245213 .". log-4C? = 3'i26i90 
.-. AB= 1758*78, AG=i 1337*18. 

15. tan I iC-B) = ^|=^tan ^ (C+B) = ^ tan 65- a.'j 

.•. log tan - ((7— B) = log 72 + log tan 65^ 22' - log 780 

2 

= 9-303861 log 72= 1-857333 

= logtanii"'a2'55" A.C. log 780 = 7-107905 

Hence, 1{G+B) =6^ ^z', .. log tan 65»..'= 10-338623 

^ — 10 



liC-B) « iio 22' 5fl 9-303861 

/. C= 76° 44' 55", and J?=53°59'5" 

Again, BC^AO?^, 
° sin 5 

log 5(7= log ^(74- log sin -4 log 354= ^'549003 

— log sin -B log sin 49® 16' = 9*879529 

= log33i*63, A.C.logsin53*'59'5"= 0*092127 

/. 5(7=331*63. -10 

2*520659 
log 331*6= 2*520614 

PP- 3 45 
J9 

* Note^ A.C. denotes Arithmetic Complement. 

21 



162 SOLUTION OF PLANE TRIAiTGLES. 

/. «-a = ao45-i468 = 577; 

H \ bo ) [i359Xi»63) ' 
/. log cos ^ ^ « ^ {log 2045 + log 577 - log 1359 - log 1263}, 

log 2045 = 3-310693 

log 577 = 2761 i7<S 
A.C. log 1359 = 4-866781 

A.C. log 1363 = 4-898596 

a ) 9-837346 

/. logcosi-4 = 9*918633 = log cos 33" 59' 35"! nearly; 

. .-. 4 = 33" 59' 25"!, ^ = 67° 58' 51". 

Now 8mC=8in^.^=i^8in67«'58'5i", 
whence (7 = 53" 54' 5" > •'• ^ = i8o« - (^ + (7) = 59" 7' 4". 

19. By question, c : J :: 7 : 4; 

.-. c — b : e + h :: 3 : 11 ; 

.-. tani((7-5)=^cot^ = f^cot4»37'47-5", 

log cot 4** 37' 47'5"= 11-091609 

log 3= 0-477121 
A.C. log 1 1 = 5-958607 

.-. logtaiii((7-J?)= 10-527337 
2 

whence = 73** ^7' 43"> 

2 
but ^t? = 85° 33' I3"| 



.-. C = 158° 49' 55"! 
5= ii»54'a9"| 



HEIGHTS AND DISTANCES. 



163 



Heights and Distances. 



Ex. 18. 



1. BC^AC.iBiiBAO; 
,\ log BC = log I20 + log tan 6o** 30'— 10 
= a-3a6539 == log mrogg ; 

.'. BC the height = a 1^*099 feet. 




xao/t. 



3. 



arm nr\0 ' 



sinao^ 




/. log AB = log AD + log sin 15®— log sin ao* 
= 1-878944 = log 75-6735; 
.*. length of ladder = 75*6735 yd. 

BG^ABBinAi 
.'. log BO = log -4jB+ log sin -4 — 10 
= i*«37535 = log 43-4045; 
.'. height of wall « 43*4045 yd. 

AO — -4J?cos35**; 

.*. log-4C7=slogAB + logcos35*— 10 

= 1792309 = log 61*988 ; 
.*. breadth of ditch « 61*988 yd. 



164 



HEIGHTS AND DISTANCES. 



17. Given BG= 40, CD =60, 




AGB=i4^ 18' 19", ADB^ 19^ 14' 5a", 
and /. BAG =^ If 3' 27" . 

smlJAG 

/. log ^ (7 = log (72) + log sin ^i) (7- log sin i)-4 (7 
= 1-881589 = log 76-1358; 
.-. S + a= 116-1358; i — a = 36-1358 ; 

... tani(5-^)=|^coti(7=4^cot 17-9' 9-5", 
a^ ^ b + a 2 116-1358 / y yof 

log 36-1358= 1-557938 
A.C. log 116-1358 = 3-935034 
log cot 17° 9' 9-5" = 10-510539 



.-. log tan 



B-A 



= 10-003511 =log tan 45** 13' 53-5"; 



/.J (5-^) =45- 13' 53-5" 

^(J?+^) = 720 5o'5o-5" 

/. jB= 117° 4' 44" 
A^ 2y^3&5f 

Again AB — BG.— — TnrBi 
° sm GAB ' 

and log BG{== 40) = i -60^060 

log QinAGB (34** 18' 19") = 9-750963 

A.C. log sin GAB {flf 36' 57") = 0-3339 1» 



.-. log AB = 1-686935 = log 48-633 ; 
.-. ^J5= 48-633 ft. 



HEIGHTS AND DISTANCES. 



165 



18. Given AB = loo, 



z CAD = 34' 30I . . ^ ^2)C7- loo'^.;' 




Bin ACB^ 
logAG^logAB + logBiaABG'-log&mACB 

= ^•^39777- 

Again, CD^AG— — ttttyj 

log CD = log ^ (7 + log sin CAD - log sin -42) (7 

s= a'ooo594 = log 100*137 ; 
/. CD = 100*137 yd. = the height of the castle. 



20, Let z JDAB = i"* 54' 10" = a, 

AG= 250, 
£2) = 12, 
GAB^e. 



12 




Then tan (5 + a)-tane 

AG AG'' AG'" 250' 

sin(g + a--^ _ 12 
' * cos {0 + a) cos "" 250 ' 

2 cos (^ + a) cos ^ or cos {20 + a) + cos a = ^ — ?E-?=: i:*3835, 

^■2 

COB {20 + o) = 1-3835 - -999448 = -38405 = COB 6f 24' 55" ; 

.-. = i (67» 34' 55" - 1° 54' 10") = sa- 45' ar5". 



166 



:i.'( tllhr. 



Now BC^ACXmhO, 

= 160-85 neaity. 



ASD DISXAHCES. 

log 250 = 2-397940 
log ten 3a* 4^' aa-^ = 9-808468 

— 10 



log i6o'8 = 7r2.o6%M 



10,% 



23. LctJLB-^(7=8, 

AB^c. 



sin 



Then, 







^ . ABD 
— -I sin 



-^/ 



(a '+8')(a'-y) 
6'c 



Hence, we find z J!4 O = 79' 55' 1 7", z ^52) = 144° 9' 57" ; 

/. A1P= ioo» 4' 43', ABP= 35" 50' 3' ; 

and /. -4P5= 44" 5' 14". 



NowP^=^5^4|g; 

smAPB' 

log AB = 2*698970 

log sin-45P= 9767483 

A.C. log sin -4PB = 0-157545 



/. log-4P= ^•623998 
.'. AP^ 420*72 yd. 



iuidP5=AB?!^44S, 

sm.ilPjS' 
log -4P = 2*698970 
log sin BAP- 9*993246 
A.C. log sin APB = 0*157545 

/. log PP= 2*849761 
PP=: 707*56 yd. 



HEIGHTS AND DISTANCES. 



167 



26. Let A be the harbour; £ 
the position of the privateer, and 
-4 C the direction in wnich the mer- 
chantman is sailing. 

By question AB = lo, 
A BA Colas'; 




b-c 



A^ 3-5 



.-. tan- (5- (7) = T-r^ cot - = ^2^ cot 6a' 30', 
2 ' h + c % 33*5 •^ 

when i (:B - (7) = 4° 26', and - (5+ O) = if 30', hj question, 

.-.5=31^56', O^asW; 

.•. /^J9(7=45'' + 3i''56' = N 76^56'E, the direction in 
which the privateer sails. 

Again, BO^AB . j^^ > whence £(7=20*907 miles; 
.*. 20*907 -r i^= 13*938 miles per hour. 



31. Let A, By denote 
the positions of the three ob-» 
jects. 

At A draw a line, making 
an 

zGAD = /.C8B=^S''53\ 

At draw a line, making 
an 

Z-4aZ> = Z-4>SB=i4**i6'. 

Of these two lines, let the 
intersection be 2>, and about 
the AAJDC describe a circle; 
join BD and produce it to 
meet the circumference in S; 
join AS, 08. 




168 EXPANSIONS, SEBIES, &C. 

(i) Find AD, from the data, AG, zs GAD, A CD. 

(a) 2: BAG, AB, AC, BO. 

(3) jL&ABD,AI)B AB, AD, a BAD. 

(4) A8, B8 AB, Z&AB8, ASB. 

(5) 08 AG, AsAC8, A80. 



36, Let z ABD = 0, 

zDBO^<t>y 
AG ^216, 
CD^5. 

rrn tan ^ AD an 
tan^ DO 5 

tan — tan ff> __ 206 __ 103 ^ 
tan^ + tan ^^ 216"" 108' 

Hence 0^ "{{0 + <l>) + {0 — if>)} is easily determined, 
and £i? =-4i? cot 0, gives the required distance = 200'2i ft. 




• • 



Expansions, Series, &c, 

Ex. 19, 

1. Let 2,cos0= x+x"^; then 2cosw^ = a?* + «"* ; 

.\ {2COQ0)^=^a^ + 6x^+ 15a:* + 20 + 150?"^ + 6a?"* + a?"^ 

= {a^ + x'^) + 6{a^ + x'^) + 15 {a^ +a?~') + 20 
= 2 cos 6^ + 1 2 cos 4^ + 30 cos 2^ + 20 ; 

/. cos^ ^ = — COS 6^ 4- ■%cos4^4- —cos 2^ + -4. 
32 10 ^ 32 16 



EXPA17SI0NB, SERIES, &C. 169 

3. Let acos5 = a5+ar*; then iiV^Bm5s=a5 — or', 

and («V^sin^)^=(a? — ar')'*=aJ*-4aj* + 5-4iir* + a?"^ 

«= («* + »"*) - 4 (a?» +ar*) + 5, 
2* sm*5 = iicos4^ — 8 cosii^+5; 

/. sin* ^ = 3 cos 4^ QOB^O + $. 

o (I o 

6. JjQt 2,cos0=:x + x'^i then 2,*/^ emO^^x-^x"^; 

••. (a V^ sin ^^ = oj* — 5a?5 + loa? — loar' + ^xr^ — ar^ 

= (a^ - £c~^) -5 (a?* — aj"5) + lo (a? - £c""'), 
a* VIIT sin* ^ = liV-i sin ^0 — lo V— i sin 3^ + i^o V— i sin^; 

,*. sin* ^ — "< W^ 5^— 5 sin 3^4- 10 sin ^}. 

7. By Demoivre's Theorem, 

cos 4^4- V^ sin 40 = (cos ^4- V^ sin ff)* 

= cos* 5 + 4 V^ cos'^ sin ^- 6 cos* sin* ^ 
-4 V^ cos sin^ ^ 4- sin* 0. 
Equating the impossible terms of this equation, we have 
sin 4^ = 4cos^ ^ sin ^ — 4 cos 5 sin^ 0. 

m 

2, cos 2^ e ' + e 

_ (e^>^^ , e-^^^) [e'^^ 4- 6-^^) ^ 2 V=T sin g 2 cos g 
{e^^-- + e-*^-^)« - <j (^ cos 0Y - 1* 



/ atand . ^ !itan^ 

V— I. ^-Tn'y .'.tan 2^ = T — j-a* 

2 - sec" 0' I - tan* ^ 



22 



170 EXPANSIONS, SERIES, &C. 

11. aco8a^=6^^' + 6^^"^ 

_ (2cosg)' + (2V^smg)' 

= 2COs"^— asin'd; 
.*. co8«d = co8*^ — sin'd. 



Ex. 20. 



iV^ . _-«v^ 



1. Let 5 » tlie required sum, and 

then a5= a + (6^"^+ €-*^') + (c^^'^H- e-'^'^O +...to n terms 
= i+^'^' + €^^' + + €<*-')*^^ 



6^^^ -I e"**"^-! 

cy cos(n— i) a? — cosna;--cosa?+ I 
"" 2 — ii cos 0? 



sin 


fw- 


-aj'"- 


sm - + siTi' 

2 


X 
2 


1 


sin 


(• 


-;) 


a? + sin 


ay 


sin 


nx 

2 


12 sm* - 

(n — i) a? 
, cos^^ ^ 

2 








. X 

2 sin - 

2 




- 


, X 

sm- 

22 


^^^^^^^^ • 





EXPANSIONS, SEBIES, &C. 171 

2. Let S^ the required sum, and ij cos 0^z-\ — , then 

2 cos n^ = «" + -u , 

z 

= i+asz+ {xzy+ ...... + (aj«)*"' 



a?« — I a?«""^ — I 






a^ — xlz + -1+ I 



^ 



a?*"^' cos (n — i) ^ — a?" cos ng— a; cos ^ 4- 1 
•'• aj*-iiajcos^+i 



3. Let iaV- I sin (9 = e*^"' - e"*^ fi « the required 
sum; then 

N.B. Proceed as above in i. 




6. We have cosec54-cot^ = cot-, 

(I 

hence cosec = cot- 5— cot ^, 
cosec 1*5 SB cot 5 — cot lid, 



cosec 2,"^ 5 = cot li"^ 5 - cot li*"* 0; 
and .•. /S5=cot--cot2*"'5. 



172 EXPANSIONS, SEBIES, &C. 

10. Since tan5 = cot5-acot2^; 

/. - tan- = -cot — cots, 

— tan — , = -J cot — - — eot - , 



tan -=-r as -—■ cot -=-; ==z COt 



I S 

.'. the sum S=^-i-:;COt-==: — 2COt20* 

2 ^ 2 



11. Let acos^ = « + -, and /9s=tlxeteqtuiedBam. 
Then 

« 



^•^ ^^ ^ cos • j^ i /^ slid V^ ^g-»iiii*V^\ 

5-^co8«coa(aJsinS). 



14 Let li V— I sin S = iSf -^ - , and /S= the required sum. 
Then 

.vr-.«..(._i)_|(,._i)^|(^-i)- 



= log (i + xz) - log ^I + -j = log 



I +XZ 

' + z 



.: e 



S^_I+£« 



0/ 



EZPAKSIONS, SEXIEB, &C. 173 

X 

Hence ^ "'~':= g aa^V-iaing 

i+a?cos^ \i+ajcos^/ 

17. sm^==e ^+ ^ 

i.a-3 1.^^.3.4.5 






+ 

Therefore equatmg the coefficients of 6^, 

••1* + ?' + ? + ^6' 

20. c«^'^-e'-«^^ = n{6<*+«»^^-e-<«+a»Vrii. 



.*. (i - nefl"^^) e*^— = (i - ne-*"^) e-«^ 
hence ^''^^ = i - «g-<»^ 

« 

or ax V=rr = log, (i _ „<,-«v=T) _ log^ (j _ ^v=ij 

+ J (e^"^ - r^^, &c. ; 

/. a? = nsina+— sinr^a4-— sin3a + &c. 

^ 3 



174 EXPAKSIONS, SEBIEiS, &C. 

22. If acoB^ = a! + -, and aco8jB=y + -: then 

X y 

a sin A x — 



b SihiB 






= i flog. (1 - »•) + log. (!-»-»)- log. (i -y*) - log. (i - sr*)} 
a 



i{- (a!» + - aj*+&c.j - (ar»+ ^ ar* + &c.) 



= - [{(y' + y") - (^^ + *'^)J + ^ {(3^ +3r») - (a;* + ar«)} + &c.] 
= (coa a-B- cos a^d) + i (cos 4B- cos 4^) + &c. 



23. We have c'ss a» + &" - aa&cos (7; (if 2 cos C^x + x"') 

ss a" + J* — aJ (a? + arO 
= (a — Ja?) (a — JaT^i 

.-. a log. ^ = log. (i - I «) + log. (i - 1 ar') 

= ^i{x+x-')-^,ix' + x-)-&c. 
a^ ' aa' 

.-. alog.J = |(« + »-)+^(«* + ar') +&c.; 
/. log.-=-co8C/+-^cosaO+— vcos3(7+&c. 

" c a aa 30' 



EXPANSIONS, SESIES, &C 175 



Ez. 21. 



1. Since ton 0s -; .*. cosd — 






Z 

n 



4. By the preceding example, we have 

sin5=:ia*^'sm- .sin— ^—. Bin — -^— sm^- — ^ ^^ 

n n n n ' 



put — + 5for d; 



-1)^+0 



••. sin( — +5j = ji""' cos- cos cos — — ...cos^ ^ 

Now if n be a mtiltiple of 4, sin [— + $) = sax0; 

.-. tan ^tan=^ t^(n-i)^+g^ 

n n n • 

wliatever (0) be. Put 

^=^, and tan ^ tan 5!r tan 2!: tan ^i^IZ^W ^ i 

4 4n 4w 4n 4/1 ' 

wlien n is a multiple of 4. 

If « = 2»», 8inr^ + d) = (-i)«8in0, 

and tan ^. tan 5? tan ^^ ~ 3) ^ ^ (- i)«. 

4n 4w 4w ^ ' 



176 BZPANSIOKS, 6EBIES, &C. 



IT 



If nss2(9n+i, and ^ = -, 

4 



Binf— +5j = Bin rwir + ^j=co8i»7rsin^=(— i)*sin-; 

sin 6 , . _ 

or tan -^ tan ST tan 2^ tan (^^Lzilz = (_ ,)• 

when n is of fhe form %m + i. 

Hence if n be of tlie form 4m, or 4m + ij ^ = i ; 
if n be of the form 4W + 2, or 4??i + 3, -4 = — i. 



7. ^=i+a; + + + 

and e =1— ojH h 

1.2, I.2(.3 
I s^ ^ 

»^ ' 1.::^ i.i&.3.4 



I 

1.2, i.a.3.4 



Now 

/ a'o^v 2^x'\f 2'X'\ 

...=cosa.= ^i-— j(^i-— j(i~— j.... 
But if the last expression, and 

(-^('-f)('-^) ■ 

be multiplied out, the results will be the same, excepting that 
in the former the terms will be alternately positive and negative, 
and in the latter all the terms will be positive, 

.-. 2<1+ + : + > 

I I. a 1.2.3.4 j 



Similarly, e" - e-* = oo? fi + ^Ui + -i^j. 



EXPANSIONS, SEKIES, &C, 



. 9, Let ABODEF be a regular 
polygon of n sides, r the radius 
of the circumscribing circle; then 
the angles subtended at the centre 
"by AB, A C, AD, &c,, drawn from 

. ZW ATT 67r p 

-4 are — , -^, — , &c. respect- 
ively ; 



177 



n n n 



TT 



/. AB—2r sin-, 

n 




. ^ . 27r ,4 n ^ sm 37r p 

AG=^2r&m — , AJJ = ar — ^^— , &c., 

n n 



and S the sum of all these lines 



f . w . . aTT , . 37r . (n - i) tt] 

= 2r Jsm - + sm hsm ^^ + + sm-^^ ^— k 

\ n n n ^ ) 

XT 1 i. C/ . TT , . ijw . , . (n — i) w 

Now let >8: = sm- + sm h + sm^ — ; 



.'. ijcos— o«sm hsm^ — hsm^^— + 

n n 



TT 



. 27r 
n 



n 

+ sm — 
n 



+ sm- + sm — + + sm^^ ^— 



/S- sm - 4- sm h 8- sm ^^ — = 0,3-- a sm - . 

n n n n 



. IT 

sm — 



hence /Sfi — cos — ) =« sin — , or 8^ ^ : 



li sm' — 
2n 



.-. S^ixr/S'^ 



ar sm — 
n 



— = ..i-..i«.. — ^ cosec' — . 
• !ism" — 2sm' — 



in 



%n 



Ex. 22. 

3. If a; = m cos d, we have 

m^ cos^ ^ — 3w cos ^ 4- 1 = o, 

^ I 

or cos^ ^-.-2-cos^+ — = o 



(0, 



23 



178 EXPANSIONS, SEBIES, &C. 

but coB^^ — 2cos^ — COS 3^ = (2); 

4 4 

.•• comparing eqtiations (i) and (a), 

m — z; and /. cos 3^ = — -^ = , 

or 3^= lao®, and 37r + lao®, 

hence the valaes of a; are 

2 cos 40*^ ; a cos 80** ; — a cos 2o^ 

■ 5. a;4 = — I =cos(ar+ i) w + V— I 8in(ar+ i) tt; 

.-. aj^cos-i ^— ±v-i sm-^^ --^ — , 

4 4 

and since the index of x is even, we have, giving to r the 
values o, I, 

aj = cos^ ±V— I sin — , and cos 2- ± V— i 2_ 
= 4- + -^ X V^, aiid - -7- ± -7- X ^--^ 

jj I- V - 



Ex. 23. 



5 = a»+J»=a»(i+^[), 

put — = tan <^, then 
a; = a' sec* <^. 



fllTl -4 

6. cos a? « cos -4 {cos c . j sin 5— cos B} 

cos .^ 

= COS -4 {cos c tan -4 sin 5— cos J9}. 
Assume cos c tan ^ = cot <^. 



APPLICATION OF ALGEBRA TO GEOMETRT, 



Then, cos x = cos A {cot ^ sin ^ • 

cos J , . 
= — : — T Ism B cos 6 
sm9 ^ ^ 

cos^ . 
= . . sin (^— 6). 
sm ^ ^ ^' 



cos-B} 

• cos B sin 0} 



179 



2, 



Application of Algebra to Geometry. 

1. Let X be the greater segment; then 

lOi : X i: X : 12, — x^ 
or a?' = 144— i7,x; 
.'. aj* + i2a?= 144; 

/. {x+6y= 180; 

/. x + 6=i 13*4164; 
/. a?= 7*4164 in., 
and i!^- a; = 4*5836 in. 

Let a? = the hypothenuse. 

Then ar» = (a?- 8)» + ao* 
=5 3?*— i6a?4-464; 
.*. i6aj = 464, a? = !i9 ; 
/. a? — 8 = 21 ; 
.'. the sides are 20, !2i, ^9. 



7. Let GB=-x, 

CA^z; 

then ^5 = (ar» 4- «")*, 

x + z + {af + z*)^^20 (i), 

GB X (7-4 = aarea of J.5(7 

= Oi> X BC+ OExCA+ OFx AB; 



5 
4 



(^). 




From (i), 

a3* + «'= ao* - 40 (a;4-«) + {x' + z^ + %xz) ; 



180 APPLICATION OF ALGEBRA TO GEOMETRY. 

40 4 

Squaie and subtract 4 times (2) ; then 



X 



.,.^m:,.,s'^m^^^i,.r. 



••• a?-|(9 + Vi7)=8*iJoi94,' 
«=|(9-Vi7)=3-048, ^ 



; .•.^5=20-^ = 875, 

4 



9. Let AB^Cy 

CA^x, 
OB = z. 

Then sc' + z^ = (^ (i). 

By similar triangles 




CM: ON^AB : PQ; 
.•. CM : CM-8 = c : s. 



Dividendo, 



CM: 8 = c : c — s; .*. CM— 



cs 



c— « 



Now Oix CB = 2areaof-45a=Cilfx^J9; 



• • OOiS ^ 






(a). 



Multiply this equation by (a), and add the result to (i) ; 
hence aj4-« = c(i + -3-) "^(""IT") • 

Similarly, x-z^c(x-^^*=c(^^f', 

_ <> f (£+ *)* + (<' ~ 3*)* 



c f 



(«-«)• 



J' al (c-s)* )' 



APPLIGATIOK OF ALGEBRA TO QSOHETRT. 



181 




(2) Using the same notationi b 
we nave 

iC» + J5' = C* (i), 

xz^H area oiABO 

= 5(a?4-«) (a); 

••. (a? + «)"-a«(ic + «) + ^ 

= c* + ^, 

Hence the sides 



11. Let Sj St, S^ be the areas 
of the triangles ABCy CBD, ACD 
respectiTely ; p^Pn p^ be the semi- 
penmeters of the triangles ABG^ 
CBD, A CD respectively. 



Let -4jB = c, AO^h, CB=a. 

8 




S. 



S, 



Then r = ^, r.=^, r,=^'; 

a* + b* 

= — u— "i; 



.'. r*=r,* + r,\ 



13. 



Let CA 
CB 
CD 
BD 
DA 



5, 
6-4, 

4; 

y- 




182 



APPLICATION OF ALGEBSA TO OEOMETBT. 



Then BD : DA :: BO : 04, by EucL Bk. vi. Prop, in.; 

.-. 0? : y = 6-4 : 5 (i). 

And -RDxjD^ + jDa*= OBx CA by Eucl. Bk. vi. Prop, a; 

•'• ay = 5 X 6*4 — 4* = 16. 

From (I), 2 = ^ = ^; 

16 X 64 3a , , 

16 X <o 30 , ^ , 

y*=— 3-^, .*. y = -§- Va = 3*5 Va ; 

.'. AB=x +y = 57 V3 = 8-06094. 

A 

14. Biaect BC= 14 in 0\ 
let OZ> = a!, 

DA = ^; 
Then 

J9^* = £i)»+D^» = (7+a)»+8», B- 
^(7* = (7i)»+2>^» = (7-a;)»+8' ; 

But BA-AC = 3-5; 

.: 3A + AC= — = Sx: 

3'5 

.'. BA=4x + ^, and AC=4x-l, 

4 4 

^4aj + 2j = (7+a;)« + 8% 
i6a;'-a^ = 49 + 64-g = £259. 

.-. AB = io'8a9 + 175 = 13-579 ; ••• ^ G= 9*079. 




17. In the fig. to (14), let AB+AC=zd, AD 
BD-DG=%n', BO=OG=x', 
.'. BD = x + n, GD^x-n. 



=i'. 



APPLICATION OP ALGEBRA TO GEOMETKT. 

Then AB*== {x + ny+p\ 
Hence aa in (14), AB = d + -j, AG^d— 



183 



d 



d' 



and x = d(i + t_^ ) ,. •*• tl^e sides =d±n{i + J^_ ,\ \ 



20. 



Let BG^x, 

CA=y, 

AB = z; 

AE=h, 

BF=h, 

Ga=i. 

3« = a!+y + 2. 




Then, 3 area of ABG = hx = hy=lz. 



8 



I , , V X f h h\ 



s 



If, N X f h h\ 



Now area of AB G = ^8{s — x)(s'- y) {s — z); 
* A = -^, {(iM?+ hk) (hk-Jd+hT) (M-hl+hk) {kl+M-Ak)f 
If tiie root be pat = D, we obtain 

a; = \ ^ > y = — VsT^: » = — ^ — ^ 



hD 



kD 



ID 



25. Let BG 
CA 

AB 

BP 



a, 
h. 



x; 



.♦. CP = a — x. 




184 



APPLICATION OF ALGEBRA TO GEOKETRY. 

Now AF*^BPx CP^x{a-x). 
Also c' + a?"-aca?cos5 = uiP* = fiB(a-a?) ; 






7,ac 



-ja? + c*«o; 



aa' + c»-JV /aa' + c'-JV c* 



4a 



■)-( 



4a 







.-. a; = — {ao* + c» - J» ± (40* + J* + c* - 4o»6* - 4aV - a6V)*}. 

26. Let ^5 = -42) = 30; 

BF^x, BE^z. 
Then 

aj»+«*=i6' = 256 (!)• 

By similar triangles, 

BF I BEx: AF x AD] 
.\ 30x^{30 + x)z; 
.-. 0^ = 30 («-«) (2). 

Subtract twice equation {2) &om (i), 
hence (a?-«f)* + 6o(a:-;5) +3o* = 900 + 256= 1156; 
/. a? -«« 34-30 = 4; 
.*• aj + « = ViJ56 + 60x4 = 22*27 10; 



31- Let P, C, and JB be 

the points of contact of the 
three circles, and HO the chord 
of the exterior which touches 
the two interior circles at their 
point of contact B ; let A^ By C, 
and J^ be the areas of the cir- 
cles described on PQ, PB, BQy 
and HO respectively, as dia- 
meters; and let 

PB^d, 
BQ^d. 




APFLICATIOK OF ALGEBRA TO OCOUBTBT. 



185 



Then we have 



TT 



4 



IT 



B 
P 

c 


\ 


M 






\ 


A O 


1> Q JJ 



4 

But E^v.HE'^v.PR.RQ^irdd . 
Hence -4-5- (7=rJS?. 

32. ■ Let ^4^', 55' be the 

?'ven lines, JfcT the given point, 
Q the leaoired line. From 
M draw MCy MD perpendicidar 
to BB\ AA' respectively. Let 

then Pif« + jlfg» = J«, 

or c?* + y» + rf* + aj» = i% 
and from the similar triangles POM^ MD Q, 

X d 

y 

or oj* +y* -%xy^ J» -4<i* ; .'. a;-y = (J' - 4^*)* ; 
whence a = - {J ± (J» — 4^")*} ; 

36. Let ^5, 5C; GZ? be the 
chords in the semicircle ABGD* 
Join -40, BB. Let 

^5= a, 5(7= i, 
05= c, ^5 = a?, 
Then (Eucl. vi. d), 

^0. 55 = ^5. 50+ui5.05=Jaj + oc. 



2 = -, or xy:=:d'; 




24 



186 



APPLICATION OF ALgEBRA TO GEOMETBT. 



But, since ABD^ A CD aie right angles, we have 

Or, squaring both sides, and reducing the result 

aj* — (a* + J" + c*) a? — 7,0^ = o. 



41. Let ABCD be the given square, ^ 
and EFHG the inscribed square, having 
its side equal to a given line. 

JjtiAB^a, EF^h, AF^x, and 

.-. AE^FB^a-^x. 

Since AE^ + AF*^EF*; 

.\ a" + (a-a?)* = i*: 

whence a; = - {a ± {2b' - a*)*} ; 




therefore the limits of & are a and 

42. Let -45 = a, BO^l, 
AO^d, Draw -D-BT perpendicu- 
lar to AC^ and let 

By similar triangles 
J^jE: BC :: AN: AC, 



^% 




also, ^^ : AD :: .42) : ^O; 
Similarly J = ^; 



APPLICATION OF ALQEBRA TO GEOMETBT. 



187 




43. Let the angle OCB=e, 
and .-. OGD^yd'-e, 
then A* = aj* + A:* — a&D cofltf, 
Z* = a?" + i* — aib? sin tf; 

.% aj* + A:*— A* = 3feccofltf (i), 

and aj*+i* — P = afec8intf (2), ^ 

and adding the squares of (i) and (2) we get 

49. It may easily l)e shewn that 

A r + r, +r, + r3 + - a (r, + r, + 7-3 + ) ; 

/. r = r, +r,+r,+ 

51. Let r be the radiosy 
and the centre of any circle 
touching the two sides of the 
square terminated at the point 
Ay r, the radius, and (y the 
centre of the circle which 
touches the last circle, and the 
same two sides of the square; 
draw OMy O'M' perpendicular 
to one of the sides of tne square; 
then OJf = AM, OM'^AM' ; 
therefore ui(70 is a right line 
passing through P, the point of contact of the two circles, 

and ^0 = rVa=0P+P0' + 0'-4 = r + r, + r,Va, 

Similarly if rj, r^, &c. be the radii of the consecutive cir- 
cles, we shall have 

»*a=(3-^Va)n; ^3 = (3 - i& Va) ^a, &c. = &c., 




188 



APPUGATION OF ALaEBRA TO GEOKETBT. 



and the sum of the areas of all the circles 
« ttt" + wr/ + wr,* + nr,* + 



-(?)' 



^ TTr 



I — (17 — 1% V«) 12 Vi^ — 16 



^ 4 U88-256/ 4 V 8 y* 

if a be a side of the square. 

63. Let 0, (7 be the cen- ^ 
tres of the spheres, touching the 
slant sides of the cone in the 
points -F, H, 

Let OE:=OF=^Ji; 

From similar triangles, we 
have 

R \ r x\ OC : (TC, 
or R^r X r v. R+r I OG\ 

R—r 
Similarly, 00 = ^^^; 




B- 



M-r 



Again BE : BC :: 0^ : 00" :: 5-r : B + r; 

.-. 5^» : EC' = {B-rY : 4Br; 

Volume of cone = i wBE' xEC^-.—. ^- ^— ( ^^' \ . 

3 3 r B-r 3 XBr-r*} 



APPLICATION OF ALGEBRA TO GEOMETBY. 



189 



54. Keferring to the figure of the last example, let 0, 0„ 
Oa, &c. be the centres of the spheres ; then r, r„ rj, &c. being 
the radii of the spheres, we have 

r= 0(7 sin a = (r + r, + OiC)Bma= {r + r^) sina + r,; 

/I — sin a\ , » 

/. rj=^r[ — -—T (i). . 

Again, r, = 0,(7sina= (r, + r, + O^C) sin a= (r^ +ra) sina + r^; 

'•*^" *^^ Vi + sin ay Vi + sin ay ^*^- 

Similarly, r^ = r Q^^|^ J , &c, = &c., 
or the radii are in geometrical progression. 



55. Let ACDB be the given 
frustum of a cone. Complete the 
cone. Let FO^ HK be 'the diame- 
ters of the sections required. Draw 
the altitude OL of the cone,' meeting 
the sections ia E, I, M and L. Let 

AL = o, 

CE=l, 

JEL = h, 

and FI^x. 

Then we have 

a : 6 :: OE-Vh : 0E\ 

.\ a-h : I V. h \ OE; .-. 0E== 




hh 



a — b' 



hh 



hh 



Again, x : h xi EI+ ^ • 5 , 

hh 



.\ x — h : h :: EI : 



a 



-b' 



.-. EI^ *-(^) ; 



.-. MJI = — i^ r-^ ; 

a — 6 

therefore, by the question, we have, since the volume of the 

given frustum = — (a' + oJ + 6*), 



190 APPLICATION OF ALaSBRA TO OEOMETBY. 

3{a-6) 9 

whence x = [ j = the radiufi of one section. 

Similarly, the radios of the other section 






66. Let -45 =-4 (7= 12, 

Then 
AD^^AC*- OD" « iii» -5*= 119 ; 

.*. area of base ABG^ - 10 Vug. 

In the triangle VAD^ 
FZ>» = F-4* + ^2>»- aiii> x AE; 




.-. ^LE?=: 



F-4^ 



TOO 



, since VD = AD. 



7,AD 2^/iig 

FE?'=F4»-.u4J^» = ioo I22L=ioofi-^V 

4x119 V 119/ 

Content of pyramid = i F^^x ^J?0=- (10 a/-^) (5^119) 

= ^V94= 161-589. 



ANALYTICAL GEOMETRY 



AND 



CONIC SECTIONS. 



I. Stbaight Line, 



Six. 1* 



1. 



If y = o, 



X =s — 



and if a;ssOy 

y = i. 

Therefore, X'^X and r-4r' ?i 



"being assumed as co-ordinate 
axes. Take 

3 
and AM= i, 

and join MN^ whicli will be the line required. 

6. Dividing the given equa- 
tion by 2, we have 

^ . y 
6 14 

On the rectangular axes X'AX^ 
1:4 F', take x' 

AN=^ 6, 

and AM^ 14, 
and join 2fMy 

MN will be the line required. 





192 



STRAIGHT LINE*. 



9. At die point of intersection of the lines, the values 
of X and y will be the same in both equations, 

6y — 20? = o, 
/. 73^=1, and y=-, 



13. Let AD be the distance from the origin of co-ordinates, 
of the given line, and let z DAX= a, then 

cota = -, and therefore cosa = -7— . 
5 V^9 



Now AD = a; cos a = 5 X 



lO 



V29 V29 ■ 



18. Let y = - a; be the equation to -4P, 

5 




The co-ordinates of the point of intersection P are 

Hence, y — i/' = A{x — x'). 
Since PQ bisects the angle ^P£, 



8TBAIGHT LINE. 198 

^ tan/9 + tana 15 14 

I— tan^tana •_/ 4\* ^3 

V 3/5 

= tan(i8o«-3iM9'43"); 
.-. ^ or tantf = tan 74* 20' 8" =3*566; 



•••y-7| = 3-566(«-f^)- 



N.B. If tana = a, ismfi^m^ 

tan 20 = = - suppose; 

I — ma c '^^ 

.•. 2ctand= I— tan*d, 

tan*d + 2ctand + c*= i + c*; 

«» + « 

22. y* — aojy secd+ aj*sec'd= aj*sec*d — aJ*s*rc"tan*d; 

.*. y = a? (sec ^ ± tan ^, 

the equations to two straight lines passing through the origin. 
If ^ be the z between them, 

^^ . _ (secg-f tan g) - (secg -- tang) ^ a tan g ^ 
^~ I + (sec^ - tan» 0) 2 ' 

.-. <l> = 0. 

25^. The equation may be written thus, 

4«(3y + »)-9(3y + «) = o> 
(4»-9)(3y + a) = o; 

.*. 42; — 9 = o, a; = ^ the equation to a line parallel to the axis of y, 

3y + 2 = Oy y«--, ,...........*... X. 

25 



194 STRAIGHT LIKE. 

25^. Solving the equation with regard to y, 

y«+ ay (oa? - i) + (20?- i)* == (4x5* -4a; + 1) 

— a* — 20? + a = 3 (aj* — aa? + i) ; 

.-. y«-(2aj-i)±(aj-i)V3 = (-^±V3)«+(iTV3)» 
which represents two straight lines. 

27. The equations of two straight Unes referred to ohlique 
axes inclined at an angle », beingy — ww5 = o, «ty + a; = o; find 
the angle between them. 

Let be the angle between the lines, and oc, /3, their respeo- 
tive inclinations to the aids of x ; then 

sin a / ^ 

-r—f V =w (i); 

sin (cD — a) ^ 

«^d -T— 7 5^=5 (2), 

sin («— p) m 

- ^ i»sina> , ^ 

whence tana-_j___ (3), 

tan^^ 2B^ (4). 

«* — COS© 

Now 0-a-fif and 

^_ toa-tanj3 
i+tana.tany8 

m sin a> sin a> 

I + wt COS a> m — COS ft» _ {m* 4- 1) sin <» 

"^ m sin (0 sin © (m* — i) cos (w 
I . ^ 

I+WCOS© fW— COS© 

w»+i . 
ar— - — tan©. 

80. Let »♦ cos (^ — a) =/) be the equation to the line. Then, 
since the line passes through the two points (r,ax), {r^a») we 

have 

r^cos (ax-a) =/> (2), 

and r, cos (a, — a) ==^ (s)- 

From (i) and (2) we get 

(r cos ^ — r, cos a,) cos « + (r sin ^— r, sin a,) sin a =0, 



STRAIGHT LINE. 

and firom (z) and (3) 

(r, cos a, - r, cos a,) cos a + (r, sin a^^r^ sin a^ sin « 

. rcosg — r,co8a, _ rsini?— r, sing, 
** r,co8a, — r, coso, r, sinai — r^sina,* 

Whence, by reduction, we obtain 

sin(g^-a,) Bin(tf-a^) 8in(a,-g) ^ 



18S 



o; 



36. Let y 

y 



^ tan a, be the equation to AB (i), 

a?tanag AC (a), 

xtana, + a BCD (3). 




Now AABO^AABI)-'^ACD^^AD{BN'-OM). 

In(3)puty = o; then a? or ^i> = ---^. 

tana. 

To find BN^ eliminate x between (i) and (3) ; then 

y (tan a -tana,) =s a tan a; 

g tang a sin a cos a. 



.'. y or BN= 



tana— tan tta sin (a — a,) 



Similarly, caf = ?-?^ «' «» «' 



sin (a, — a,) ' 

.-. A ABG = - ^-^?^ L '^^ , - _«E«£_l 

a sin a, l8m(a-a,) sin (a, -a,))' 

which by reduction, 

_ g* ( sin (g - a,) cos' a, ] 
2 (sin (a— a,) sin (a, — a,)) 



196 CIBCLE. 

89. Draw the ordinates AL^ BM, CN from tlie angdir 
points Ay B^ (7 of the triangle, whose abscissae, a, a\ a" are in 
order of magnitude ; then tne area of ABO = trapezoid ABML 
+ trapezoid BCNM- trapesoid A CNL 

„l{(a'-fl)(6 + y) + (a"-a')(y + y')-(«"-a)(5+*")}- 

IL CiSCLE. 

Ex. 2. 

I. The equation may he put under the form 

»■ + 4^ + 4 + y* - 6y + 9 = 4 + 9 + 3 = I ^> 

hence, A« — a, i — 3; and = 4. 

6. Comparing 

aj* — a (cos 5 + V3 sin tf) r — 5 = o, 
with r*- 2r (A cos 5 + i sin ^ + A* + A* - 0* = o, 
the general polar equation to the circle, we have 

A cos d B cos 9 ; /. A = i, 

A; sin d s= V3 sin ; /. Jfe = V3» 
and jfe* + A»-c' = -5; 

For the centre, 

h 
tailtf=j = V3; •*• 

and r' = A*+A* = 4; 

II. Let y = mxy be the 
equation to the straight line. 
The origin of co-ordinates, when 
the equation to the circle is 

^+y'-3^+4y=o> 

is a point in the circumference. 
Now when y = o, 
x^i^AE, 
a? = o. 






3 



y = -^ = AF; 




CIECLE. 



T:. 



.'. m = tan HAS =tm UFA = ^: 

4 

.*. ^x = 4y is the required equation. 



•■ 14. Let £FC be the tan- 
gent at P whose equation is 

a^h ' 

then AB = a, 

' AC=h; 

radius = e. 



Now BP = 



CP = 



a" 



OB' 
I' 



GB' 

and BPx CP 



= AP' 






^ + J5 = ''' ^' 



- + - 



,a • 



197 




21. If -4 be the origin, 
y* = 3ra? — a^^ 
is the equation to the circle ; 

y^mx 
is the equation to the chord AP. 

Now^P=^lBcos^= — ^^. 

(i + m»)* 

\4^=u4(?cos^ = 



i + m"' 




NO = ANtm0 = 



mr 



Hence fa. !— )'+ fy — ^^y = _!l^ . 

and/. (i + w*)(a;» + y»)-i;ftr(a? + wy)=:o, 
is the required equation. 



198 CIBCLE. 

23. Take the point in the original circle for pole, and let d 
be the diameter of this circle; then r = dcos0 is its polar 
equation. 

If now the diameter d' of one of the other circles be inclined 
to the initial axis at an angle a^ d' = d cos a, and 

r = (? cos a cos (^ — a) 

is the equation to this circle. 

In like manner, the equations to the other two circles are 

r==d COS fi cos{0 — IS)y and r = (?cos7COs(^ — 7), 

At the point of intersection of the first and second small 
circles, we nave 

cos a . cos (^ — a) = cos )8 cos (O-^/S); 
.*. cos^ + cos (^— lija) = cos^ + cos(^— a)8), 

hence = a + fi, and r = rfcosacos/8. 
At the point of intersection of the first and third small circles 
^ = a+ 7, and r = c? cos a cos 7. 

In the polar equation to a right line, r cos (X — ^ = p, substi- 
tute these values of and r; thence we obtain 

(if cos a cos )8 cos {X — {ol+/3)} =jp =* rf cos a cos 7 cos {X — (a + 7)} ; 

.•. X = a+y8-|-7, and jp = c? cos a cos)8 cos 7. 

Hence it appears from the symmetry of these values that the 
same straight line passes through the intersection of the second 
and third circles as through the intersections of the first and 
second, and the first and tnird. 

27. The co-ordinates of the centre of the circle being A, ky 
the square of the distance of the centre of the circle firom the 
line a? cos a + y sin a = a will be equal to 

(A cos a + Aj sin a — a)' ,, . , . . , 

■i — -— r-- L = (A cosa + isma — a)*. 

cos" a + sm' a ' 

But this distance, the line being a tangent, must be equal to 
the radius; hence 

± (A co8a + A; sin a — a) =r (i). 

Similarly, (Acos/8 + A; sin/8 — J) =r (a), 

and (A C0S7 + A; sin7 — c) s=r (3). 



PARABOLA. 199 

firom eqnatioiiB (i), (a), and (3), we find 

1 _ ^ (sing — sin 7) —a(giny8 — ainy) — c(8ina — sin/S) 
"" Bin(a— /8) — Bin(a — 7) + sin(/8— 7) ' 

, _ a (co8/8— CO87) — h (coaa — co87) + c (cosa — cos/8) 
"" Bin (a — /8)— sin (a — 7) + sin (/9— 7) 

Substltating these values in (i) we have 

_ g sin 03 — 7) -f & sin (7— g) + c sin (a — /8) 
"" sin (g — /8) + sin (7 — a) + sin 08 —7) 

fl sin 08 — 7)4-3 sin (7 — g)+c sin(g->)8) 

— ~ — ^— — ^— — — — — ^— ^-^— ^— ^^^-^— ^— ^— — — — — — — — ^ 

4sini(^-7)8ini(7-g)smi(g-^) 

29. Compare 05* — a?y+y* — oa? — ay — o, with the g;eneral 
equation 

(a? - A)' + (y - A)* + a (a? - A) (y - A) cos o> « o% 

where oi is the inclination of the axes, and c the radius. 

I 2W 
2C08tt)=— I, .•. cos ft) = as cos , 

« 3 

^A + aA; cos « » a, 2% + 2% cos o» s a, 

and A" + i:* + i;ftA*cos» — e*«o; 

.'. »A - A == a, ai — A = a ; 

.*. A = a = i, 

c*=3!:u»" — a" = a"; .*. radiussa. 

« 

Parabola. 

Ex. 3. 

1. i + Jia? + 3y'»o; 

is the equation to a parabola whose latus rectum = - ; co-ordi« 



200 PASABOLA. 

nates of Tcrtex, A = — ; Jc = o; the axis of the curve coincides 

with the axis g(x; and the concavitj of the parabola is towards 
the negative direction of a;. 

4. y« + Jwy+»"+y-3aj + i=o. 

To take away the term involving ocy, let 

a; = ^cos^ — tt sin^, y = ^ sin^ + wcos^. 
Then 

y« «s t* sin" ^+ a<w sin ^ cos ^ + tt" cos' 
+ %ay «s at* sin ^ cos O+itu (cos* 6 —sin* ^) — aw* sin^ cos 6 
+ »■ = t* cos* tf — lift^ sin ^ cos ^ + u" sin* 
+y=3 +e8in^+wcosd 

-3a;=5 -3<cos^ + 3ttsin^ 

+ 1=' +i; 

/. o = f (1 + sin 7.0) + 2tu cos 20 + 1** (i — sin 20) 

+ * (sin tf — 3 cos ^) + tt (cos ^ + 3 sin ^ + !• 
Let the coefficient of tu = o, then cos 2^ = = cos 90** ; 

Hence a<* + aV^«*— V^^+ i = o> 

...(,--i-)-=-V.(..3^), 
the equation to a parabola, of which 

4 • . 16 ' 

the axis of the curve is parallel to the axis of u, and its con- 
cavity is towards the negative direction of the axis of u. 

7. (y + c)* + (a?4-c)* = i^c*; 

.-. y + aJ-ac = -a(y + c)*.(aj4-c)*; 
/. y* — 2xy 4- a^ — 8cy — Sea: = o. 
Now proceed as above in (4), 



PARABOLA. 201 

9. The equations to the tangents in terms of their inclina- 
tions to the axes, are 

y-mx + '- (i), 

and y = 7nx-\ — -, (i)^ 

and since they are drawn from the same point in the directrix, 
the values of y in (i) and {2) must be equal, when x^^a; 

/. — waH — = — maH — j. 
mm 

, m — m 

m — m= ■- — 7—; 

mm 

.-. mm =— I, or m = ; 

m 

•*. the tangents are at right angles to each other. 



11. Let QBq, PAp be the parabolas ; Qq being a tangent 
to the interior parabola at any point P{x\ y'). 

The equation to the interior parabola being 

y^^Aox (i), 

the equation to the exterior one will be of the form 

y» = 4a(aj + A) (a). 

Also the equation to the tangent to (i) is 

y = -^(»+»') (3)- 

Eliminating x between (3) and (jx), we have, 

the roots of this equation are the ordinates QMy qm, therefore 
by the Theory of Equations 

QM+qm _ , 

the ordinate of the point of contact of the tangent Qq, therefore 
the tangent Qq is bisected at the point of contact. 

26 



202 PARABOLA. 

13. Let y* s Jiv be the equation to the parabola, 
and (a?-A)* + (y-Ai)*«r" circle. 

Eliminating x between these eqoationSi we get 

(^-A)V(y-*)* = r«, 

or y^-(aW-P)y'-ii?%+Z*(A' + *'-r")=:o. 

Now the roots of this equation are the ordinates of the points 
of intersection of the curves ; if therefore y^, v,, be the ordinates 
on the upper side of the axis, and.— ^3, — y^ the ordinates on the 
lower siaei tiien by the Theory of Equations 

(yi+ya)-(y3+y4) = o> 

or y,+ya=y3 + y4. 

16. The equation to the tangent to the curve 

y» = 4(KP, 

is y=5maj + — , where w = tan^, 

and the equation to the tangent to the circle is 

y = »ia? + c(i +m") , 
and these two equations represent the same line ; 

.-. c(i+i»»)* = -, 

or c.sec0 = a cotd, 

I. . /) -c±(c«+4a»)* 

hence, sin 9 = - — ==-^ 

' 2a 



19. Let (a5-A)* + (y-i)» = c», 

be the equation to the circle of curvature required, 

radius of curvature = -ggi = ^ g^ — ^ = 4VM. 

Since ^(? = M = 55} .\ B'0^BS-^4a; 



PABAfiOLA. 



208 




.*. BO = 4 Vaa and is the centre of the circle of cnryatore; 
hence, 

.-. (aj-5a)«+(y+aa)* = (4Vj^a)'; 
/. aj*+y»-ioaa? + 4ay-3a* = o is the required equation. 



22. The equation of either tangent in terms of its inclina- 
tion to the axis, is of the form 



y = /3aj + -g> where )8 = tantf; 



I y 1 X 

p* a /8 a 



(I). 



but when sin^.8in^=m, cosec"^.cosec*^= — 



m 



t > 



•*-^=(^+0(r-+') 



+1 



X JL X 



=(;l+;^)+fe-0'' 






X 

a 



tOi PASiLBOLA* 



a* 



or (a5-a)»+y' = — , 

fn 

wUch is the equation to a circle whose centre is the focus, and 

radius b — s a cosec 6 . cosec ff. 
m 

(2*). From (i) we hare 

.*. -5 — ■^ = cot^ — cot^ = ^^^-— ^^ = w; 

fi P a ' 

/. y* — 4aaj = aW, 
or y'==4aaj+aV, 
which is the equation to a parabola, 

L^Aa. A = — ail?. i = o. 

24. Let z ASP— ^i then the polar equation to PT" is 

5? = cos^+cos(5-<^) (i). 

Simihirly,ifz-45e = X, 

2a 

— =cos^ + cos (^— X) is the equation tp QT. {a). 

m .1 

At the point of intersection 

cos {9 — ^) = cos (5 — X) ; 

... tf-^ = «(<J«x); /. d^\{j>+\), (3); 

2a /^ + X\ /<^-X\ d> X 

=COS [-^ 1 +COS l-^^-- — J = aCOS — COS-, 

\ 2 J \ 2 J 2 2' 



• • 



ST 



COS ^ COS - 
2 2 



whence, bj the polar equation to the parabola 

ST' = _£L. ._i- = 8P. SQ. 
cos'-^ cos'- 

2 2 



PARABOLA. 205 

Again, from (3), + X = a^, 

but X — ^ = a by hypothesis; 

which being substituted in (i), we have 

%a >, , I 

— = cos a + cos - a ; 
r % 

I 
aasec-a 



I + sec - a • cos ^ 



for the required locus, which is an hyperbola, whose eccentricity 

I 
= sec - a. 

2, 

29. Let )8 be the constant angle between the tangent and 
the focal line. The equation to the tangent at* any point 
x\ y' is 

y = ~7^(aj+y) (i)- 

The equation to the focal line is 

-r + tan /8 . / x n 

y= ^aa, ^ -^^-^^^ y-aatan^g ^^"^) — ^^^^ 
I — — TT tan p "^ 

Eliminating y , a;' by means of (i) and {%) and the relation 
y* = 400?', we have 

yiy— (a?— a)tan/8}{aj — a + y tanyS} — a?fy— (a? — a) tan/8}* 

^a{x-'a + y tan/8}* = o (3), 

whence, reducing, 

{y* + (»- a)*}{y tan/8 — a; tan*/8 — a} = o. 

If y" + (a? — a)* = o, then y = o, oj = a, these are the co-or- 
dinates of the focus, but not of a point in the locus of (i) and (a), 
for these values do not satisfy both (i) and (2), hence the re- 
quired locus is 

y^x tan /8 + a cot /8, 

which is the equation to a tangent inclined at an angle /9 to the 

axis. 



206 



PARABOLA. 



83. 



Let AM^Xy MQ^y; 
AN^x\ MQ^y'. 




Smce SG^SP, the line SQ bisects PG; 

/. QM^'^PN, and GM=^GN=NM. 

Hence y^-y\ and x^AM^x' + a. 

Now y'*s=4aaj'; /. 4y* = 4a (a? — a); 

•*• y^^a[x^c^j the equation to a parabola whose latos 
rectum = a, and vertex is at S. 

35. Let h be the abscissa of the given point from which the 
normals are drawn; then the equation to the normal is 

y^-^{x-h) (X). 

and the equation to the parabola is 

y* = 4«» (a); 

therefore, eliminating the variable parameter a by means of (i) 
and (a), we have 

or y* + i;fta;(aj — A) =o, 

for the equation to the locus which is an ellipse, whose minor 
axis (coincident with the axis of the parabola) is A, and major 
axis is h hj%. 



PARiLBOLA. 207 

41. The equations to the parabola and circle referred to 
the tangents as axes are 

8/-®*- (■). 

and «+y— 2 {xyy sin-© = c. cot-© (fl). 

Now for the points of intersection of the two curves, the 
values of x and y in equations (i) and {o) must be the same, 
therefore, eliminating y, we have 

x{h + h'^Q, (aJl)* sinift)} - a {i (A)*+* (A)* sin^©} 

+ h {i— c cot- ©} =s o, 

and since for the point of contact x has but ^ne value, this 
quadratic must have its roots equal ; 

.•. 2^^\h + k'\-2fjhk sin-jr^ — ccot-js=4A^(Ar + A*sin-j ; 

/. |A + * + a(W)*sin|UA-ccot|) 

= i|* + a(AJfe)*sin| + *sin«|l, 

t AA;sinQ> 

whence ac = 



h+k + 2 ihh)^ sin - Q> 

2 



48. Let (a;", y") be the co-ordinates of Q ; (»', y') those of 
P\ and therefore —, , — ^ those of jp, then the equations to 
PQ and Qp are 

y-!''--^(«-«0 (I). 

•"*»-»"--7(?^")(— "> W- 

In both these equations let a; = — a, then from (i) we have 
or since y* = 4aaj" and y'" = 4aaj', 



208 PAKABOLA. 






• y'+y" ' 

and from (a) we have 

y Kd^ — xx J 

.•• DBxDr=' Afl*. 

61. See solution of 33. 

56. The fbcuB of the parabola being the pole, we have 

5^=1+003^ (I), 

r 

and ^=1-008^ (a), 

r 

from (I), {(^)-i}' = cos'5 :(3)> 

from (a), 0) - ij' = sin'^ (4), 

whence (^ -i)V(y-iy= I ; 

•'• (r ■" W ■*■ V " W ""(a^* 

64. Referring the parabola to the tangent QF, and the dia- 
meter at Q as axes, let (A, h) be the co-ordmates of the point B\ 
then the equation to the line BEF is 

and when this line meets the parabola 

k' = Ix, 

whence a? = -y = .EF; .*. JJ^= ^ "" y J 



PARABOLA. 209 

The equation to Qq is 

y = -j-aj (i). 

Eliminating x by means of (i), and the equation to the 
parabola y' = to?, we have 

y = — -7^ = the ordinate of q ; 
.-. liQ : Bq:^EF : ER. 



66. The equation to the given line shews that it passes 
through tlie focus, and is inclined at an angle of 45° to the 
axes. 

Let P8p be the given line, and let tangents be applied at 
the extremities P, p ; these tangents will intersect at a point T, 
in the directrix, and the line 8T will be perpendicular to the 
given line ; 

I — cos ^ ts/% — 1 

and 8p = . . a ; .'. PSp = 8a, 

and 8T^ {8P x8p)^=2^/ax a. 

The area of the parabolic segment = - AP2J?; 

3 

.-. area of segment = 5^rx^=i^a'. 

68. In the triangle ABCy let AB=c, BG^r, and 
z ABG^ 0. Draw CS perpendicular to AB. Then 

. BD ^ , X 

tan A = -j^ tan Q (i), 

and JRZ? = r.cos^; .*. -4i? = c-r cos^; 

27 



210 



.*. i$aA 



ELLIPSE. 


r Bin^ 
c — r COB ^ "" 


tani« 


r Bin 6 

• MM 


a cos - ^ 

2 


• " c — r COB 5 " 


Bin- a 

2 



/. r 8in"-tf = c — r COB^: 

2 



C 20 

/. r = 



Sin"- tf + cos a 



/. the locus of (7 is a parabola. 



Elupse. 



Ex. 4. 



35 35 

72 48 

Hence the co-ordinates of the centre of ellipse, are 

3 4' 

the semi-axes are a = [25 j ^ j -5 [ 35 j ^ 

3. To transform the equation, so that the rectangle of the 
co-ordinates may disappear, we must change the direction of the 
axes. 

Let a; = ^cos^--w sin^; y^t sin0 + u coaO. 



ELLIPSE. 211 

Then 

^x^^t* cos*(? "lotu COS ^ sin ^ + 5tt" sin* ^, 
-h 2X1/ = a«» cos ^ sin ^ + 2tu (cos* — sin* 0) — au* cos 5 sin tf, 
4-5y* = 5^* sin*^ + io^m cos ^ sin ^ +5w*cos*5, 

— X2,x = — 12< cos ^ + iriw sin 0, 

— iay= — la^sin^— lawcos^; 
.*. o = ^(5 + sin 7,0) + ntu cos 2,0 +u' (5 - sin a^) 

— 12^ (cos ^ + sin ^) + law (sin ^ — cos ^. 

To ^, give such a value, that the coefficient of <w may = 0; 
.•. cos!^^ =o = C0S9O°; .•.^ = 45*', 
and the equation becomes 

3 (<* - a V^^+ z) + 2u' = 6; 

.\ - — ^^—^H — =1; .•. A = V^, A; = o; a = V*, S = V3- 
^ 3 

6. Transfonning the origin to a point a, yS, and equating 
the coefficient of x and y to zero, we find 

a=:i, ^ = -1, 

and the equation of the curve when the origin of co-ordinates is 
removed to the centre, becomes 

3a * + 2xy + y'* - a = o. 

Turning the axes through an angle 0y and equating to zero 
the coefficient of ay, we get 

— 6 sin5co8 5 + a(cos*tf — sin*^ +»sin^cos5 = o, 
whence tana5 = i, and .'. 5 = Q7r; 

• « /J \/2—i , z, s/^ + x 

.-. sm'^=-5^ — 7 — , cos*5 = -^^ — -. — , 

2sl2 ^ %isl% ^ 

and the equation to the curve then becomes 

{% + ^]%) «"*+ {% - Va) y* = lift; 

therefore the axes are {% — V^)* aiid (a + V^)*> 



312 ELLIP8E. 



and A = {a*+i8*)* cob 67** 30' = (i - 4-)*; 



i\* 



Jb«(a» + )8«)*Bin67*3o', or k = ^i + ^ 



8. Transfonning the origin to a point (a, /3), we have, after 
equating to zero the coefficients of x and y, 

a = -a, ^ = 0, 

and the equation to the curve then becomes 

i6a?'* + 7y'* + i6a;y-36=o (i). 

Now transfer the axis of y through an angle of 30% aad 
we have 

^ V3' V3' 

which values being substituted in (i), we have for the trans- 
formed equation 

i6a;"» + 4/^ = 36; 
therefore the semi-axes are ^ and 3, 

11. The centre of the ellipse being the origin of rectangular 
co-ordinates, the equation to the circle will be 

{x'-aeY + {y — rY = r*, 

and when the circle meets the axis of y , 

jc = o,. y = J ; 

air = J" + aV = a*, 
or J : a :: a : ijn 

13. Let >SP=A, PH=h, SE==l; 25 = A + ifc + ?- Then 



ELLIPSE. 213- 

Now 8^-[{a + ex)+{a — ex) + %ae] = a (i + c), 
5 — Z = a(i + 6) — aa€ = a(i — e) ; 

, P8H^ PH8 i-e 
.". tan tan 



15. Using the same notation as in (13), 

^ ^ p-A)(a-fe)(«-Z) U 



p_ '^^^ 



4{s(»-A)(a-A;)(«-Z)}*' 

45 4a (i +6; 

18. Let jc', y' be the co-ordinates of P; the equation to 
AQO will be 

y = ^ (a5 + a) = wi(aj + a) suppose ......(i). 

To find the intersection oi AQO with the ellipse, eliminate 
X from between this equation, and a'y* + hW = a*&'. 



Then ay + J» T^ - aV^ a» J», 






y = o; 



w 

a^'*/ 



• • ^ "" ^ wV + &» a»y'» + i V a»i» "■ a 



.'..' 



Hence AQ : ^ = CP : y\ 

AO : a= OP : x; 
.-. AQ.A0 = 7,0P'. 

21. The equation to the tangent is, generally, 

and at L. a?' = — a«, and y = — ; 

a 



214 SLLIP8E. 

or ysa + ex. 

If the ordinate y, or NR cut the ellipse in P, we have 

/8Ps=a + ea?; 
.-. 8P^NR. 

26. Let A, ib be the co-ordinates of the intersection of the 
tangents, then the equation to the chord joining the points of 
contact is 

a% + h^hx = o*J% 

and since the chord passes through the focus, when x = ae, 

.'. 6*A.ae = a*J*: and /. A = -^ 

e 

the equation to the directrix. 

The equation to the perpendicular to the chord is 

3 •■ o ha 1 

and when 0? = -, ys=^=k; 

therefore the perpendicular passes through the intersection of the 
tangents. 

28. Let A Yy AZ be the lengths of perpendiculars drawn 
from the extremities A^ A' of the major axis upon the tangent 
atP(aj',y'). 

Then -4r=-4rsintf = (~+a)8intf, 

AZ = -4'rsin ^ = (^ - a) sin ; 

.-. AY.AZ^ ^, (a»-a;'») sin- tf « J,.^' sm-0^h^(^^\^^e 

,, sin' tf T, • /I 



ELLIPSE. 216 

29. Let Xf y be the co-ordinates of P\ then 

cb +0? a — X 

2ay 20^ 

If fi^F he the perpendicular from the focus 8 upon the 
tangent at Fj then 

sin(?=^ = -^^P^sint = -sm^, where tani/r = « — ; 

. , - J»aj 

.". smy>= , ; 



/. sin ^ = 






Vo^y » + i^o* * 



Hence tan tf == = - tan d> : 

cwty li ^ 

/. !%tand:s:6tan^. 

32. The equation to the tangent is 

y = wa? + ( J* + o*m') , or (y — «ia?)"=: J* + a*w*, 

or (fic* — a*) m* — aa?y m +y* — J* = o, 

where 9n = tand. 

If therefore x, y, be the co-ordinates of the intersection of 
the tangents, the two values of m in this quadratic will be 
tan 0y tan ff. Hence we have for the equation to the required 
locus 

(y»-J>)=c(a:»-a*), 

which, c being a negative (quantity, represents an ellipse whose 
centre is the centre of the given ellipse. 

33. Let be the centre of the circle, z PCT^0, z OT^fi. 
Draw CD parallel to Tt, then z.ACD = fi. 

firp 

ISow PL^CL''CP^2rcoB{0-ff)-CP, ~~ = cos/9; 



216 ELLIPSE. 



.'. ar = 



CPcosdcosfi* 



CF COB cos p CF cos 

a* (i + tan tf tan /9) ^ j, 
= CF ^^• 

But tan^tani8 = ^; 

s half the chord of curvature at Pin the direction of 0, 

^^""CP "^^^^ CP GP"' 

/. CL. CP=a*'\-l*f a constant quantity, 

38. Let Q be any other point in the curve; x and x-^h 
the abscisssB of P and Q respectively. 

Then G^^=*{(a; + A)» + ~a*-(a; + A)»} 

= e* (as - A)*- ae»a!(a + *) +e<a* + b\ 

and (JP* = ^ (a» - e'se') = (i - «») (a» - «V) 

.-. (?^ - GP^ = e'{{x + hy - ao; (a; + A) +ae} =c»A*; 
••. OQ iR always greater than GF. 

42. Let a?', y' be the point at which the tangent RZ touches 
the ellipse. Then we have for the elimination of x\ jf the 
equations 

and ay» + JV» = a'J», 
whence f , ^-^ — 1 +JV = a"&% 

whence aj'=— ^; 
.•. the A BEZ is isosceles, or PZ= BZ. 



ELLIPSE. 217 

43. Let (A, k) be the given point. The equation to the 
curve is 

aY-^'h^x'^^oJ'i^^^^ (i), 

and the equation to the chord is 

y — Aj = m(a;— A) {%). 

At the intersection of (i) and (2) there is 
{a^mJ" + V)x^ — 2a*m {mh — ^x+a' (A?" - zJikm + rn^l* - V) = o. 

The abscissa x of the middle point of (i) and (a) will be 
equal to the semi-sum of the roots of this equation, hence 

_ a^m^h — a^mk . ^ 

^" aW + 6» • ^3;- 

Also y being the ordinate of the middle point 

y — A? =w(a; — A) (4). 

Eliminating m between (3) and (4), we have for the equation. 
to the required locus 



y* — hy a^ — hx _^ 
15 ^ 7^ ^> 



6» ' a 

which is therefore an ellipse, the co-ordinates of whose centre are 

— and - and whose semi-axes are 
a % 

-^ (a»Jk« + 6«A»)* and — {a'k^ + J»A»)* 

46. Let (a;', y') be the co-ordinates of P, and (a?, y) those of 
the centre of the circle, ; then 

the area of Ai8PH*=a«y' = y(i +6)a.........(i). 

Again, since the line from P through bisects the angle 
flPH, we have 



but sinPffiff=-^,, and cos PHS^ "^ '^' 

28 



a — eaj ' a — eai' 



218 ELLIPSE. 

Also tAn 0H8=-^ 



ae — x^ 



whence ^ =7 — —^7 k (2). 

ae — x (i + e)(a — a?) ^ ' 

Therefore from (i) and {%) we have 

# 1+6 , a 

which yalnes being substituted in the equation 

wehaye (i +6)y*+ (i -e)a5*=aV(i-e) 
for the required locus, which will be a concentric ellipse. 

48. Let O be the foot of the normal at P, then 

Sa _8P_PY_8Y_8Q 
EG' HP'TZ^ HZ' QZ' 

therefore Q the point of intersection of 8Z, HY\b in the normal 
atP; also 

PQ_PY_8a_Qa 

HZ' YZ' 8H' HZ' 

therefore PQ is bisected in Q, 

Hence, if («', y') be the co-ordinates of P, and (a?, y) those of 
Q^ we have 

^ 2. %' 

therefore substituting in the equation 

ay* + J V» = a»iS 

we have \ , ^ ,x r + ^ = i * 

for the equation to the locus, which is an ellipse, whose major 
axis is a (1 + e"), and minor is i, or a (i — e")*. 



ELLIPSE. 



219 



50. Let Xj jfi be the co-oidinates of B^ 




y-y' = TrT («-«') equation to PJ8, 



Now y = - a/ 



a 



^^''^-T^ 






JVaj' 



= — I. 



Hence the line BCEia perpendicular to PD. 

61. The equation to CP being 

y = waj, 
the equation to its conjugate CD will be 

a «* 
when CP meets the directrix, we hare 

a . ma 

« = + -, y = + — , 

6 € 

and when CD meets the directrix, we have 

a 5» 
a; = + - , y = . 



220 ELLIPSE. 

Therefore for the perpendicular to CD^ we have 

ma ma* f a\ , . 

y— ^ '-¥-[''- 1)'"" W' 

and for the perpendicular to CP, we have 

y + — = — [x — ) (ij), 

^ mae m\ ej ^ '' 

If y =: o in equations (i) and (2), we get 

, , a b* a/ h'\ 

^ ' e ae e\ a^J 

/ \ a J* 

(a) a: = = ae. 

^ ' e ae 

Hence, these perpendiculars intersect in the nearer focus. 

53. Let ^ be any diameter of an ellipse, {x',v') the co- 
ordmates of P, and therefore — x\ — y' those of p. ^ake R any 
point in the ellipse, and join RP, Rp, and let the co-ordinates of 
R be {Xf y). Then if w, w' be the tangents of the angles which 
JBP, Rp make with the axis of x respectively, we shall have 

w = ?^^^, w' == ^ . ^, . and /. mw^ = ^^^,, , 
a?- a?' x + x^ a"-a?*' 

also ay + J V = a*i", and ay* + b'x'* = a'b*, 

whence y' — y^ = — 5 (a?* ,— a?'*) ; 

J' 
a*' 

which shews that if pR be parallel to a diameter, RJP will be 
parallel to the conjugate diameter. 

69. EE' bdng parallel to the tangent at P, and the focal 
distances making equal angles with the tangent, the angles 
EE'P, E'EP are equal ; 

.-. PE' = PE=a, (vide Ei. 49) ; 
.". a-ex' + E[E^a; 
.-. ffE = €x'==SE. 



Again, from Trigonometry, we have for the radius of the 
circle circumscribing 6k8EGy 

SE 
!2sin 



5=-^% (I). 

'^ "im (7 



ELLIPSE. 221 

and for the circle circumscribing A CHEy 

jB = — : — 7y • ••••• (2). 

ij sin (7 ^ ^ 

Equations (i) and (2) show that the circles are equal. 

60- a » + h'^ = a» + J*, 

and a' J' sin 7 = aJ ; 

2ah\i 



\ sin 7/ 



Now sin 7 is least, when -= — is greatest, i.e. is =a* + 6*, 

. sin7 -° 

at which value a'= &', 

aoJ I + sin 7 {a-\-hY 

^ a« + J»' •• i-sin7 (a-J)»V 

7.-7 X 7 i 

cos - + sin -^ 1 + tan - i + - 

2& rj 2& « ^ 7 '1 

.'. — — - or — — = • /. tan - = s • 

7.7 X 7 6' iit 8 

cos - — sin - I — tan - i — 
2 7, % a 

61. Let a;', y' be the co-ordinates of P. 
Then a» = aj'» + y'», 

y» = --«/'» 4. — a;'« • 



•••«'^"*'"-(^-S-'^--(?-02^"^ 



a'^^V' ^x'^ y'^_GN PN _0N GP _PN CP 
a'-b^ a» b' GT GT GP' GT GP'GT 

cos /3 . sin i8 

= cos a -; — ; — : — ^ — 7R — sin a '^ 



sin (a + 90® - /3) sin (a +90* - fi) 

_ cos (g + 13) 
"■cos(a-)8)' 

66. The co-ordinates of P being x\ y\ and those of i>, 

M It 

then «" = ± I y, and y" - ± "^ • 



S2S ELLIPSE. 

Now COS a » -Ti sin a =S; •'• «'* aia za = aajV, 

a:' v" 

oosi8 = y, smi8 = ^; /. V* sin 2/3 = zx'Y ; 
.•• o'* sin aa + y* sin 2)8 = aa^y + Jj f? f y')(± — j = o. 

70. Let P, Q^Bhe the points of contact of the sides -4,i« 
^a^3, A^A^ of the ciicomscribing triangle, and let j?, y, r be tie 
lengths of die semi-diameters parallel to these sides, then 

A,Q q' A,P^p A,B^r' 
^ A,P . A^B . A^Q p .r .q _ ^ 

whence A^P. AJt .A^Q^A^Q. A^P. A^B. 

75. By fbrmnla (la) we have 

I _ I - e* cos* ^ _ (i - ^') +^* s^^* ^ 
r*" i» b' ' 

J I i-e»cos«^ (1 - e*) + 6' sin* ^ 
and ^ j5 ^^ ^^5 ; 

1 I _ e* (sin* g - sin* y ) 

•'' r* r'* ■" 6» 

= ?^'(sin*d-sin*^). 
a*6* ^ ' 

78. Let x\ y* he the co-ordinates of P, SiF a perpendicular 
firom 8 on the tangent at P, 

tan O/-^ - ^p- SQ, cos r (ae+eV).cos T" aj'e ' (a+ea;') ^ ^ 

""ea?''aV o^yaer sin j9 ^ping ^(^~^') 

I — c cos ^ 
I — e cos tf 
esm^ 



ELLIPSE. 



228 



iJT. 



-f 



'r' 




k Hence tan ^^FJ^ = iH^^^i^ =i±i^ 
r e sm (7r+ a) — 6 sin ^ 



tan <^ = tan (iSP'^- >SP^ 

I + e cos 5 I — 6 cos 5 



— 6 sin ^ 



I — 



e sin ^ 

I - e* cos" ^ 



ae sin 6 
i-e" 



6"sin"d 



82. Bj the polar equations to the ellipse and circle, we 
haye 

j.^ aa&*cosd J >ii> ^ ii 

-4Q = -T — 7-i — xix ra> and -ajB = aa cos ^, 

^ a»- (a*-J*)cos*d' ' 



j,^ 2ai" sin ^ 

^^■"a»-(a"-J")sin»5* 



jl'j8 = aasin d; 



^^ ^^^_ a'--(a'-6')cos»g , a»-(a'~i»)sm'g _ aHy 

•'•^(2"*" ^'(2" J» ■*■ J" "" i» • 



83. Let iffG = p, z CSQ^e, 

then aP8C=-%0, and ;&P=-?iln^; 

I — e cos ij^ ' 

.•• p*^aex8P= ^^ ^: 

"^ I — e cos aa ' 

or p' {i - e (cos» d- sin* ff)} =« a"e (i - e^) ; 



224 



ELUPSE. 



or (i— «)a'+(i +e)y* =a*6(i— e*), 

hence, -r-7 — —r + -r-r x = i> 

is the equation to an ellipse whose centre is S^ and whose 
eccentricily is {— t~) • 



PF 



91. PL^PO cos CPF^ 2p . ^ 




Now P^ -Ti hj formula (13), 
8>nd 77D=-"a:/ > from formula (9) ; 

LfJr ao 

PF 

Also, Pr=PO cos VJPO=^ip pg 

4 

~ a ~ C^ • 



HYPERBOLA. 225 

92. Let {x, y) be the point of concourse of the tangents. 
Then y'' = J' — j «'* is the equation to the ellipse (i), 

CI/ 

and .y'=y + m(aj' — a?), 4... tangent (2). 

Eliminating y' we get after reduction, since the roots of the 
resulting quadratic in x* will be equal 

(a:* — a*) m' - WJyw + (3^* — i*) = o. 
Now if m, m* be the roots of this equation, « 

w + m^= a^a > and mm' = f ^^j- — ^), 
a?" — a' \a;' — a'/ 

(m — m') = * (i + mm')* 
Hence, eliminating m and m', we haye 

.•. f (aj» + 2^» - a» - 5") = 4 (ay + h^x" - a'J"), 
which is the equation to the required locus. 



Hyperbola. 



Ex.6. 



1. y>-y+i-aj3=£; 

4 4 



(y-j)' a,' 

• • ^^^^^^^^^^ ~~" ^^ — X. 



I I 

4 4 

Hence it appears that A = o, ^ =- ; a = J=-. 

6. Let aj = < costf — w sin tf, y=^sin^ + wcosd. 
Then xy—t^ cos 5 sin ^+ tu (cos"tf - sin' ^ - w* cos 5 sin 5, 

oj* tan a = — ^ cos' tan a + aft^ cos 5 sin 6 tan a — w' sin* d tan a ; 

29 



226 HTPESBOLA. 

/. - J* cot a » <• f-sina^— cos'^tanaj — tt'T-sina^+sin'tftanaji 

ifcoaa0 + 8ln2^tanas=o; 

and /. tana^ss — cot a= tan r+ — 4- aj ; 

which values express the inclinations of the axes of t and u 
to that of X* 

Now sin a^ = sin f— + a j « cos a, and cos a^ = — sin a ; 



.'. sinad— (i + cosa^ tana = cosa + 
and sin ad+ (i — cos a^ tan a = cos a + 



sin* a 
cos a 

sin* a 
cos a 



— tana = 


I— sma 
cos a ' 


+ tana = 


i + sina 

• 

cos a ' 



- I + sm a ^ I — sin a h* . 

/. «• X — ■ ^ X = — cot a. . 

cos a cos a a 

Hence, h^o^ h^o\ and the ratio of the axes, or 

a . a 

. vx <5<>s sm - , . 

g /i-sma Xf^_a 2=tanp-?^). • 

6 Vi + sma; ^«+sin^ ^^ ^^ 

a a 



9, y* — 3a?y + a" + I ==o. 

The centre being the origin of co-ordinates ; 

A = o, i = o. 

Li the given equation substitute for x and y as follows, 

a/ sin (6o - g) + y^ sin (6o - y) 

Sin DO 

_ aj' sin 5 + y ' sin 6 
y gin 6o^ 



.0 



and eqnate to zero the coeflScient of x'y\ Then, after reduction, 
we obtain 



HYl^EEBOLA. 227 

azxd the reduced equation becomes 
of which the aemi-axes are 

11. Let -5 + Tj = I, and -73 --^^ = i, he the equations to 

tlie ellipse and hyperbola; then G8^=^a*''l*==a'* + b'* since 
the curves have the^ same foci and centre ; and at the points of 
intersection 






Now if d, ff be the angles which the tangents to the two 
curves at the point of intersection make with the axis of a;, 

^ b^x >,, V*x 

tan^as — 5-, tan& = -77-; 

.•• tan Stand's r-?j--^ = -i> 

aV* y* ' 

or the tangents at the points of intersection of the two curves are 
at right angles to each other. 



or 0^ = 0* + - i, 

hence CP» = aj» + y » = a» + i (a* + *')*, 

hence PY^ = CP» - CF* = a» + ^^ 

or Pr= (7/S. 



228 ilTPEBBOlA. 

19. Let x^ y, x'^ j/ be the co-ordinates of I^ and J) re- 
Bpectiyelj ; then the equation to the nonnal at P is 

Similarly the equation to the nonnal at i> is 

y,— ^a'.+(x + g)y'... W. 

from (i) and (a) we obtain 

but the equation to a line through G perpendicular to JPD ia 

y -y 

•\ X, F aie co-ordlnate8 of a point in this line; or the normals 
at P and D intersect in the perpendicular drawn from G 
upon PD. 

21 • Find the area included by the normals to a hyper- 
bola which pass through the foci of the conjugate hyperbola. 

Let 8' be the focus of the conjugate hyperbola; S'FK a 
normal; x, y the co-ordinates of P; then the equation to the 
normal B'PK is 

t <**y / / \ 

y-y=-j4^'^ -^); 

and when a;' = o, y' = ^i + |-!)y= 0/S' = (a» + 6»)* ; 

.-. y = 



also when y* = o, a?' = Cfir= (i + -5) a? ; 

jj a?' .V* . i' o*+2&* 
a' 0' a' + 6' a* 4-0 



• • a? 5^ 



and the area included between the four normals 



HYPERBOLA. 229 

27. The equation to the hyperbola, and circle, referred to 
the asymptotes as axes, are 

ajy = , and aj+y — ay^ sm- = r cot-, 

-where = the angle between the asymptotes, and therefore 

tan- = -. 

a, a 

Therefore for the points of intersection of the cn^^es 
a. + ?^±i!-(a» + i')*sin- =rcot-, 

and in order that the circle may touch the hyperbola, the two 
values of a? in this equation must be equal ; 

.-. (a" + i»)*sin- + rcot- = (a*+J*)*, 
whence r^{a^-\- i')* ( i — sin - j . tan - 

= ^{(a. + jf_5}..... (I). 

Now if the semi-latus rectum 8L produced meet the asymp- 
tote in F, we have 

I 

a 



fifF=-(a» + i»)*, 



but 8L^^; 
a 



... iF=^{(a» + jf-5} (3); 



.•. from (i) and {o) r = LK 

30. Let a;, y be the co-ordinates of the point of intersection 
of the ellipse and normal; the equation to the normal through 
the given point A, A;, is 

y — 7c = m(aj — A) (i). 



S80 HTPEBBOLA. 

and the equation of the tangent at (x, y), is 

• ^^ ^^ 9 

irX 

which being sabetitated in (i), we have 

for the equation to the locos, which is a rectangnlar hyperbola, 
passing through the given point, and having its asymptotes 
parallel to the axes of the ellipse. 

32. The equation to the tangent of the hyperbola, in terms 
of its incUnationto the major axis is, 

y = «ia? + (a^«-J")* (i). 

The equation to a peipendicolar npon this tangent from the 
centre is 

y^-^a' (^)- 

At the intersection of (i) and {p) we have, by eliminating m 
between (i) and (2), 

(«*+y')'=o'«"-&y, 

which is the equation to the required locus. 

33. The equations of two right lines meeting the curre 
being 

y + maj + n = o (i), 

and y + w'a; + n' = o (2). 

Eliminating y by each of these equations and the equation 
to the hyperbola, finding the value of x in the resulting equa- 
tion, and equating the radical in each with zero, we have 

o*w' - J* - n" = o, and a»7n'» - J' — V = o. 

The values of n and n* in (i) and (2) being substituted in 
these equations, and the equations arranged by the dimen- 
sions of wi, w', 

2xy h^ + y^ T ,^ 2xy , S*+v' 

^ - a a ^ — r-^ = o» and W'- ^ ^ . m'- a l , = o» 



SECTIONS OF THE CONE AND (lENEBAL PROBLEMS. 231 

The values of w, tti being the roots of either of these equa* 
lions, let mtri = — i, then 



V^=i, ory« + a;» = a»-i% 



is the equation to the required locus, which is a circle whose 
radius = (a» - Vf. 



Sections of the Cone and General Problems. 



Ex. 6. 

1. Comparing the equation 

J _ %d^\xx a sin sin 6 sin (ga 4 6) . 

^ COS a cos* a 

with y» = — a; + -ja^, 

, 5" , sin ^ sin (2a + ^ , . 

we have, — , = ± \ '- (i), 

' a' "^ cos' a ^ ^' 

the upper sign bein^ used when the section is an ellipse, and the 
lower, when it is a nTperbola. 

Substituting - ir for 2a in (i), we have 

ja I 

"5 = ± a sin ^ cos ^ = ± - by hypothesis; 

.*. 7,6 = 150® or 30**, 
and ^ = 75** or 15*^; 

.*• for the ellipse the inclination is 

^-(45^ + 75^) = 6o^ 
and for the hyperbola 

i8o«--(i35-+i5") = 3o<'. 



232 SECTIONS OP THE CONE AND QENBBAL PROBLEMS. 

. 8. Prom Ex, (i), wc liave 

oh : %a » Vain sin (aa + ^ : cos a (i), 

and d=i8o*»-(a + S); 

/. sin d = sin (8+ a), 
and ain(aa + ^«sin{i8o°-(S-a)=sm (S — a)}; 
.•• 8a1)Btita1ing in (i), we have 

minor axis : major axis = {sin (S + a) sin (S — a)}* : cos a. 

5, a« = A* + ifc»- ihk (cos" a - sin" a) 

= A" +i* - aAAj + 4Ai sin" a 

=:(A-Ar)" + 4«;sin*a (i).* 

From tte extremities of the axis-major let fall the perpen- 
diculars AF^ A'O upon the axis of the cone, and through C/ the 
middle point of Am! draw a plane parallel to the base of the 
cone, cutting the plane of the ellipse m BB\ which is its minor 
axis, and the cone in the circle MBQ; then 

BC*^MGxCQ=-AGxAF.., (2), 

but A'G^ksma, 
and AF^ h sin a, 
which being substituted in (a), we have 

— = AA;sin»a, or &» = 4^sin»a (3), 

4 

and from (i) and (3) we get 

7. The general equation is 

ikZsinasin^ sin^sin(2a + ^ ^ , . 

^ cos a cos' a ^ -" 

but for the parabola, (aa + ^) = w ; 

.•. sin (aa + ^ = o, and sin =sin rja ; 

therefore when the section is a parabola, 



tzd sin sin a 
y" = 

^ cos a 

= 4^ sin' a . a? ; 



SECTIONS OF THE CONE AND GENEEAL PEOBLEMS. 233. 

/. the latus rectum = 46? sin* a (i), 

and when the plane passing through the directrix of the section 
and the vertex of the cone is perpendicular to the section, the 
expression for the latus rectum is 4a cos 2a; 

.'. a cos 2,(1 = 2, sin* a = I — cos 2a ; 

I 

.*• cos 2a = -. 

3 

13. Let the perpendicular from <?, the foot of the normal 
meet the focal distance 8F in i, and let the angle F8G = ^. 

The conic section being referred to its principal axis and 
directrix as co-ordinate axes, and (a?', y) being the co-ordinates 
of P, we have for the abscissae of 8 and (r, 

ep, and ep(i 4-e) 4-eV respectively, 

and for the length of /SP, 

Now, PL=8P-8L 

= 8P-e'{p + x')coB0 
= 8F- e ySPcos d 
= ySP(i-ecos^ 
= semi-latus rectum. 

16. Let {x\ y') be the co-ordinates of P, then the equation 
to the normal is 

At the point G^y = o; 

.-. AQ = m-{-{n-{'i)x' (i).* 

The equation to PK^ perpendicular to the line y = ^ a?, is 



X 



w 



At the point K.,y — o; 

... AK=t±^ (2). 



X 

30 



284 SECTIONS OF THE COKE AND aENEBAIi PBOBLEMS. 

From (i) and (a) and the equation to the cnrvei y* =^ nmx + tiaf, 
we haye 

0K» — ; wissmsssemi-latus rectum, 

X 

18. Let PP, QQfhe any two chords which moving parallel 
to themselves intersect in 0. Let be the orig^ of co-ordi- 
nateSi and the chords the co-ordinate axes; then the equation 
to the cnrve will be of the form 

ay* + bxy + €a^ + dy-\-ex +/= o. 
Let a? 8 o, then 

f 
and the product of the roots being - , we have 

a 

Simikrlj, POxP'0=^; 

c 

.-. QO.QO : PO.P'0'=i 'X=c : a, 
and is therefore invariable. 

21. The polar equation to any conic section, the focus being 
the pole, is 

I + e cos r ' 
also BP being produced to meet the curve in p^ we have 

» _ SL SL . 

™ I 4- € COS (-n- 4- ^) I — e cos ^ ' 

which proves the proposition. 

22. Refer the curve to the tangent as axes, its equation 
will be of the form 



i%+l-'J+f^^=° w- 



SECTIONS OP THE COKE AND GENERAL PROBLEMS. 235 

^ Let the equation to a straight line drawn through the 
origin be 

y^'^^r (a). 

Thus the distances from the origin of the points of intersec- 
tion of (i) .and {%) will be the values of r found from the 
equation 






I m 
or 



(a + I-^)+'*'" = ° (3). 



If / and r" be the roots of this equation, we have 

Also the equation to the chord of contact is 

f + |-^=° (4)." 

Hence for the distance {r) of the point of intersection of (a) 
and (i), we have 

f'-k-'' '''7rh-^% (5). 

From (4) and (5) we have 

T^ r r 

which proves the proposition. 

25. (i) Let the point of bisection C, of the line /Sff, be 
taken as origin of rectangular co-ordinates, SH being the axis 
of X. 

Assume 80 = 1, and 8F' + HP^=^(f. 

Then, 8I^=y^+{x + b)% 

.-. <^=:a(^ + iB» + i*), 

whence y^ + of^ J* 



236 SECTIONS OF THE COKE AND GENERAIi PKOBLEMS. 

is the equation to the required locus, which is a circle, whose 
centre is C7, and rad. ( J*] . 

(2) In this case assume the point 8 as the ori^n of rect- 
jmgular co-ordinates, and let {x,y) be the co-ordinates of P. 

Let SP^nc, and HP^^c, 8H^b. 

Now fiP* = y« + a:» = nV (i), 

and c« = y» + (i-a:)» (a), 

and eliminating c between (i) and (2), we have for the re- 
quired locus 

which is the equation of a circle, the co-ordinates of whose 
centre are 

28, Let 8 be any point in the circumference of a circle, 
and P the point of contact of any tangent to the circle of which 
C is the centre. Let Y be the extremity of the perpendicular 
from 8 on the tangent at P. Join CPand draw iSlAT perpendi- 
cular to PC produced. 

Assume 8P=r, 8Y=p, CP-=^c. 

Then. 8 Y=PN. 
And 8P^=^SC-+ CP--\-Q.GPx CN; 



is the equation required. 



^ 2,0 



Ex. 7. 

2. The formulae for transformation are 

which being substituted in the proposed equation, we have 
4»Y-a.2^.{±<i^ + ^y} 

= rj* . a (a;' - y ) {a?'" + y'* + xt/'} 



LOCI. 237 

Loci. 

Ex. 8. 

3. Let AB be the given base, through D the middle of 
livhich draw the perpendicular D Y"; then taking DBX and DY 
for axes, and denoting the vertex of the triangle by (a?, y), half 
the base by a, and the difference of the tangents by m, we have 

m=-^^ y-^ ^^ . 

Hence the equation of the locus is 

the equation to a hyperbola. 

4. Let the base be taken as axis of a?, and a perpendicular 
through the extremity A as axis of y; (y, x) the co-ordinates of 
the vertex, and c the length of the base AB^ then if one base 
angle be a, and the other 2ia, we have 

, 2 tan a , . 

tan 2a = (i), 

I -tan" a ^ '' 

but tan a = ^ , and tan %a = — ^ . 
Substituting these values in (i), we have 

I _ Q,X 

c — x x'—y^' 
therefore the equation of the locus sought is 

y' = 3a;* — 2CX^ 

which belongs to a hyperbola whose tranverse axis is two-thirds 
of the base. 

6. Let C represent the base AB of the triangle, then, the 
altitude being constant, the locus of the vertex, C, is a parallel 
to AB, Hence, taking AB, A Y for the rectangular axes, and 
putting a?', y' for any point in the locus of G, we have for the 
equation of BC 

therefore the equation to the perpendicular to BC at its middle 
point is 



238 I'OCi. 

y-|=-7-(»— ^j W- 

Similarlj the equation to the perpendicular toAGia 

y-^=-|'(«'-|) w; 

therefore for the intersection of (i) and (2), 



whence cx=^^i 



• • M/ ^'^ ■" 



is the equation to the locus, which is a straight line perpendicular 
to and oisecting the base. 

9. Let AB be the given base, CD the given altitude, the 
centre of the inscribed circle, 

let GD=^c, AB=^2a. 

Assuming the middle point of AB as the ori^n of rectanga- 
lar co-ordinates let (a:, y) be the co-ordinates of O; and a;' the 
abscissa of C\ then 

tan -4 = — : — ,, tan— = — ^7 — . 
a'{-x 2 a-\-x 



a'—x' ' % a^x* 



Also tan B = : , tan 

2y {a + x) 



a + oj'"^ (a+ aj)' — y* 
c _ ay (a — a?) 



(I), 



and ?= , ^ v> — ^ (li), 

a - a? (a - a?)* - y* ^ ' 

Eliminating a?' between (i) and (ij), we have 

g^i y 

c * y a^^QC^^ 
or c (a* - aj* - y') = 2y (a* - a?*), 
^or the expression for the locus. 



LOCI 



239 




12. Let BCP be the position of the 
"ttimed page. 

Assume AP==r^ aPAG^O^ then 
since the triangles ABE^ PBE are equal, 

*^ve have AE=-. and the angle AEB 

a. right angle; 

/. AE^AQ.Qo^Oj 

and AE^ AB .Qoa{^ - e\^ AB .sm0 \ 

.•.— = — sm Q . cos B. 

4 ^ 

or r* = o* sin a^, 
is the equation to the locus, which is a lemniscata. 

13. Let CN^Xy NP=y^ be the co-ordinates of the point 
jP, and let C8=Cy then 

8P= {PN* + SN')^ = {y' + (c + xy]\ 

HP = {PN' + HN')^ = {y» + (c - a:)»}* ; 
therefore, by the hypothesis, we have 

••• (y' + c» + aj")*-4cV = c^; 
.*. y + 2x^y' + »:*.= ac»aj* - acy ; 

is the equation to the locus, which is the lemniscata of Ber- 
Bouilli. 

16. Let the two given lines be t^ken as axes of co-ordi- 
nates ; and let m and n be the intercepts of the moveable line on 
these axes, then 

40* =s w' + n' — amn cos 2a. 

Now, let (ic, y) be the co-ordinates of the middle point of 
the moveable line, then 

m n 

hence c" = a:* + y' — 7,ocy cos aa 
is the equation to the required locus, which is an ellipse. 



240 LOCI. 

Now change the direction of the axes bj means of the 
formnka 

" _ a;' sin {aa — 0) — y' cos {aa — 0) 
*" sin aa 

— g^'sin^-f y'cos^ 
^ "" sin aa * 

and equate the term involving x', y' to zero, and we have, after 
redaction, 

sintf = sin (aa — ^ ; /. — a (a), 

hence the equation to the curve referred to its axes is 

a:* sin^ a + y^ cos^ a = c' sin* a cos" a ; 

.*. the semi-axes are c tan a, c cot a. 

Equation (a) shews that the direction of one of the axes of 
the curve bisects the angle included between the given lines* 

18. Assuming the two straight lines as co-ordinate axes, 
let (cc, y) be the co-ordinates of the vertex ; the equation to 
the parabola will then be 





©'- ©'= ■ 


> 


and it is 


eanily found that 






^ ~ (A'+ k')' ■ 








A'A» 

(A' + *«)»' 


- 


whence — = — = — r 
^ i/ ax* 

• AA;= "' • 




* 




I0> 




whence &= ^ ^ ; and . 






aji^T 


a?3y3 



(I). 



LOCI. 241 

And sa1)8titating in (i) we have 

7* i"TT iisf^lisJ 8 loj 

^x^y3 aj3y3 ^ ^aj3y3 aj3y3 ' 



,9 



5 a i i * 



4 1 » i 

.•. aj3y3 + 053^3 =: a* 
is the expression for the locos. 

19. The equations to the parabola, and one of the tan- 
gents, axe 

y"=4^' - (i), 

and y'-y = w(aj'-a;) (a), 

where {x, y) are the co-ordinates of the intersection of the 
two tangents. 

Eliminating y' between (i) and (iz), we shall obtain a 
quadratic in x\ the two roots of which must be equal: this 
condition will, after performing the requisite reductions, give 
us the equation 

m^x — my + a = o. 

Bepresenting the two roots of this equation bj m, m\ we 
shall have 

97^ + 97i' = £ , and 4mm' =5 — ; 

X X 

^ ' Ql^ x^ X 

But by the hypothesis 

(m — «»') =tan)3 (i +mm'). 
Hence eliminating m and vi, we have 

y' - a?" tan*i8- aoic (a + tan* )8) - a" tan' )3 = o 

for the equation to the required locus, which is a hyperbola of 
which a principal diameter coincides with the axes 01 a?. The 
co-ordinates of the centre are 

ysso, a? = — a(acot'i8+ !)• 

31 



243 LOCI. 

The origin being removed to the oentre, the equation W 
comes 

y* — tan'/3 • «■ =» — 4a*cosec*/S. 

Now tan A = - = tani8. 
^ a 



21 . Let the centre of the ellipse be the origin of rectangular 
co-ordinates, (a;, y) the co-ordinates of P, and (a;', y') those of 
Q; then the equation to AQ is 

y = mx + Cy 

and at A we have c = ma ; 

/. y = m(a? + a), 
and at Q, y' = m(a?' + a); /. m = — ^^ — ; 
hence the equation to ^ Q is 

y"5qrs("+"> W' 

and similarly that to ^'Q' is 

y = -^t^(^"^) W- 

Eliminating a;' andy' between (i), (pi)^ and the equation to 
the ellipse, we get 



fD' = ~ and y'=«^. 



Substituting these values in the equation 

^ a"y'» + iV" = a^h^ 
we have for the required locus 

which is the equation to an hyperbola, whose centre is C, and 
transverse axis %a. 

22. Let AX be the given line, Q (A, h) the given point, 
'^the point of contact of the circle and given line, (a?, y) the 



LOCI. 



243 



co-ordinates of the centre of the circle, and (a?,, y^) those of any 
point on the circumference ; then the equation to the circle is 

{yi-yy+i^i-^Y^r- (i), 

but, passing through Q, it becomes 

(*~y)'+(A-«)^ = r« (a), 

and being tangential to AX, we have 

^=y (3); 

.*. from (a) and (3) we have 



(«"*)'= «*(y--)> 



for the equation to the locus, which is a parabola. Transferring 
the origin to the point {h, 0) the equation becomes 

oJ" + i = iky. 



Ex.9. 

1. For sin put - , and for cos 0, - , and the equation 
to the curve referred to rectangular co-ordinates becomes 

The following table gives the corresponding values of 
X and yi 





I 


2 


3 


4 


5 


6 


values of x 





a 


<ia 


2a 


>aa 


— 


values of y 





a 


pOBBi- 

ble. 


00 


impofr- 
Bible. 


impos- 
sible. 



Hence it appears from (i) that the curve passes through 
the origin, from {7) it bisects the semicircular arc A QB, from 
(3) there are possible values of y for all values of x less than 
2a, from (4) there is an infinite ordinate at B, or BM is an 
asymptote to the curve: from these values we thus obtain 
an infinite arc proceeding from A to meet the asymptote 
BR. 



244 



Locr. 



Agaiiii firom (5) for anjr value of 
X greater than aa^ y is impoesible, 
or no part of the curve is lound to 
ti^e right of the acfymptote, and from 
(6) no part of the curve is on the 
left of A. 

Also for eveiy value of x there 
are two of y, equal and opposite, hence 
there is a branch below AB similar 
to the one above it. 

The curve is the cissoid of Dio- 
des, the equation to which has been 
found in Els. 23. 




6. When a: = o, y = — 2^ and the curve cuts the axis of 

y atD; as long as a;<i, y is negative, and becomes infinite 
when x= j; therefore the ordinate BJE^ corresponding to 
a; =» I, is an asymptote, and we thus get the branch D£. 




When X is between i and a, v' is positive, and becomes 
very great both when a? is a little less than a, and a little 
greater than i, and gives the portion MOF\ 

When a; = a, y is infinite, so that the ordinate FF* is an 
asymptote. 

When X lies between a and 3, y is negative, and gives the 
portion FH. 

When a? = 3, y = o, and the curve cuts the axis at H. 
When a;>3, y is positive, and gives the portion HK\ and 
when X is very great. 



LOCI. 



246 



y= 



X 



= o. 



y=- 



therefore the axis of a; is an asymptote. 
When X is negative, the equation is 

(a?+i)(aj + a)' 

therefore y is always negative, and diminishes as x increases, 
and becomes o wnen a; = 00 and gives the branch DL. 

(Hjmers^ Conic Sectuma.) 

9. When y = o, a;=oo, therefore the axis of a; is an 
asymptote. 

When y = J, aj = o, therefore the cnrve passes through B. 

When y <bj x is possible, and when y > &, x is impose 
sible; therefore the cnrve extends from the asymptote up- 
wards to J?, and no higher, and thus the branch BFP'. 




Again, when y^^a, aj = o, and when y^^hy x^o; 
therefore the curve passes through A and I) if CD = h. 

When y < — a, a? is possible, hence there is a branch AX 
extending from A to the asymptote. 

When y>'-ay or < — J, x is possible ; therefore the curve 
exists between A and D; the double value of x gives the 
same result along CX\ 



MECHANICS 



AND 



HYDROSTATICS, 



STATICS. 

Forces Acting ik the same Plane. 

Ex. 1. 

1. Let AB, ^(7 represent the two forces in direction and 
Ynagnitade; then will the diagonal AD of the parallelogram 
ABGD represent the resultant both in mag- 
nitude and direction. 

Let X s the resultant; then 

AD" ^ AB* -^^ Bl> - Q,AB . BD COB ABD, 

or aj* = 2" + 3* - a (a X 3) cos 135° 

/. a? = 4-635. 

3. Let P, Q be the two forces, . B their resultant « Q 
the less. 

Then jB» =P» + Q» + %PQ cos 1350 ; 

.'. o = P»-aPQcos45**; 

P _ 2 cos 45° Va 
^11 







FOBCES ACnm IN THE SAME PLANE. 

6. Let ^ be the required z. Thea since 



/. cos ^ = — 



^ «-^^ = . 



4V3 



p . 



cos30''=<;osi5o''; 



/. ^ s= I5o^ 



247 



JI 




5^: 



6. Let P= 8, 5 = 10, 
i2' = — jB, 

required Q^ and the 

Now ii' : P= sin (36^ + ^ : sin 36°; 
.•. sin {36''+ff) = ^ sin 36** = sin 47^ 17'. 

And C : ^=:sind : sin (36^ + ^; 

. ^ sin 11° 17' 

sin 47'' 17 ^ 

8. Let P=(V3+i) a, Q = V6a, jB = 2a; 

^ = Zbetween Pand Q, ^^ jL between Pand B. 

Then jB" = P»+ ^ + iiPG cosd; 

.'. 4=(4 + «V3) + 6 + a(V3+i) V6cosd; 

Again, 0»=^ P» + iJ" + aPB coa^; 

.% 6=(4+iiV3)+4+«x2(V3 + i) cos^; 

The angle between Q and JK = 360'' -- (^ + ^) = 105**. 



248 



FORCES ACriKG IK THE SAME PLANE. 



10. Let Xbe the resultant of all the forces resolvedparallel toAxj 

Y Ay, 

R 




Then X^P, cosa + Pa cos6o*'+P3 + cos90*'+P^cosi5o* 
= 1 X cos o® + 2 cos 60"* + 3 cos 90^—4 cos 30** 

5= — 1'466 I X cos o** = 

ax cos 60® = 



I* 



3 X COS 90 = o* 



— 4 COS 30® = — 3'466 

r= P, sin o** + P, sin 60"* + P^ sin 90** + P^ sin 150^ 
= isino'* + asin6o** + 3sin9o''+4 sin3o'* 
= 673a ; I sin o® =5 o 

2 sin 60°= 173a 

3 sin 90^=3* 

4 sin 30*^ = a* 

6732 

/. S = (X» + Y'f = {(- 1-466)* + (673a)*}* = 6-88. 

If ^ be the inclination of the resultant B to Ax^ then 

Y 6*7^2 

tan^ = -:^= ^rstanioa** 16'; 

X 1-466 

.-. ^= loa** 16'. 

11. Let ABC he the triangle; a, 5, c the middle points of 
its sides. Let P„ F^, F^ be the forces acting at a, J, c in the 
manner stated in the question* Then because the sides of the 



FORCES ACTING IK THE SAME PLANE. 



249 



triangle are bisected perpendicularlj, 
the lines F^a^ Fjby F^c being pro- 
dnced will meet in the centre of 
the circle circmnscribing the triangle. 
We may therefore suppose them to 
flct at 0. Then 

F^ : Jr, : F^ \\BG\AG\AB 

:: sin^': sin £ : sin (7 

:: wiF^OF^ : wiF^OF^ : fsaiF^OF^\ 
therefore the forces keep each other at rest 




15. 



. i?+y= 



i?-p = 



p*+y* + aRp' COS a = P», 

.•. aj5>' (i + COS a) =5 P* — rf* ; 
••• 4«>' = (-P*-rf") sec'-; 
|p3 sec* 5 -efa /gec» 2 _ I^l*^ /ps _ rf. gj^. a\ 

.•.i> = i|rf+(p»-rf'sin»?)*sec2}; 
l>' = ,-f^+(i"-rf'Bin»?)*sec?}. 



i a 

sec- 
2 



17. Let 

^ the required resultant 



Then 



AB = d cos a, 
-4(7=rfcosd; 




32 



250 



FOBCES ACnNG IN THE SAME FLAKE. 



aiid£'s<2^{co8'a + co8'0+2co8aco8dcoB (a— 0)}, a maximum. 
Differentiating with regard to 0, we obtain 

— 2 sin COS 5— a cosa {sin ^ cos (a- ^ — cos d sin (a — ^} = o; 
.•. 8ina^=acosa sin(a — 2^ = sina(a — ^ — sina^, 

2 sin 20 = sin za cos !Z0 — cos 2a sin 2^, 

P _ 2+cos 2a _ I +2 cos'tt _ sin* a+3 cos'a _ i +3 cot*a 
"^ sin 2a "" sin 2a "" 2sinacosa "~ 2 cot a ' 

.•. cot2^cota+ i = ^(i+oot*a),cot2^ — cota = ^ ^ ; 

2^ 2 cot a 

.'. cot (a — 2^ = 3 cot a. 

20. Let ADy BD represent the direction of the forces, and 
DB the direction of the resultant. 




Then 5 = VP» + ^« = V3* + 4* = 5. 

Let DB meet AB in O; and let w4C = a;, Z w4GZ> = tf. Then 
since the moments about C must be equal, we have 

PxAC.BmDAC==^ Qy.BC. mi DBG, 

or 4».^ = |(io-aj), 

whence a? = 3*0217. 
To find 0. . If J2' = -jB, we have 
JS' sin go® • //i «\ 4 • ^ » ^ r^tf 

.-. ^ = 83<» 7' 48". 



FOECES ACTING IN THE SAME PLANE. 



251 



Ex. 2. 

3. The forces as in the figure. 
"When there is equiUbrium the ten- 
sion in the string BG = tension in 
the string AC=F; and these two 
tensions are balanced by the weight 
JP X V3j which we may regard as 
their resultant. 

Therefore if ^ be the angle 
A CBf we have 





3P»=P» + P» + aP> cosd, 
whence cos^=- = cos6o**; 

6. The forces as in the figure. 
The tension in jB(7=Tr. 

Therefore A G produced will bi- 
sect the angle BGJV. 

Letz.BAG^0; 

then jLAGB^'!^ + e and ^^45(7=-- ad: 

.'.AB: ^a=sing + d) : smg-ad), 

or a : I = cosd : cos ad; 
.'. a cos ad = cos d, 
whence 4cos*d — cosd«=a, 
which equation solved gives 

COS d= ^ = -8430703 = COS (S^** 3^' sly 

and therefore the remaining angles of the triangle are known, 
and therefore BG may be found in terms of AB. 

9. Let .4GZ)J5=ai, AG=6, GB^y; 
.-. i)J?=8, AB=: 14. 
l{ AjE=x, BF^y] 



252 



FOBCES ACTING IK THE SAME PLANE. 



r the tenflion of string CD, AGE^ 0, BDF^ <f> ; 





thenP: r=sin(90^ + ^ : sin(i8o^-^ 
s C08 ^ : sin =s I : tan 9. 

Similarly T: Q = sin^ :cos^ = tan^ : i; 
.*. P : ^ = tan^ : tan& = y : x. 

Now AC'^AE'=CE'=-DF^=^DB'^BF'; 

y* - ic* = 64 — 36 = a8. 

But y + aj = i4-7=7; 
•*• y — aj =4. 

Hence 2y= 11, aa? = 3; and .'. P : $= 11 : 3 



12. Besolving the forces parallel to 
the plane, we have 

iTT+iTTcosa^- TTsina, 
3 3 3 

a being the inclination of the plane ; 

.*. I + cos a = Ji sin a, 

a, cos" - = 4 sin - cos - , 



ft T 

whence tan - =: - = tan ^6'* 33' 54" ; 
.'. a = atan-'('5)=53«'7'48". 




FORCES ACTING IN THE SAME PLANE. 



253 




17. Since the string is 
equal in length to the arc 
of a quadrant, the angle 
PGQ will be a right an^e. 
Therefore taking the mo- 
ments of P and' Q about C> 
we have 



Px T co&POA = g X r cos QGD = ^ x r sin PGA, 

p 
whence tanPC4= yr. 

19. The reactions B, E pass 
through the centre of the sphere 
whose weight is W\ hence the 
centre is kept at rest by these 
three forces ; 

^miWOBiBmAOB 
= sin)8 : sin (a 4-/8), 
and E : Tr= sin ^OTT : BmBOE = sin a : sin (a + /S). 




Ex. 3. 

1. Produce AB to E, making PjE?= AB, and join CE, GA. 
The resultant of BE and BD is the force represented by the 
diagonal BG\ therefore the resultant of AB^ BD and DG ia the 




resultant of BG and DG, and is therefore represented by AG. 
Hence a force represented in magnitude and direction by GA 
will keep the parallelogram at rest. 



254 



FDBCE8 ACTING IK THE SAME PLANE. 



6. Let each of the forces equal P, and JR their lesnltant; 
and let £, be the resultant of the forces 
in the directions BC, CA\ then we 
have 

5/=P" + P" + aPP cos 9o*> = aP*; 

/• £| n P»J% and its direction CBt 
bisects the rieht angle ACF^ and is 
therefore parallel to AB. Now B is the 
resultant of B^ and P in the direction 

AB\ .-. 5 = 5,-P=:P(Va-i): and 
E its point of application is determined 
bj taking moments about Ei hence 

Px DE^B^ X CiT; .-. DE^ CE^z. 

8. The beam is kept at rest hj three forces R^ R^ the 
reactions at A and P; and W the weight of the beam acting at 
its middle point Q. 





BesolTing these forces horizontally and vertically; then 
taking the moments about A^ we obtain 

iJ sin a = P' sin )ff, Pco8a + P'cos)8= TF, 

and R x AB cos ABE^ WxAGcob BAE; 

.-. P'xcos(/8-^=i Wco&e. 

Eliminating B from the first and second^ we get 

Ii'(^^coBa + coBfi)^W; .\ Rsm{a + fi) ^ Wsina. 
Vsm a / 

Hence ^2i^ = :5 = «i^i^±^; 

^ _ sin {13 -■ a) p__ Wain/S p, _ TF sin a 
2 Sin a sm ^ ' sm {a + /3)' sin (a + /8) 



FOBCES ACTING IN THE SAME PLANE. 



255 



9. The sphere A is kept at rest "by three forces : viz. its 
0"WTi weight Ty , the reaction B of the plane, and the reaction M 
at G of the sphere A'. Besolving these forces parallel to BJE^ 
•we have 

Jlf cos (a — ^ = TTsin a. 

Similarly for the forces which keep the sphere A' at rest, 
we get 

Jlfcos(a+^) = Tr'sina'. 




.^ 



_^^^ cos (a — ^ cos g + sin g tan _ TTsing 

w nence / § . /i\ or ^ • . /» ^^ tx7» • » • 
cos (g + ff) cos g — sm g tan W sin g 

cot g 4- tan __ W 
•'•cotg'-tan^" F''' 

W 
.'. tan^ = -w— T=- (cota + cotg) — cotg 

_ TTcotflt'-Trcotg 

w+w 

11. The beam -45 is kept at rest by three forces; the 
reaction JR at A^ the reaction B' » »' 

at P, and the weight W at its 
middle point Q. 

Let AB^yr^ r being the 
radius of the bowl, 

AP=^x, z.AFD^0. 

Besolve parallel to AB and 
take the moments about P; by this course the solution is not 




256 



FOBCES ACTING IK THE SAME PLANE. 



complicated by the introduction of the nnknown reaction S into 
the equations. 

Then -Bcos&= TTsin^, 



and ^ sin ^ = PT ( a; — 2_ J cos 5 ; 
/. »tan5=5 foj — -^Jcot^; 



/. tan*^ = i-3!l. 

Now cos^ = -x>i = — ; .'. sec*^ 

AO ar' 



7f • 



.. a-3!:^ 



.'. 7, 



7,X 



8ec«<? = ^, 



aj* — 2 rjB ss ar* ; whence x = i'838r. 

12. Let z BCA = ^, BAE^ 4>. 

Since the beam is uniform, its centre of gravity 
will be at its middle point. The beam is kept at 
rest by its own weight, the tension T of the string 
JB(7, tne reaction B of the wall. Let GA = x. 

Then, resolving perpendicular to the string, » 
and taking moments about (7, we obtain 

JS cos ^ = TTsin ^, 

Bx=iWx- AB sin 0. 

C08^ _2 Bm0 ^2, 2__3 

X "s'sin^ ■"3^5""5* 
. Butcos^ = 5l±^:i3!^^. 

Eesolving vertically, jTcosfl^ W; 

2X 8 




FOBCBS ACTING IN THE SAME PLANS. 



257 




14. Let AB represent the beam resting on the vertical prop 
CD at G, and on the horizontal 

Elane AD. The beam is kept at rest 
V the reaction (If) of the plane 
AD ; its own weight ; the tension of 
the string AD ; and the reaction (jB) 
of the prop CD perpendicular to the 
beam. 

Now since CD = 3, and AD = 4 ; 
.-. AC=5. 

Let the z CAD = 0. Then, resolving the forces hori- 
zontally, 

r-fisind = o; .•. T=MQm0=^^B. 

5 

Taking the moments about A, we have 

BxAC=tWxAgco&0, 

or 5-B = 3IFgos6=: ^ ; 



17i Let be the inclination of the beam AB to the horizon, 

z.BAC=2<f). 

Besolving the forces which keep the 
1)0101 at rest perpendicularly ix> AO, 

Pcos <f) = TFcos {2<f) + 0)> 
Taking moments about Ay 

PxABcob6^Wx^AB cos 0. 

Hence, since P= - TT, 

cos^ = cosd; .*. ^ — 0\ 
and .'. cos 5 = Hi cos 3^ = a (4 coa* 5 — 3 cos ^), 

5 = o, and 8 cos" 5 - 7 = o ; .'. cos ^ = f | j . 

83 




cos 



258 



FOBCES ACTINQ IN THE SAME PLANE. 



Hence 6 = 90**, or cos"" f ^ j . 

N.B. The point (r, when the beam is at rest, will be in the 
vertical through (7. 

21. The lines and forces as in the figure. 

Let zA4C=a, AACE^e, W 
equal weight of the beam, and AB = 2a. 

Then, resolving the forces verti- 
cally, 

B-TBiae- Tr=o (i). i 

Taking the moments about B^ 
Bx2a co&a— Txaasixia cos e 

— Wx acosa = o {2,). 

Eliminating B between (i) and (ij), 
we get 

^ . , TTT J^sina.cose . i ^^ 

cos a 2, 




.'. r(sinacos€ — cosa sine) = 



FFcosa 



T= 



IF cos a 

ij sin (a — e) * 



Otherwise. Resolve horizontally and vertically, and take 
the moments about O. 



23. Let AB be the beam resting against the vertical plane 
at Ay and upon the bowl at B, 

Let ECF a horizontal dia- , 
meter of the bowl produced meet 
the vertical plane at D. Let 
the radius of the bowl be equal 
to r; AB=2,l; CB==a, and 
W equal the weight of the 
beam. 

Let zBCE=4^; and incli- 
nation of the beam to the hori- . 
zon = 0, 

The beam is supported by 




PORCES ACTING IN THE SAME PLANE. 



259 



its weight acting at G\ the reaction R of the vertical plane; 
and the reaction R in the direction of BC. 

Resolving vertically, we have 

W 
iZ sin 6 — 1^=0, or ^= . . . 

^ sm9 

Taking the moments about -4, we have, 

R.ABBiTL{4)-0)'- TF.^(?cosd = o; 

sin(^ — ^) cosff _ 
sm ^ 2 

.'. tan^ = Djtan^ (i). 

Let Cm be a vertical line meeting AB in w ; 

sin BCm _ cos __ 5w _ AB — Am 
Qin BmG" cos 19" BG" r 

2,1 —a. sec 



y. 



/. ijZcosd = r cos^ + a {i). 

27. The forces as in the figure. 

Since the arms of the lever are 
uniform, their weights are propor- 
tional to their lengths. 

The lever is kept at rest by the 
weights of its own arms and the reac- 
tions of the circle at the points of 
contact. 

Let 6 be the inclination of the arm 
AC {= !za) to the horizon. 

Resolving the forces in direction of the arm A (7, we have 

B'-{a + b) sind = o (i). 

Taking the moments about the point of contact of the same 
arm, we get 

B xc — b{ccos0 + b sin 5) ^a{c — a) cosd = o (3). 

Eliminating B by means of (i) and (2), 

c{a + b) sind = Jccos^ + J* sin^ + ac cos^ — a' cos d, 

bc + cM — a^ 




whence tan = 



bc + ac^b' 



260 



roKBB AcriMa nr the same plane. 



31. Let AB be the beam 
leaning upon the prop CD. Let 
AB^2a,^ CAD^a, /.ACD=fi. 
The beam is sapported bj its 
weight {W) acting at its middle 
point g'i and the reaction R of 
the prop at right angles to the 
beam. 



Taking the moments about A^ we have 

R.AD-W.AE^o. 




Now AD : Cfl = 8in)9 : sina; /. AD^^^^ 



Bin a 



am a 



/. J2 = 



W. a sin a cos a TT. a sin Q,a 



bsinfi 



2b Biafi 



and the resolved part of this pressure in a direction perpendicu- 
lar to DCf or the strain upon the prop is 

Wa sin 2a . ^r^T^ W^« sin 2« cos (a + /3) 



Tib sin /3* 



ib sin/9 



32. Let .^^(7 be the isosceles triangle resting with its 
angular points in contact with the bowl ; AB ^ AG =^a^ 
AD^h; radius of bowl =r ; G^ the centre of gravity of the 




triangle, and therefore DG = ^h. Let measure the inclina- 
tion of the triangle to the horizon; z DAO^ a, ADO = fi. 



FOBCES ACTlNa IN THE fiULHB PLAKE. 261 

The triangle is kept at rest, (i) bj reactions at its angular 
points, which act in the directions of normals to the bowl at 
Ay JB, C, and therefore pass through the centre 0, {%) by its 
own weight acting vertically at O. The reactions at B and O 
"being equal to one another, may be replaced by their resultant 
( Q) which bisects BC in i). Let B be the reaction at A. 

Besolving the forces horizontally, and taking iftoments about 
Oy we get 

, Bco&{a-0)=Qcos(j3 + e), 
B X GA sina—Qx QD sin^; .•. zB sin a = Q sin^. 

TT cosacot^ + sinot cosyScotd— sin/S 
Hence ; = -, — 5 ^; 



X /I ^ /COS B cos a \ 

/. COtd='^-r -r-3 z = 

% Vsm p a sm a/ 



acot)8 — cota* 
Now OD^=^OB'-BD^-^r^-{d'''h^)y 

• o {4A'(r«-a* + A*)-(4Jl*-.4a»A*+a*)}* (4r*A»-a4)* 
sin p = -* x = T . 



Also cosa=» ^—7 ^ = — r, 

am am 

sma = -^-5- T — ^; 

(4r^A» - a*)* (4r»A* - a*)* (4r»A» - a*)* 
,.eot^=.3i4g^!^*. 

34. Let AB be the beam moveable about Ay TF its 
weight acting at its centre of gravity g. 

B the point of contact of the beam and sphere, of which the 
centre is u, and weight Q (=4 lbs.) 



262 



FORCES ACnNa IN THE SAilE PLANK 




The sphere is in equilibrium from the reaction (JR) of the 
plane at the point of contact ; from the pressure (P) of the beam 
at B upon the sphere ; and from its own weight. These three 
forces aU act through the centre ( C) of the sphere. 

The data are 
Aff^42vx.=^a suppose; BC= 9 in., z BAD = 15*", 

•'• ^^"^^^5^" 9-3177; 52^=18-3177, 
and from the right-angled triangle ALB we have 



BL 



LB 



<jg=tani5«; .'. ■^^=t^^^^ = 68-3751 =i suppose. 

For the condition of equilibrium of the beam, taking the 
moments about Aj we have 

P. AB^ W. Ag 00845**; 
/. P=Tr.|co845^ (i). 

For the condition of equilibrium of the sphere, resolving the 
forces in the direction AD^ we have 

Q sin 30° — .P sin 15** = o, 

or ii — P sin 15® = o. 

Therefore from (i) we have 

2" W.J sin 15® cos 45*^ = o, 



FORCES ACTING IN THE SAME PLANE. 

or, substituting for a and b tbeir values 

TF= 2, X 6S'ci*7 ni X oo 

^'^ 4:^ X -2588 X 7071 

= ^3 75^^ ^ jy.yg iija^ nearly. 



263 



37. Let B be the point at which the rope 
must be tied. Let zBCA = 0, From A draw 
AD perpendicular to J5C7, the direction of the 
man's pull (P). 

The moment of P about A = Px AD 
= P4J?co8^ = P. a sin ^ cosd= -P. a sina^, 

and this is greatest when %0 = 90® or 5 = 45°. 




a 



Hence AB^BC sin 45** = -r- • 



38. Let^(7=aa, CB^ib, AAGB-=^gd*, 
Z CAM^ the inclination of (7^ to the horizon. 




The arms being uniform, their weights are proportional to 
their lengths, and may be supposed to act at their middle 
points J?, F respectively. 



264 



VOBCBI ACriKa IN THE SAME PLANE. 



Taking moments abont A we liave 

aa X AM^ ib x AN. 
Now AM^a cio%di and -4-^= CL — Ci?= i sin tf — 2a costf; 

/. «• cos 5 = i' sin 5 — aoJ cos ^ : 

.*. tan&»i — Ti — » or ^t»tan '-^-T5 — ^. 

41. Let -45 be tlie beam; O 
its centre of ^vity at its middle 

Eoint, and 6 its inclination to the 
orizon. 

The beam is kept at rest by 
the forces, 

(i) the reaction at the hinge .4, 

(a) the weight {W) of the beam at G, 

(3) the tension (= J W) of the string BC. 

Take the moments about A^ and let BA 0= 2^. 
Then Wx^AB co&0 = - WxAB coad>; 

/. cosd = cos^; .'. 5 = ^. 
Hence when AC \s vertical, + 1:^^ = 90°, and /. 0= yf. 

44. Taking moments about C we have 

14 CB cos 5+ %CA cos d = 6GN. 





Fig. I. 



FORCES WHICH DO NOT ACT IN THE SAME PLANE. S65 

Hence in the ist case (fig. i), 

Q 

9 cos ^ = 6 cos - ; whence = 53® 27' 23". 




Fig. 2. 

And in the 2d case (fig. 2)9 

9 cos 5 = 6 sin - ; whence 6 = 68** 5' 10", 



Forces which do not Act in the same Plane. 



Ex. 4. 

1. Let Bj S^ iT denote the reactions at the props ; then 

B + E+E'^W. 

Taking the sides containing the right angle for axes of x 
and y, ana resolving parallel to the axis of z^ we have 

3^= Wx, and 4^"= Wy. 

Again, E : B =4:5, 

J2 : ir' = 5 : 3; 

.'. JB* : -B" = 4 : 3 ; 

/• Wx = Wy and x = y. 

Also jB^i+^ + ^I^TF; .-.jB^i-TF; 
\ 5 5^ 12 ' 

5 ^ ^Vi2 /' 



5 ' 3 5 

,-, x = 1 =y. 



34. 



266 



THE OENTSB OF GRAVITT. 



8. The weight of the triangle acts at 
its centre of grayity G. Join Cu-^ and pro- 
duce it to meet AS in 2>. 

LetCG = a; DB^p^DB. 

The weight of the triangle acts verti- 
cally; reaolye it into two vertical forces at 

the resolred force at C= IF. Tvyj = - TT. 

CD 3 




(I)- 



Similarly the resolved force at 2) = W.-rrFi^ 2 Wi let this 

CD 3 ' 

be again resolved into two forces acting in the directions of the 

strings at A and B\ then 

tension ofstring at J5 = -Trx- = i W fa), 

3^3 ^ ^ 

and ^=:-TFxi = -Tr (ct). 

3^3 ^^ 

Equations (i), (i^), (3) shew the truth of the enunciation. 



The Centre op Gravity. 

Ex.5. 

1. Bisect BCy CA, AB in E, F, H respectively, 
will AJEy BFy Cff intersect in G, the centre ofgravilj. 



Then 




Now GB* + GG^= 7,GE* + %BE^ 



^^GA^ + IbC^ since GE 



^GA SLndBE==lBC. 



THE CENTBE OP GRAVITY. 267 

Similarly GC*+ GA^^l GB*+^AC\ 

and aA'+ GB'^^aC' + ^AB'; 

U-'^{GA*+GB'+GG*)^^{AB^ + AG^+BC% 
or 3{GA'+GB'+GC')^AB* + AC^+BC\ 



5. Let^5=:5, BC='4y CA=^7.\ 

the weights at -4, J5, G are proportional to 3, 2, i, respeo- 
tiyely. 




Let ^ be the centre of gravity of A and B^ 

Then BE : EA^s : 2.. 
Comjxmendo, BE : BA^^ : 5; 

/. 5^=3, 

CS» = J?(7" + BE* - a5C. 5(7 COB B 

= 16 + 9-2x4x3 ( ^ "*" '^ "" ^ ) 

5 
/. (7jB« 1-67333. 

If 6^ be the centre of gravity of 5 at J? and i at G\ then G 
is the centre of gravity of the 3 weights at A, B^ G; and 



GG : EG^K : 1; .\ GG=^ GEx^^^^^^^^==v 

^ ' 6 12 



3944- 



268 



THE CEMTBE OF QRAYITT. 




6. Let Q be the centre of gravity of the weights P, Qy -B, 
whose distances from O are 

p, y, r respectively; TOrodace 
PG to meet QR in J?, then 
E is the centre of gravity of 
Q and B ; hence 

P: + 5 = ^0^ : GP; 

.\ PiP-^Q + B^EG'.EP 
^AQGB : AQPB. 

Similarly 

Q : P+ Q + B^ABGP : ABQP, 

and ^ ; P+ C+£ = AP(?C : AP^G- 

Hence P : Q : B=^AQGB : AiZfl^P : APGQ 

= jrsin)9 :jwsin7 :^jsina 
sinB sin 7 sin a 

7. Let P, C jB express the weights of the sides AB, BC, 
CA respectively. Join PQ^ QBj BB ; in PQ take E the centre 
of gravity of P and Q, and in PB take F the centre of gravity 
of P and i?. 




Then PE : ^(2= g : P=P(7 : AB = PB : 5(2; hence 
the line JEB bisects the angle PB Q, 



THB CENT&E OF GBAYITT. 



269 



similarly, the line FQ bisects the angle PQR] therefore 
their point of intersection is the centre of the circle inscribed in 
the A PQR. 

Now the centre of gravity of P, Q^ R is in the line -EB, 
and also in the line FQ^ and is therefore at O the point of their 
intersection. Hence the centre of the circle inscribed in the 
A PQR is the centre of gravity of the perimeter of ABC. 

12. Let-45 = a, CZ> = i ; produce -4 C> JRZ) to meet. 

Bisect AB in E, and join VE 
cutting CD in H, 

Let G^, i^, ^ be the centres of gra- 
vity of the triangle VABy trapezoid 
AJby and triangle VCD respectively ; 

EH^h. 
Then GF: Gg-^LVGB : trap^^i? 

and Gg^VG-Vg 

^l(VE^Vn)^^h. 
3 3 

Also VE : VH^a : b; 

.'. VE : JSff=a : a-b; 

.-. EG = ^EV=-.^. 
3 3 «-* 

H^ce MF, Ea - OF. \ (^, - „-^) = \ {^) . 

Through F draw NFM perpendicular to AB and CB ; 
then NF : FM^EF : FH=^a + 2,b : 2a + b. 

17. Bisect the diagonal BC=p in E; join AE^ EB; take 
EH=^-EA, EK=-'EB; draw HL, KM perpendiculars on 




270 



THE ONTBB OF OKA.yiTr. 



BC. Join HKf and l«t <7 be the oentie of grraTity d the ^atin- 
btenl ABCD', dmw GitTpeipendicuUr to BC. 




Then qoMiLABCBxNG^AABCxSL- ABGDxM; 

Let the axis of a? be horizontal, and that of y Tcrtical; let 
CH^ 5, HG = y be the co-ordinates of G, the centre of gravity 



C H 




ThenififC4 = d,^0J? = a,wehave 
J (a + 6) = a. - cos ^ — ft . -cos {i8o* — (a + 0)} 

= — cos ^+ — COS (a + ^), 

y (a + J) = a . - sin 5 + 6 . - sin {iSo** - (a + 0)] 

= — sin 5 H — sin (a + ^). 



a: 



THE CENTRE OF QBAVITT. 



271 



\ 



/ 



-• Hence C& = */W+^ 

^ [{a'cosg + 5'cos (a + g)}' + {a» sing +i' sin (g+g)}' ]* 

2,{a + b) 

^ [(X^+h^+ ga'&' {cosg cos (a + g) + sin g sin (g + g)}]* 

2,{a + b) 

_ (g^ - ^ 4- 2a'i^ cos a)^ 
"" aia+b) 

y 

22. Let L, M, N he the forces represented by magnitade 
nd direction hj PA, PB, PC respectively, 

P4 = A, PB=h PC=l; 
z CPA = a, CPB^P; 

I ' ?, ^ the co-ordinates of the centre of gravity of the A ABC; P 
)emg the origin, and PC the axis of k. 

The centre of gravity of ABC coincides with the centre of 
jravity of 3 eqniu weights (w») placed at the points A^ B, C; 
iience 
^t yrnx = mh cos a + vnk cos P + w?, 

ymy = mA sin a — inA; sin /9. 

^ To find the direction of the resultant R of the forces Z, if, .W; 
we have 



C' 




5 sin g = i/ sin a — if sin ^, 
5 cos g = Z cos a + 3f cos )3 + ^; 



272 



THE CENTBE OP GRAVITY. 




^ Xflina — Jfsin^S hBina^JesinB y 

Zco8a + ifcofl)8 + -Ar hco&a + kco8 fi + l x* 

Hence the resultant passes through the centre of gravity of 
ABO. 

23. Let ^^be the given chord in the circle ABC; A, OB 
anv triangle inscribed in the circle. Bisect 
Ab and x>; and take I) as origin of rectan- 
gular co-ordinates. 

Let x\ y* be the co-ordinates of the point 
0\ (a;, y) those of the centre of gravity G\ (fiyO) 
those of the centre of the circle. 

For the circle we hare 

»" + (y'-i8)» = r« (i), 

and, hj similar triangles, 

I , "I , 

a= - as, V— - y I 

3 3 

therefore substitating in (i) we get 

9«*+(3y-/9)» = r«, 
ora. + (y-D = (iry, 

for the required locus, which is a circle whose radius is - r. 

26. Let V be the vertex, and ABO the base of any tri- 
angular pyramid. Bisect BO in 0; join -40, and divide it in 

Q so that OQ =^- OA; join 

QV, and take QG^- QV; then 

4 . 
O is the centre of gravity of the 

pyramid. 

Join GA, GB, and GO; also 

ro. 

Now 
aVO'+^BO'^^VB'+VC', 

and BO=-BO; 




THE CENTBE OF GBA.VITY. '273 

.-. a VO* = VB* +rC*-- BC ; 

2 

also stAO'=AB* + AC'--BG'', 

2. ■ 

and VA' = AQ'+VQ'-2AQ.VQ COS AQV, 
V0''= OQ'+VQ' + aOQ.VQ COB AQV; 

.'. VA*-¥aV0' = AQ'+20Q' + iVQ'', Baux>AQ=!iQO. 
Hence VA*+VB*+rG'--BC' 

= 60Q'+i(^rdj'^-AO + —GV' 

= -(aB' + AC'--BC'] + — GV', 

.: 3{VA*+ VB*+ V0')'=AB' + AG*+BC'+i6GV'. 

Similarly, 
3{Ar' + AB' + AC') = VB'+VC' + BO' + i6GA', 
S{BA* + BV' + BC*) = Ar' + AC*+VC* + i60B', 
-and 3{GA'+ CB* + OP) = AB" + AV* + BV + i6aC', 

add and divide hj 4, then will 

4{GA*-\- GB*+ GG'+ GV*) 

= AB* + AC + BC* + AV + BV* + CV\ 

32. The base of the pyramid cut off will be an eqnilateral 
triangle ; and if a be each of the eqnal edges of the cnbe, the 
diagonal of one face,^-or a side of the base of the pyramid, — 
-will be equal to a V2> and the diagonal of the cube =a<^^; also 

the altitude of the pyramid =—r-. 

that cut off. 

- — 2 = diat. of cent. grav. of pyramid 

35 



=s dist. of cent. gray, of cube from the angle opposite to 



274 THE CENTBE OF GBAYITT. 

Let X s diet of cent gray, of remaining portion from the 
same angle, the volnme of the pyramid = tt . 

. 50^ ^V^3 «V3 

'• 6 ^ a 8^' 

^^ 5g_ 4gV3-«V3 _ 3<^V3 . 
^'T 8 8 ' 

40 20 

33. Let a denote an edge of the cube, 

then aV3 = a diagonal, 

and a : a Vil = - • side of base of pyramid = -j- , 

.•. 2. — T- = 5^^ = dist. of cent. grav. of pyramid from the 
4 ^V3 8V3 , ^ ^^^ 

angle cat on; 

•'• ®V3^ o/-^* 0-7- =dist. of cent, of grav. of pyramid from 
angle of cube opposite to the one cut off. 

Let X = dist. of cent. grav. of remaining portion from the same 

angle, 

volume of pyramid = - . — 5^ . — 7- = -5 ; 
^^ 38 aV3 48 

.'. volume of remaining portion = ^^; 

45 

•• 48 '"■"'' a 48*8 V3' 

. 47^_^V3 7^ _ 185a 

••48 a i6x8V3~i6x8V3' 

whence x = ^ Y^ a. 

376 



THE CENTRE OF GRAYITT. 275 



Ex.6. 



1. The area "being sTmmetrical, the formula to be used is 

jydx ' 
and from the equation to the curve, we have 

fxydx _ J{a^ - a^)^ xdx 

Now fV-«*)*db = ?(a«-af)*+l%in-5 = '!^ 
Jo a^ 'a a 4 

between limits a? = o, a? = a; 

/ (a* — a*ra^ = — -(a* — »•)* = - a^ from a? = o to a: = a: 
Jo 3 5 

3 4 3*^ 

5. Since y = a verar* -+ {Q,ax - a*)*; 



/. /a^do; = ajx vers"' - da? +Jx {%ax — af)^dx. 



poa _ ^ • itaa 



Now I a? vers""' - da? = ^5 — and / a?(iiaa? — a?*)*{& = — ; 

Jo ^ 4 ^ 4 

Similarly I ydx = a I vers"* - dx+\ {aax — afydx 

Jo Jo A Jo 



. I • 3 • 
= a X wa4- - ira* == - Tra*. 

2 2 



- J ^'^ 7 ii 7 

Hence 5 = -^-zr = - ircfi -«- - ira* = > a. 

I V ^ » ^ 



276 THE CEMTBE OF ORAYITT. 

7. a:=jl€^ + €-}; /.€^+- = — ; 

» 

.. € — — 9 •• « Z > 



whence y = c log ^ ^ -r . 






,. |!-,„g {£±(^^»fe. ^log. . - f - ^ log. , 

Again, 
m£±Kz^'| & . « bg {5±.^} - (^ - c,' 

between limits = ^ log, i* - ^= ~ (5 log* ^ - 3) J 

4 4 4 

•'• ^ = ^(34 log^a- 15)^^(5 log,a-3) 



16 V 5log.2-3 /■ 



THE CENTEE OP aRAVITY. 



277 



8. Let t7 be the centre of the 
circle of which ABC is a sector. 
From G draw the straight line 
CD bisecting the arc AB. 

Let CD = c, z.DOA=:a. 

Then if P be any point in the 
area of the sector, 

CAB mi CP^r, 

ADCP^e, 

we shall have 



a ra 



J-1 




r* cos .dO.dr 






CO&0 dO 



J —aj a 



rdOdr 



- c» r de 

1 J-o 



3 



a - g sip QC _ 2 /radius x chord\ 
2a ~ 3 \ Arc 



)■ 



10. The area not being symmetrical, the formulas to be 
used are 

•:__ !!xdxdy ^ -- _ Hydxdy 
"^"JJdxdy' ^ JJdxdy' 

Litegrating with respect to Xj we have 

" fxdy * 
and from the equation to the circle 



j\a'-ffdy |(a--y»)* + ^sin-| 



between limits =-a} -. = — . 

3 4 S'T 



278 THB GBITIB OP CKA.YITr. 



^^y- l^idx f/««_«»^*/fa «/_. .^i^^ 



a; 



/3^ /(«*-a^*<fc ?(a'-a!*)* + ^8iir'| 

I fl^ 40 

between limitB = - <^ -^ T" ~ "^ * 

3 4 3*^ 



Jairo*af <fe - jaa?dx _ 5 3 

~ (infill dx-Jaaedx ImM-- osc* 

3 * 

Now tftf^f-Zifinx; .'. x=o and x^^. <^® l*™^** <*^ 

integiati<m; 

_ _ 641R' 8ot* _ 8« /j\ 

•*• '^~^d^^^~5fi^' 

„ . , lf^{4mx-a's^)dx 
. . - (fydxdy a Jo '^ 

Jo ■ 

i6m^ 3am^ 

30^ a^ 

18. For the centre of gravity, we have 

/^^ (j) 

"^ Jy'dx 

"^ from the equation to the revolving circle 



THE OENTBE OP QEAVITT. 

whicli being Bnletitated in (i), tto liaTe 

I (fi* — x?)xdai — a'sc* — oe* 

and between the limits x=o, x=a; 



22. Let AD=:a, BE=h, AB=h; G the centre of gra- 
■vi^ of frustum FD. 



hhol&t GA=p, GB=q; .•.p-q = h. 
Then p = — and y = — , 

Hence x^AG=p-^(^±n±r\^t±M^:^ 

_p-g ^ j>+gg _ A / g' + a£' \ 
3 P + S sU'+iV' 



1 



280 THE CENTBE OF GRAYITT. 

24. Let a be the angle between the bounding radii of 
sector ; 

I I r^ Bind COB 0d0dr -a* einO cos ddO 
then i^-^—-; =±—L^ 

I ff^Brnddedr ^a^^^l Bm0d0 

\ sin QtOdO 

^ ^ jjjxdxdydz ^ Jjxdx.dy.z 
///«My<*5 Jlzdosdy 

_ !!{r'-y*-x')^xdxdy ^ 
ffedxdy 

But /(r» - y* - oi?focdx = - (r* - y')' taken between the 
limits a? = o and x = (r* — y") , 
and bj the question we have 

Jjzdxdy = ^ vol. of sphere = ^ wr^. 

Therefore 5^ ^^(^-y')^''^. 
But 

/(r»-y»)^e7y=|(!l:i^^ 



8 



r 



from y = o to y = r = -2_7rr*; 

i6 8 

Now the solid being symmetrical with respect to the three 
co-ordinate planes, we have 

^ = y = i=*|r. 



THE CENTRE OF QBAVITT. 281 

28. The formnlsB for the centre of gravity are 

- _ j^fxdxdydz - _ JSSy^^ydz ^^ fffzdocdydz 
""^ Sjjdxdydz ' ^^ JJfdxdydz ' ^~ SJJdxdydz ' 

ffjxdxdydz^ffxzdxdif^ from z=^cx to z^mx; 

s= (f» — c) ffa^dxdy = (w — c) fyaMx, 

from y = — Vmoj — aj* to y=s4-Vaa» — a*; 

= a (m — c)l aj* (aoo? — afydx. 

Similarly, we find 

ffjdx.dy.dz=^2{fn-c)j aj(aaB-ir*) &:; 

' ** j^{2ax--af)ixdx ^ 4 ' 

JJJydxdydz = ffzydydx = - (w - c) /y 'oxfo, 

from y = — (!jaaj-ic')*toy=4-(aflKB-a*)* = o; 

.-. y = o, 

JJJzdxdydz=iJJ^z*dydx^'' {ni^--(^)jiK?dx.yi 

from y = - (iiaaj - a*)* to y = + (aaa?-a*)* 

= (w» - c») /a:» (liaaj - a»)*c& ; 

c 

••• ^= r V~ ^ — s^"^'^'^''* 

a(m— c)t x{2ax — af)dx ^^"a" 

36 



1 



882 



THE CENTBE OF GRAYITT. 



29. Let BAD be the 
aemiciicle, of which the cen- 
tre is (7, and radioB a. 

LetuxeAP^a, GM^xi 

a 
.\ sr»a.C08-: 

a 

a • B 

/. /axfc = a /cos— &=a*sin-: 







xds 



• » sD "~ 






Tra TT 



& 



82. Since y = a vers"' - + {mx — a*)*; 






((ZOOJ — of) 




- _fxda _^ 



Also y = 



r{zax^)^dx 






(i?a)*7r-ijl (iia-aj)^c& 



Jx'^dx 
from aj = o to 0? 



= 2a 



= 9ra — 



4a 



34. 



2 (aa)* 

Since y = sina;; 

^y ^ da , ^ ,i 

• -^ = cos a?, and -=- = (14. cos*aj)* 



dx 



dx 



XHS CEirrBE OF QaAYirr; 283 

- , I Bin 05 (i + cos*a?)* <fa 
Hence y^i^^^ = -^ , 

If « denote the length of the curve between the limits a? == o 
and x^^TTf then will 

« a*+log(a*+i) 

35. From the equation to the generating curve we have 

^ {zax-af) y* ' 

or yda^adx; 

f , / aocdx - c' 

. ^Jj^J^ 2 Ic 

I adx 

m 

36. Let y =s oa; be the equation to the generating line ; 
then 

dy==€Klx^ c?y*+ciB*= (a*4-i) c£c*; 

.\fxyd8==fa{a'+i)^ai'dx^ "^^^'^^^ , 
and I yds^l a(a*+ i)*fiwfo?ar: — ^ ^; 

JO Jo a 

. -_ Jo^^ _ a*(o» + i)* . o»(o' + i)* _a« 

• • flJ — * _ — — ^■^^— — — -A- —————— ^ , 

r J 3 2 3 

J ^ 

38. Since if==a vers"* - + {ciax — ar*)' ; 
fxi/d$ <= Jy {2ax)' dx = (aa)*-l-a!'y /as (aa — »)*<&[; 



28i 



THE dSMTBE OF GSA.VlTir. 



- (ao* {a (aa)* (a^) - 1 (aa)*} = (aa)* (^ - 1) ; 

8^1 



— fxgda ta 



TT — 



,r-± 



3 J 



_ 2a ( i5ir — 8 



aa f i5Tr — 8 | 
I5t3'r-4J 



89* Let AB be the ^ren physical line, and C the 
giren point; CA^a, CB—hy OP=x. 

The density of the elementary portion PQ yaiies in- 
verselj as the sqnare of its distance from C; 

/. dm s= ~ dx, ■ 
p being the density at the distance of unity. 

Then /iwfoi = pj a? — = plog-, 

- (xdm , 5 /I iN oft , 6 
jam °a \a i/ 6— a °a 

42. The equation to the generating circle is 
and for the centre of gravity we hare 



\ pxy^ 

x=^'^ 

/•a 

J Q 



pj^^dx] 



c 

T 
I 

i 

1 



P 
Q 



J 



THE CENTRE OP GRAVITY. 



285 



hntpccy»; .\x = -^ • 

But /(a« - a:*)' ajda? = - i (a« - a:»)3 = i o^- 
from x = o to x^a; 

O T fi 

and fta^'-afydx-a'^x aV + - o^ = — a^; 

3 5 ^5 

from xsso to x=sa; 
6 15 16 



Ex.7. 

1. Let ABC be the triangle, G its cen- 
tre of gravity, and the centre of the 
inscribed circle. Join AG and produce it 
to meet the base BG in P. From G draw 
a perpendicular to BC^ this perpendicular 
win pass through 0; if not, let it fall other- 
wise, dA GQ. From draw O-AT perpen- 
dicular to BG\ and from G draw uR per- 
pendicular to AB and therefore parallel to 
BG^ for the triangle ^^(7 is right-angled 
9,iB. 

Then GQ==-AB, 

3 

and GB = BQ=^^BP^-BG^i, 

3 3 

and by Trigonometry ON=^BN= -. — ■ = i, 




or BN=^ BQ. 



perimeter 



Therefore the point Q coincides with N; that is, the per- 
pendicular from </ on BG will pass through 0, or GO^ is 
perpendicular to BG. But by tne property of the centre of 



278 XHE CENTBB OF aBA.yiTr. 

a '2 a 

I a'w 4a 

between limits = - «^ -5^ -r ~ "^ • 

3 4 3^ 



12, 5 = 






« 4 - W a?' CMC* 

lltrfo? dx - jaa^dx _ 5 3 

fzm^x^dx-faxdx im^x^ — -ax* 

3 ^ 

Now aV=y* = 4iiia?; /. a? = o and »=-^i *^® limits of 
integration ; 

• /Y«^_3I ^ SS •••••••• •••••••••• 111* 

• • * 15a* • 3a> 5a' 
Again, y-fr^^- ^ 

Jo 

c^ "" 30* _ i6m^ , 8m' _ ^ 
"" 3!2m' 8m' ""30* ' 30* "" a ' 
3a^ a^ 



18. For the centre of gravity, we have 



^Jx^ , (I), 

fy^dx 



and from the equation to the revolving circle 

y» = a' - a;', 



THE CEKTBE OF aitAVnT. 


■which heing Buhstitnted 


m (I), we 


lave 




- £'«- 


-^■)a* 


1.. 


4 


"£<»• 


-^)das. 


a"a;- 


3 


and between the limits x 


= o, x = a 


; 




a'- 


if ^ 
1.. °* 


3 


= i- 



22. Let ^i>=:a, £:£=}, ^S=A; G the centre of gra- 
vity of frustom FD. ^ 




Hence x = Aa=p-^ {^±M+^)^^+IS^l^ 
3 f + y sla' + iV' 



then X 



280 THE CEIITBE OF GRAYITT. 

24. Let a be the angle between the bounding radu of 
sector; 

r fr^eindcMddOdr -a*j sin^cos6l£W 

rfr*Biaed0dr -a^-j aia0d0 

j^ eta I — cos 2a fz ,a 

_^^:c'^!, ^, = ^aco8»-. 

8^ r . aM ^^ i-cosa 4 a 

- fjfxdocdydz jjxdx.dy.z 
26. » - jjj ^y^ = //«(&rfy 

ff ( y* - .V* - ^)^xdxdy 
" fjzdxdy 

But J{r*-if'-x')^xdx^'{r*-y^)^ taken between the 

3 

limits 05 = and a? = (r* - y*)*, 
and by the question we have 

JJzdxdy = ^ vol. of sphere = g tit*. 

Therefore 5 =^i>^3^. 
Bat 



( 



/(r« - y')Uy = {^^^^^^ + 1 »-} • y (»^ - y') 



o r 



from y = o to y — r^ Ts'^ ' 

Now the solid being sjnmmetrical with respect to the three 
co-ordinate planes, we have 



THE CENTBE OF QBAYITT. :281 

28. The formolse for the centre of gravity are 

- _ jjjxdxdydz ^ _ jjjydxdydz ^ _ jjj zdocdydz 
IJJdxdydz ' 2^"" fjjdxdydz ' ""^ JJJdxdydz ' 

fJJxdxdydz^JJxzdosdy, from z^cx to z = mx; 

= (w — c) Jfa^dxdy = (m — c) Jya^dx, 

j&om y^ — ^a^ax — of to y = + Vr;&aa? — a?*; 

Similarly, we find 

jjjdx.dy.dz-2{m-c)\ x{2ax-a^) dx; 

r{2ax^af)^afdx 5^ 

.^.xJ^ ^^^^a, 

j {2ax - sc")^ xdx -^ "^ 

fjjydxdydz = Jfzydydx = ^ («» - c) /y 'axfo, 

from y = — (aoa; — a:")* toy = +(aaaj — ic')* = o; 

.-. y = o, 

JJJzdxdydzssfJ^z^dydx^- {m*'-<^)Jafdx.yj 
from y = - (aooj - a")* to y = + {%ax — afY 



a(«»— c)J a!(aaaj— af) <M5 *^~^ 



36 



THB CEirrBB OF QJLAYTTTm 



S9. Let BAD be the 
of which the cen- 
tre is (7, and ndiiifl a. 

LetaroAP-i«, CM^x\ 

/, 9aBa.cos-; 

a 

.\ fxds'^afooB-ds^a^Bm-'; 




r; 



acds 



« = 



I'ds 



ira IT 



32, Since y= a vers""*- + (aaaj— a:*)*; 






2a 






--_fxds _ 



« = 



Also y = 



-j^flfds^ 






jda 



Jo \X/ 



dx 



(M)*7r — al (aa — a?)*<fe 



Jx'^dx 
from a; = o to 0? 



=2a 



= wa — 



4a 



34. 



a (aa)* 

Since y = flina;; 

- dy ^ ds , , - ># 

^ = <^s a?, and -^ = (i + cos*aj) . 



■ 

I 



ij: the centre qp aRAvirr; 283 

. , I sina?(i + cos*a?)*db 
Hence y = » = ^-^- . 

I (i + cos'a;)*da? 

— Jo 

If 5 denote tlie length of the curve between the limits aj = o 
and a? = w, then will 

« ri* + log(ii*+i) 

35. From the equation to the generating curve we have 

^ (aooj-ar*) y» ' 

or yda^adx; 

J yds r J ^ ^ 

36. Let y^ax he the equation to the generating line ; 
then 

dy^adxy rfy*+(fe*= (a*+ i) e&*; 

and I yds— I a{a^ + i)^xdx^ —^ —; 

• /M ai^ " ^» ^ * ^1 ^ ^ — ___ 

• ••«/—• _ — •^— ^— — ^— T- — • 

r , 3 ^3 

38. Since y = a vers"* - + {7,ax — «»)* ; 

Jocyds =^fy{zaxydx= (aa)* ■l-«*y Jx {m — aj)*diBh 



292 



USfEB* 



2 -.1 

3 ^ a ' 

•••y = §^ = i X extreme ordinate -BC 

10. Let c denote tlie distance of*tlie centre of the ellipse 
from the axis of revolution; then the length of the path described 
by the centre of gravity = irc ; and the area of the ellipse 
s irab (a, b^ being the semiaxes) ; therefore 

volume of ring = irab x wc = ir^abc. 



MACHINES. 

Letes. 

Ex. 10. 

1, The weight of the lever acts at its middle point, and, 
consequently, at the distance of half a foot from the fulcrum, 
and on the same side of it as the given weight. 

Let X = the weight required. 



Then a;x i =- X4+a x lo: 
/. a; = aalbs. 



5. Let AB be the rod, and 
C the fulcrunx upon which it rests. 
Then if AO =^ a, its weight will 
be axj and will act at its middle 
point Therefore 



TV 



ta 



ax 



ab 



. oaf ab^ 
naX'\ = — ; 

2 2 

whence aj=(5' + n")* — n; 

,-. AB^{h--n)-^(yJr w')* the length of the rod. 



J 



LEYSfi. 



293 



8. Let 078 the dist. of the falcrumy or centre of gnmty of 
the system, from the end wheife the smallest bodj is attachea* 

m, _ gPxq+3Pxaa + 4Px3a + >.. +{n+ i)Pxna 
men a:- P+aP+3P+ + (n + i)P 



= a. 



-n(n+i)(n + a) 

i V 

^(n + i)(n + a) J 



2 

= - na. 
3 



11. Let W be the weight of the beam, which being uniform 
\nll have its centre of gravily at Oj the middle point of AB. 



B 



6Z6 



4w 



TT 



1 

gioii 



Taking moments about C, we have 

* Wx CQ + 5X CA==^iJox CB; 

7 7 

14. Li the lever A CB let C4 = i8 in., CB = la. 




Then, taking moments about (7, P x (7Jf = ^ x CAT; 
.*. P : Q=s CB cos 30® : C4 COS45® = ^x-V3 2 3X-V2 



294 



LBYEB. 



17. Let ACB be the leyer, and ^ 

let the angle BOD ^0. Let aa; 

= length of shorter arm in first 
instance, and ay == length of longer 
arm. Then, on the first supposition, 
taking the moments about C7, we 
have 

XX Cg^jfX CD, 

or aj*=y* cos tf, 

and on the second supposition, we 
have 

y* = 4aj* cos^; 

2 = 4^ cos ^: ••. cos* 5 

cosS ^^ ' 



9 



C D 



t 



• • 



a' ' 

.*. the angle between the arms = iiko^ 

also ic* : v*=- : i; 

.•• a? : y = I : V^. 

20. When there is equilibrium we must have 

P : W^BGsixifi : ^Osina, 

a, /? being the inclination of the forces to the arms A C7, BC 
respectively. 

But, by the question, we have BC^AG^ and GE being 
parallel to the forces 

BGE^fi, and AGE^a\ 

••• P : W :: BmBGE : sin^CE 

Therefore, componendo et dividendo, 
p+ PT : P- W^ sin 5Gff+ sin ^C£? : sin BGE- sinAGE, 

F+W tani(5(7^+^C®) 



tan ^{BGE^ ACE)' 



LBYER. 295 

but BCE+ACE^ACB, 
and BOH-- AGE ^ 2DCE\ 

.\ P+ W : P- Tr=tan-(-4CB) : tmDCE. 

23. Let 2; = the true weight; a, h^ the arms of the beam 
upon which the weights 9 lbs. and 4 lbs. respectiyelj, are sus-» 
pended. Then 

X : g^a : h (1), 

and X : 4 = b : a {7,); 

/. 0^ = 36; .*. a; = 6 lbs. 

Let z be the number of inches in the longer arm. 

Then 6 : 4 = « : 45 — « ; 

.*. 6 : 10 = « : 45 ; 

6 X 45 . is . 

.-. z = ^^z=%h in. = ij ft. <i m. 

10 ' ^ 

The arms are therefore a ft. 3 in. and i ft. 6 in. 

27. When the weight w is put into the scale of the shorter 
arm, let p be the weight of the goods put into the scale of the 
longer arm. 

Then » (w+ i) = v).m. .*. » = — : — w. 

Also let J from the shorter arm balance w from the longer, 
then we have 

J . w = K? (w» + I) ; .*. J = w. 

Li the latter case the seller loses q — w^ — . 
former •• gains to-'p^ 



m+ I 
Hence, in any such pair of sales, 

the seller's loss = - 



m m+ I m(w + i) ' 



296 WHEEL AND AXLE. 

/. in one hundred sales his loss = — f r , 

m{fn+i) 

i.e. the seller's loss = — 7-^— — r per cent. 

m \jn -T 1) 

30. Let O be the centre of gravity of the beam, at which 
point its weight W^ 3 lbs. or 48 oz. is supposed to act. When 
the fulcrum is at (7, let Pat -4 balance Tr at O. 



c o 



I, "" I — 

AC^x, ^G^=i8in.; .•. (7<? = i8-ir. 
Then Pa:=Tr(i8-a;); .\ ^= p^ g^ - 

Let Xii x^ x^f be the values of x when the values 

of P are 4* 8^ iiK, &c. oz. respectively. 

^, 48X18 12X18 ^o. , 

Then Xj = -^— ; — o- = = lO^ inches, 

4+48 13 '^ ' 

_ 48x18 _ 6x18 _ 3 



8+48 



48 


X18 4 
+ 48 


XI8 

5 


= i4f 


&c. 


&c. 




&c. 



Wheel and Axle. 

Ex. 11. 

1. P : W :: rad. of axle : rad. of wheel; 
.*. 10 : 555 = rad. of axle : 2 yds. ; 

.-. rad. of axle = -* = i^ in. ' 

555 '' 



WHEEL AKD AXLE. 



297 



4. Let X = tlie weight required. Then 
rad. of larger wheel x 48 lbs. + rad. of smaller wheel x 50 lbs. 

^ rad. of axle x oi^lbs.. 



or2^X48 + 3X5o=fa;; or|x 

o o 

.'. a? =5^64 lbs. 



= 320; 



Pullet. 



Ex. 12. 

1. By the formula we have 

3Pcosa= TF, 
2a being l^e inclination of the strings; and by the question, 

i^Pcosa = P, or cosa = - = cos6o*'; 

.•. the angle between the strings is i!Zo*. 

7. Let P=7lbs. and ^ = 5lbs., acting as represented in 
the figure, support the weight TF at 2>, ^(7 = a, CV^b, 




38 



298 PULLET. 

Taking moments aboat C^ 

Px CM+ Qx CN= Wx CD; 
••. 7a sin d+5a sin (60*' — 0)= Wb. 

Hence by the question, 

Wh 
7 sin^ + 5 sin {60^ — 0) = = maximum. 

Differentiating with regard to 0, we obtain 

7 €08^ — 5 cos (60® — ^ = 0, 

hence tan ^ = ^^ = tan 46° 6' 7"'6. 

5 ^ 



8. Let t^ be the tension of the string passing round the 
lowest pulley w, 

tt be the tension of the string passing round the second ^ 
pulley w'=^2,Wf 

t^ be the tension of the string passing round the third 
pulley %o" = 2*w. 



2 ^ 

I fW+w . A W+w . w' 



'"~ a" ■^■i'^'''a'^'*' ■'■"a ' 

= a-P- i (4* - 1) w. 

6 



1 1. Let v> be the weight of each polley. 

t„ l„ t^ ti the tensions of 

the iBt, and, 3rd 8th strings reapectively, 

or the portions of the weight aapported at 
(7„ G„ Cj 

Then 
t, = P =P+(i_i}«,, 

(,= aP+«) =4p+(a- i)w, 

(3 = 2 (3P+ W) + W = 3'P+ (2' - l) W, 



= 2TP+{2'-l)w. 



Tr=f, + (. + + (g = (P+w)(i+a + 3». 

= (2«-l)P+{3''-9)w. 

If P=o, w : IF=i : 3*-9 = i : 047. 



.. + jj) _ Sw 



Inclined Flake. 



3. Let P=5)M, 

reeolve these forces parallel and per- 
pendicular to AB. 

Then 

P C03 6 = TF sin a, 
5 + P sin c = TT COS «, 




300 



INCUNED PLANE. 



Square both equationB and add 

^ + a-BPBin€+P*(co8*€ + «in€) « IT' (sin* a + cos* a) ; 



••. sin ess FTR = ^ 4 — -^ = 75 — T = 'i:& Binii 



a^P 



2x5x4 8x5 



3^' 



., S + Psinc 4+1 

Also cosa = Tfr ^— — = 

W 7 



cos 44** 25'. 



_5 



We may also find a from the equation sin a = ^ cos e. 



5. Let be the inclination of the plane to the horizon, 

Tbe the tension of the string parallel to the plane. 

Then W: r= i : sin ^, by the property of the inclined plane, 
and TiP—R : r, wheel and axle; 

.-. TT : P=5 ; rsind; 



•'•^='S»^"(^)- 



7. Referring to the figure in 
the margin; suppose the weight 
drawn up the plane AD by the 
power P acting over the pulley 
at A ; then 

P: W^AO X AD, 



I 




and while P descends through a 
space Wa, IF ascends upon the 

)lane through the same space; 

ut the space actually described by W in this time in the 
direction of gravity is la ; and the space described by P in the 
direction of gravity is Pc ; therefore, when the velocity of the 
power and weight are estimated in the direction in which they 
respectively act, we have 

virt. vel. of P : virt. vel. of W= Pciha- sin A CB : sin ADB 

= AD:AC=: W:P 



IKCUKED PLANE. 



301 



Again, referring to fig. (a), let 
jp. Whe the position of the power 
and weight when in the same ho- 
rizontal line. Let p descend to 
P, and W ascend to w ; therefore 
A> = Ww. Join wP meeting 
n^ in ff ; and draw torn, Pn per- 
pendicular to j^W, Now by smii- 
lar triangles we have 

wg:Pg = vm:Pn=^AC:AD^P: W; 

therefore ff is the centre of gravity of P and w. Hence the 
centre of gravity has moved in the horizontal line Gg ; and this 
is true whatever be the space described. 




fig. 3. 



SCKEW. 

Ex. 14. 

1. The formula is 

P: Tr=dist. between two threads : circum. described by P; 
.•. P : TTss a : i^TT X !Zo = I : aoir 

= I : j^o X 3'i4i6= I : 6^83. 

5. If ^ (7 = circumference of a transverse section of the 
screw cylinder, 




then e?= CB^ACtaJi^o^^ZTrx 9 x-j-=2.7r x 3^3- 
Hence P : 15 x iia = a-n- x 3 V3 2 a-n- x 48 ; 

.-. P= i^iilil^ =105^3- 181-865 lbs. 



302 



FRICTION. 



Fkiction. 




Ex. 16. 

1. Let the diagram be as in 
the margin. 

Besolving the forces P and 
Wm directions parallel and per- 
pendicular to An^ we obtain the 
whole pressure perpendicular to 
the plane 

= TFcosa— Psin^. 

Therefore friction = fi ( TTcos a — P sin /8). 

Therefore for the pressures in directions AB and BA re- 
spectively, we have 

P cos ^— TT sin a — /i, ( TFcos a — P sin^S) = o, 

and Pcos/8— Trsina + /4(Trcosa--P8in)8) =o, . 

hence the limits between which P must lie are 

p= TT ""'!^^"; , and P= TT ^^;-^*^"' 
cosp+/Asmp cosp — /48mp 

3. Let MKL represent the isos- 
celes triangle on the inclined plane; 
O its centre of gravity. 

Pirst suppose that the triangle 
begins to suae. Let B be the re- 
action of the plane perpendicular to 
itself, /t the coefficient of friction, and 
the inclination of the plane when 
sliding commences. 

Resolving the forces parallel to 
the plane, we have 

fiR^ TTsin^ (i), 

and resolving at right angles to the plane, 

^=Trcos^ (2), 

from (i) and {%) we have 

fjL = tan ^. 




FRICTION. 



303 



Next suppose that the triangle rolls over the comer K; then 
the vertical through O will pass through K when A has re- 
ceived the proper value; draw OHkI right angles to we plane; 
then 

a' 3a' 

.-. tan A = tan JS'(?5^= i -=- ^ = -^ . 

Hence if /t be less than ^^ , sliding will take place before 

rolling. If fL> ==, rolling will take place before sliding ; and 

for rolling and sliding to take place simnltaneouslj, we must 

, GH 

have fi = ^g., or 

/.= i^ = tan30^ 

4. Let AB be the rod, G its 
centre of gravity, which (since the rod 
is uniform) will be at its middle point; 
JB the reaction of the plane CJB; R 
that of the plane CA. 

Let the angle BAG ^6. Then the 
forces acting on the rod are, its weight 
(W) acting at <? in the vertical GE^ 
B! 2X A^ R at B^ and the friction. 

Resolving parallel to -8(7, we have 

W-E-fiR^o (i), 

Resolving parallel to (X4, we have 

JS-/t'JS' = o {1). 

Also, taking the moments round G^ we have 

jBsin^+/A5cos^-JK'cos5+/i2'.sin^ = o (3). 

Prom equations (i) and (a) we have 




i? = 



I +A^/Lt 



' > 



iZ' = 



i+fifi 



1 9 



304 



FBICTION. 




and substituting these values in equation (3), we get 

8. Let and 6 represent the inclination 
to the vertical of the string and beam re- 
spectively; and let Z = length of the beam, 
c the length of the string, and /t the coeffi- 
cient of friction. Suppose the beam to be 
on the point of sliding downwards, then it is 
maintained in its position bj the reaction (R) 
of the plane in the direction BR; by the 
friction llB acting in the direction Bu: by 
the weight of the beam acting verti^Uy 
downwards at its middle point, or centre of 
gravity (the beam being uniform) ; and by the t^ision of the 
.string in the direction AG. 

Besolving perpendicularly to the string, we have 

Bco&0 + fiBame=WBiad. 

Taking moments about c, if CB=^x^ 

Bx= Wx " I Bin 6: 
2 ^' 

cos 5 + M sin ^ 2 BinO 2 

• f, . ■ -»» „_ - 22 ^ • 

X I *sin0 c * 

zi . • zi ^ Bin{d>—0) . ^ X ^ • /i\ 

/. cos^+iitsm&= 2- = a . , ^ = 2(cosg— cot A sm^ ; 

'^ c sm ^ ^ T- / > 

/. fl + 2 COt0 as cot ^. 
C . ? 

Now sin ^ = I sin 5 ; /. cot' ^ = cosec' ^ — i =— cosec' tf — i ; 

/. c» cot» = Z» cot' tf + P - c» ; 

.\c»(cot»d-ayxcot^ + /t») = 4c»cot*^ = 4Z*eot»tf + 4?-4c*; 

.'. (4?-4^-AtV)tan»^-a/AC»taii^ + 4Z*-c^ = a. 

If the beam be on the point of sliding downwards, we must 
in this formula write — /4 for /jl. 



FRICTION. 



305 



9. Let CFED be a section 
of the cylinder through its cen- 
tre of gravity O; CD its major 
axis inclined at an angle of 45^ 
to the horizon and equal to %a. 

The forces acting upon the 
cylinder are the reaction {B') of 
tne plane BL acting in the di- 
rection EH] the reaction {B) of 
the plane AL acting in the di- 
rection FK\ the friction ijiB) of 
the horizontal plane acting at F 
in the direction FM\ and the 
weight of the cylinder acting at its centre of gravity G in the 
vertical QM. 




Besolving the forces horizontally, we have 

R-^^o 



(I)- 



Taking the moments round G, we get 

B: X QH-JR X GK+^ X GM=' o 



(i). 



The tangents at E and i^ intersect at right angles in X, a 
point in the minor axis produced. Let 08=Xf SF==y; 

then since Z -4 = 45®, -y- =tan45®=i; /. 2 = «-, 



a^y 



X a' 



•yrr x" y^ 0?* / . aVN aj* / b*\ 



• • X ^— 



a' 



(a» + ft*)* ' 



OK = GN sin 45" = e*a; sin 45** = 



aV 



>J% (a' 4- h*f 



= GH. 



Also GM=MA^-AL = -(AG'+ GL'f 



I f/oV. /**\'l* ifa*/ h*a^\i 10', 



_ (a» + £')* 



39 



306 



FBICTION. 



Now since by (i) B! = fiR\ therefore by substitution &c. in 
(a), we have 






= o; 



►'. fJL = 



e' 



e»+i + i-- 



=!«•. 




11. The forces as indicated in 
the diagram. 

Let /i, /t' be the coefficients of 
friction respectively between the cy- 
linderSy and each cylinder and the 
plane. 

Por the upper cylinder we have 
TT— %B cosa — 2jFsina = o (i). 

The other two equations of this cylinder are identical. For 
one of the lower cylinders 

Tr--B' + ^cosa + i^sina = o (a), 

jF'-jBsina+Fco8a = o (3), 

F-F^o (4). 

From (3) and (4) we have 

F sin a . a 
M I + cos a a 

and from (i) and {%) we get 

Also, from (i), (3), and (4), we get 

T^ TFsina W. a 
jP = — 7 — ; r = — tan - ; 

> a (1 + cos a) a 2 



FBicnoK. 807 

"Now in order that the points of contact may all slip together 
-we mnst have 

Wisa- 






DYNAMICS. 

The Collision ob Impact of Bodies. 

■ 

Ex. 1, 
1. Let* V = common relocily after impact, then 

^^^a+^^96 + 35^I3J^6g ft. per second, 
A+B 19 19 '9 ^ ' 

velocity lost by -4 = a — v = 8— 6^ =1^ ft. per Becond, 

velocity gained by J? = v — J = 6jJ — 5 = i^z ft, per second. 

3. Here -4 = 8, .5 = 5, i = 9; /. 1^ = 3x9. 

XT Aa + Bb 

.*. 27 X 13 = 80 + 5 X 9 ; 

..a- g - g -30-a. 

7. Given ^ = '^, a = ^, e = i; 

3 4 



808 THE COLLISION OB IMPACT OF BODIES. 

B B ^ 

_ Aa'\-Eb^Ba + Bh _ {A--B)a + ^Bh _ i'''^J'^'^ 
•*• ^^ A^-B " A^-B "" 75 

3 

--i il = ^. (I), 

7 35 ^^' 

andt;- ^_^^ ^^i.^ ^- ^^ ...(2), 

3 

.'. from (i) and {%) we have 

u : i; = a9 : 36. 

10. Let X = the ntimher of bodies. 

The velocity communicated to the second body 

%Aa a 



A + ^A 2,^ 

and the velocity oommtmicated to the third body 

_ 2 X 3A a^ a 
"^^A + i'A 5"*!^*' 

Similarly, we find the velocity of the last body = -^^ij ; 
••• ^ = g- , by the hypothesis ; 

.'. a? — 1 = 6; 

/. a? = 7, the No. of balls. 

11. Let the weight of the intennediate ball be a; lb., 
-4 = jjlb., J?«=81b., a = 9ft., € = i, 



THE COLLISION OB IMPACT OF BODIES. 



Telocity commmiicated firom Atox'^ 



.0? to jB=: 



2Aa 

^ / 2Aa \ 



x^-B 



4Aax 



{A + x){x+B)' 



.*. Z 



4Aax 



{A + x){x+JB)' 

.•. (» + «)(« + 8) = 2x^x90?, 
X* - ^6x + 13» « 169 — 15 ; 

.'• a? = 13 + V153 = ^5*369 or '63. 



309 



16. Velocity of^ after impact = a-(l±iL^ii±i) 

^ -4 + jB 



Given j4=^ 
a=7 

a 
*~3 J 






II 

6 



II 



II 



77 - 126 49 



^aw;^ ^^ 7? oft^y impnf t - (^ + ^^ (^ + ^) _ j> ^ 



v = 



± 


6 


— 51 = 


5x 

,2- 


5x 
6^ 




- p'3 






II 
6 




55 


5x 

=2^ 


63 - 3°« 
5 


_ 1575 - 

11 X 


■9«4_ 
15 


II X 






II 


15 


_3I7 

55 


= 3'945- 













310 



18. 



THE OOLUSIOM OK IMPACT OP BODIES. 

A a'-Bb + eA{a + b) 
• A+B ♦ 

Aa-Bb-€B(a + h) 
^ A + B 



therefore the lelatiye velocitj 

€{A+B){a + l) 



= !;-.»=: 



A + B 



= €{a + i). 



Hence the required distance = relative velocitj x time 

= -(^5 + 16) X ^ = 3 X41 = lasft. 

20. Let 0y ff he the angles of incidence and reflexion re* 
spectively. 

Then tan^ = -tan5, and ^ + ^ = 90**; 
/. tan(90^-^ = 3tand; 
/. tan d = -J- = tan 30''; .*. 6 — yf. 

23. Let ty i be the times of describing CE^ ED with the 
velocities t?, t;' respectively. 




^, t CE ED CE v' 

Jhen 7 = — r-^— ^=l5»7^X- 
_ sin (90^ — y) sing _ tang _6 
"" sin (90** — g) sin g* "" tan g' "" I 



THE COLUSION OB UIPACT OP BODIES, 311 

xeB^Hy^ ^. ^ be the angles of incidence and reflexion 
Then tand' = itan^ = V'3tan^, since e= tan 30°,. 
8intf' = |8in5 = Vasin5, t;' = » sin45»; 



'. cos 



^=y|co8d,' 



3 3 ^3' 



/. sin d = - = sin 30° 



sin ^ = ^ = sin 45" 



-; ^ = 3o», d' = 45». 



*^,J^A • ^^ * ^ *^® magnitude of the intermediate body ; 

(1 + e)la^^ ^"^ '^ ** ^^^ communicates to it a velociV 

"" A+x ' * ™P™gi°g on 5 at rest communicates to it a 

velocitjr = ^^M:^, t^t A impinging on 5 at rest 
communicates a velocity = ^^ "^ ^^ ^? j 

. (i + e)* Joa; _ (i + e) Aa 
" (x+A){x + £) A + B ' 

v^f!>ncea^+{A + B)x + AB={A + B)x + e{A+B)x; 

.-. a^-e{A + B)x = -AB; 

whence x = 4±^|e + fe» ^^ ?! . 
and for the expression under the vinculum to be possible, e must 

/. the limits of e are i and ^(^^) , 



812 



THE COLLISION OB IMPACT OF BODIES. 




28. Let A be the point from which the ball is projected, 
fand C the points of first and second reflexion; JiOAB^d, 




Then a^ + a0 + a^ = i8o**; .•. ^+0+^=:9o^ 

Now tan0 = - tantf, and tan^5=-tan0 = -jtan^. 

But tan^-tan{9O«-(0 + ^)} = cot(0 + ^)=^jj^-^^^ 

i-itan»d 
I — tan ^ tan ^ ^ 



tan^ + tan^ itan^+itan^' 

Again, let f,, f^ be the times of describing ^1^, CA\ and 
^1)^8)^3 1^^ velocities of the ball along AB^ BG^ GA respectively. 

TKPT,!!f-?^ Va_sin^, . t?,_sin^, 
t?a Bin a t?3 Sin 9 ' t?3 sin ^ 

, *, _ ^J5 _^ CA _ -4jB t>3 _ sin (^ + '^) sin ^ cos ^ sin tf 
^3 Vx ' Vj "" ^2 * Vj "* sin (5 + ^) sin^ " cos -^ sin ^ 



THE COLUSION OB IMPACT OF BODIES. 



313 



tan 9 cob' 9 . i + tan*^ 

■ ss €* X ^ 

tan '^ cos' '^ 1 4- tan* ^ 



,€ + €* + e»+i 



1 + 



= €* 



l(—t—) 



1 + 



H-€ + €» 



€ + €* + €» + €♦ 



29. Let uda=:a, CB^b; CD =^r the njiixiB, z CAD ^0. 
Since the elasticity of the body is perfect, z ODD « CDA. 




Hence 



AD AC 



a 



DB CB b' 
and ADxDB^AOx CB-h CD'^ah + r^; 



.\ ^i>«=:|(aJ+r») 



Now cos 9 s 



AD'+AC^-CD* 
2ADxAC 



aa*& + (g - &) r* 
2a {ab {db + r*)} 



4' 



*\ 



^ ^ 30. I^ Let the l)ody be imperfectly elastic, and € its elas- 
ticity. 

40 



814 



THE COLLISION OR IMPACT OP BODIES. 




Let BE, EChe the given 
planes, E the given paint 
Detween them. 

Draw -4(7 perpendicular 
to the plane ECTj and produce 
it to O, so that 

JG ; Ca = € : I. 

Draw OD perpendicular 
to the plane BE, and produce 
it to Fy making 

OD : DF=^€ : i. 

Join -^J' cutting the plane 
BE in jB, and BG cutting 
the plane -EG in JV; AF wiU. 
be tne direction of projection 
required, and the path de- 
scnbed by the body will be 
ABNA. 

Draw BZ perpendicular to the plane BE, and NL perpen- 
dicular to the plane EC; then 

tSLnABZ: tanJVBZ= tan GFA : tajiBGF^BO : DF= e : i, 

hence, if the body impinge in the direction AB, it will be 
reflected in the direction BN. 

Again, tsmBNL : tsjiANL = tsjiBGA : isuxNAG^AC : 
CG = € : I, which shews that if the body impinge in the direc- 
tion BN, it will be reflected in the direction NA, and will 
strike the body A. 

21^. If the body be perfectly elastic, 

AC^CG, GD^DF. 

31. In the expressions for u and t?, make h = o, 6 = i, and 
we have 

-4'b vel. after impact = ^ ."" p^ ; jB's vel. = -j — %, ; 

^ A-{-B ' A + B^ 

therefore by question, 



(ttS"^'' = ^^ •'• 4(^-:B) = ^ + ^; 



.'. A : B = 5 : 3 



(I). 



THE COLUSION OB IMPACT. OF BODIES. S15 

Now, if in the expression ( j , p ) x a we put A for J?, 

B for (7, and . . p for a, we shall have 
' A+JB 

J5's vel. after striking (7= ^^^ x . ^ ; 

.\ £'8 momentum = (^^j^y^^^ \;j x «^«; 

^(^-g) . _Aa 8B B-0 ,. 

•*' {A + B)(B+G)^^ 4 ' •• ^ + i?^£+(7""^'*'^^^' 

from (i) and (iz), =-p — ^j, whence 25=4(7, 

i.85 ^"^ 
3 

or J? : (7=a : i.... (3). 

Similarly we find the velocity of G after striking D 

C-D 2BG nAa Aa 
"" G+n^B+C^A + B" 4 ' 

which, combined with equations (i) and (3), gives 

G-n 1 ri n f \ 

therefore from (i), (3), and (4) we have 

A I B I G : D = Jo : 6 : 3 : I. 

If there be n balls, we have, as before, 

A-B J _Aa A-B i A n + i 

■2-^x^a-— ; .-. ^j— g--; •'•5""^^' 

B-G ^f %Aa\ Aa B-G A+B i 



- U p (jAa\ _ Aa B-U _ 

+ G^'^\A + Bj^ir' •'• B+G" 



B+G \A + BJ n' " B+G %nB n-i' 

B n . ., , n — I 

/. 77= — r— ; similarly -t\ = • 

G w — 2 "^ jD w — 3 

Generally, if m^, m^+x be the masses of the r^ and (r + 1)*^ 

balls, we obtain — ^ = . 



316 



THE OOLUSION OB IMPACT OF BODIES. 



83. Let fnx8iiia»m3 = «..s=iii^ be the masses of die n 
spheres, a, Vj, v, ••• v^^ their velocities, lespectivelj. 



Then «,» 



m^a 



«i, + m, 2* 



* frii + ma + m3 + f7»4 4 3 4 



hence ««-.= . 

• n — I 

Now, the time which elapses 



I 



before the %Sl sphere is put in motion ^- , 



3d 



a a 



4fl. -\^^*t 



a a 



nth 



a a a a 



_ n(n — i) i 



a 



36. Let -40 be the direction in which the body A is 
moving; BOj that in which B is 
moving; zAOB^ol. Let AO, OB 
represent the velocities of A and B 
respectively ; /. AO = BO. Re- 
solve velocity AO into two others, 
Apy pO, respectively perpendicular 
and parallel to ffa direction; then 
the resolved velocity Ap is not 
affected by the collision of the 
bodies. 

Take therefore 

ON{^Ap) =asina, 

which will represent A^% velocity after impact in direction Ap. 




THE COLLISION OB IMPACT OF BODIES, 317 

. The Ibodies therefore impinge directly on each other, A with 
veL = — acosa; -B with veLss — a; 

therefore ^*a vel. in direction OB 

— acosa — a + €acosa — €a ar . . / % i ^r% 
= =--{i + 6+(i-6)co8a}«OjB; 

therefore j^'s vel. after impact 

= {02?+ OJV»}*= r£l(i + 6+ (i-c) cosa}*+ a»sin»al*; 

r^i 1 A • X — acosa — a — eaco8a + €a 
V = Jffs vel. after impact = — ' ■■ ; 



a« 



.•. «* = — {i — 6 + (i + e) cos a}*. 
4 

The expression for jd's (vel.)' may be put under the form 

a^-lcos'-H-csm'-V +4a*fsm-.co8-j , 

which is to he a maximum ; therefore 

a Jcos* — + € sm*-^ J 26 sm- . cos — ij sm - . cos -V 

.a a / act . aa\ 
— qsm-.cos- cos sm — =o, 

which, reduced, gives 

tan*-f2 + € — c'lssi + e, ortan'- = : 

.*. a = a cot"' (a — €)*. 

41. Let a be the velocity of A before impact, 

u after impinging on J?, 

v j8 after impact of J^, 

V) impinging on (7. 

rp, A — B %Aa 

Then «=3-^xa, t;^^-^, 

_ B-0 jA {B- (T) a 

^~^T0'^^~ {A-^B){B+C)' 

6 i : I 



318 THB COLLISION OR IMPACT OF BODIES. 

Let jET, £* be the positions of B and C when at rest, L the 
point where A OTertakes B\ HK= c, KL » x. 

e + x 



Then time of -4's moving over HL 



u ' 



5's , =-+-. 



.*. x^c 



T-r C X C X ( u\ fu \ 

Hence -+- = -+ — ; /. a?(i =c — i); 

u u V vo \ wj \o J 

(A-^B \ f {A^B){B+0] 
\ %A V • t %A {B^ C) ] 



{A^B){B-^G ) {A + B)[B^C) 

'^^ {:A-B){B+G)-2A{B^C)^^sAC-B{A + B+0)* 



3AG-B{A + B+0)* 



43. Let A J B, be the bodies; a, h, their respective velo- 
citi^ before impact, and t; their common velocity after impact, 
then 

Aa + Bb = {A + B)v; /. ^(a-v) = £(t?-5) (i). 

Let Z = vel. lost hj A = a — v; 5r = vel. gained by jB = t; — i, 
then a-h=^l+ff; .'. {a + v) — l^{h'+v)+ff {2). 

From equations (1) and (a), we have 

A (a»- v») -^Z(a- v) = J5(v»- J») + Bg (t? - J), 
or ^a» + J?5»-(-4 + 5)v' = u4Z(a-v)+^(v-J) 

^Al^+Bg\ 
which proves the proposition. 

44. u = m's velocity after impact, 
and V = w"s velocity after impact, 

then u^a- ^^ + ^);'(r^^ ^^^ (r-fe)m(a-?) 



THE CX)LLISION 0£ IMPACT OF BODIES. 319 

W + W 

= (J-a) + (i + €)(a-J) (a). 

Now hj the Third Law of Motion 

ma + m'b = mu + w't?, 
and from (ii) a + w = t; + J + (i — e) (a — J) ; 

.•. «ia' - WW* = 7» V - w' J* + (i -€)m'(a-5) (v-i), 

or wV + mw' = ma' + m'b^ - (i - e) w\(a - J) (v - h) (3), 

and from equations (3) and (i) we have 



tnu^ 



-h mV = r/ia" + mb* — ^ ^—- — 7 —• 

m-i-m 



45. Let a, i8 be the angles which the directions of the velo- 
cities of the bodies, before impact, make with the line joining 
their centres at the moment of collision^ then 

^ _ {A - By a'cos^ a -4B{A-B) a eoaa.b cos fi-\- 4B*b' cos' fi 

^ {JTBT 

+ a* sm* a, 
. (A-'Byb^ cos'y94- 44 (A -5)a cos a. Jcos^S + 4ui'a» cos*a 

+ J»sin»/8. 
Multiplying (i) by -4, {2) by -B, and adding, we get 
w ». r>. (^ + 5VUa'cos*a + 5J»co8»^} 

^~+^=^ {A+By ■ — ' 

+ j4a«sin»a + J5J»sin»i8 

which proves the proposition.. 

46. Let a? = J?'s velocity after impact of -4. 

^ Aa + Bb-Bia-b) . 

Then «« — — — . . p^ ^, smce €=1, 

A + B 

_ Aa + Bb + A{a^b) 

^"" ~'aTb ' 

.'. cc — w = ^ — A \, i. = a-6; .'. a + w = a; + J. 

A + B 



320 UKIFORMLT AGCELERATED MOTION 

Also Aa + M^Au + Bx; .-. A{a-u)=^B{x-h), 
^(a«-tt«)=J?(a«-&"); /. Aa^ + Bb* =^ Au* + Bsc'. 

In like maimer we may prove that Bx^ + C(f = Bv'+ Ow*; 
.\ Aa* + Bb^'^C<f = Au* + Bv'+Otc'. 



Unifobmlt Aocelesated Motion and Gkayitt. 

Ex. 2. 
1. Here<=ii8ecA ^ ^ ^^ ^ o i^ 

^ .n.n^V .•.«=-fl^^«l6-IXiai«I948-lft, 

♦ v — fft:=^c^V2 X ii=s354*aft. 

3. Given 5 = 450 ft., gr=3raft.; 

/. V « {%8Xff)^= (900 X 3%'Q,)^=^ (28980)* = 170'2 ft. 

Momentum = 170*2 x 140 = 23828 lbs. = 10 tons 12^ cwt. 

5. Here «= 1320 ft. ; /. ^ = f -^ j == T-^ j , 

or f = 9*05 sec., the time for which the hodj has been fiedling ; 
.*. space described in the last second s - (2^ — i) 

«i(i8-i-.i) = 275-4ft. 

8. Let t OS whole time of descent ; 

.'. the altitude of the tower = -gf^ 

and the space described in [t — i) seconds = -^r (<— 1)«; 

.'. the space described in the last second = -^(2^—1), 

1 I ' 

and, bj the questioUi -^<* = 3 -5'(2^— i); 

2 2 



UNIFOBHLT ACCELERATED MOTION AND QRAYITT, 321 

... «»=6<-3; 

,•. <= 3 + /^6 = 5*449 sec. ; 
.'. the altitude of the tower = -gt* = 477*9 ft, 

10. Let g^ g be the measures of the force of gravity at the 
two places; t the time of a body describing a feet, and v the 
velocity acquired at the place where g is the force of gravity. 



Then t—l — j , and v=(2ay) , 



m 



= ^— + 



T m'T^ 



14. Let jB be the given point ; then 

time up J5(7=time down GB) c 

.•. let %t be the given time, and let -4J5= a, 

then BC^CB = ^; « 

and the velocity of projection = vel. acquired in falling ^ 
down CA; 

41 



- c 



• -p 



322 UNIFOBMLT ACCELERATED MOTION AND aRAYITr. 

time down G4 = — = •} —[ ; 

/. whole time of motion (T) = a -1- (a + - gi^) > , 

15, Let A be the top of the spire, C the top of the a 
steeple, B the bottom of the steeple, P the point where 
the bodies would meet. 

Let ^J5 = a = 190 ft., 08= 5 = 150 ft., J?P= a?, 
then -4(7s=4oft. = a — J, AP=a — x, GP^b — x. 

Now the time of descending through CjP= [— ^ — J ; 

and the time of ascending through BP= time down AB 

— time down j4P= (^) — ( — ;^ — ) : but the time down CP 

V-ffJ \\9 J 

must be equal to the time up BP\ hence 

[ h - x \i f a\i _ f a - x\ i 

or (a -aj)* = a*- (&-»)*, 

whence a (aJ — axy = J ; .\ <ib — ax^ — ^ 

4 

whence x = h = 5(i ) 

4a \ 4a/ 

61 OlijO 

16. Here « = 125, F= 90, 

and from the formula Vt + -yj?* = *, 

we have, by substitution, 

90<-fi6-i^= 125, 

whence <= 1*15 sec. = the time in which the body projected 
downwards will reach the middle point of the line. 



UNIFOBMLT ACGELEBATED MOTION AND GBAYITr. 323 

Also from the formula Vt — f^ = 8, we find 

t' = 2*57 sec. = the time in which the tody which is pro- 
jected upwards will reach the middle point. 

.*. ^' — ^ = 1*42 sec, the time which must elapse before the 
body is projected downwards. 

17. Let a? = height of tower in feet, 

then V (vel. of projection) = Vaj' (I'Saj). 

Hence x—Vt — gt^—2, *^3'6gx — a^, since ^==2 sec, 

05-2 ^3-63^0? + 3-63'= (3-6 - 2)5^ = v6g; 
.-. X = {^3^+ V^ V = {^^ g^^og^ 321-9 ft. 

18. Space described in j? sec. = Vp + - j$>% 

space described in (p — i) sec.-^ V{p — i) + -f{p — i)' ; 

.•. space described in the j?*** sec. = F+ -/(2p — i) *=P. (i), 

and similaxly, 

space described in the j*^ sec. = F+ -/(aj — i) = ^ (2) ; 

"^~ p-i' 

. Y^p {P-Q)i2p-j) ^ Q{2p-l)-P{2q-l) 

2{p-q) 2{p-q) 

20. Let X = the distance, from the upper extremity, at 
which they ^111 meet, 

then at + -g^ = x. (i), 

2 

and ct — gf^h^x (2); 

h 
.*. (a + c) ^ =s A, t^ — -— . 



324 UNIFOBMLT ACCELERATED MOTION AND GBAYITT. 

Substituting in (i) we get 

a + c 7, (a + c) 

_ A 2a* + 2ac + gh 

2 {a + c)' 

22. Let w be the velocity of projection, € = - ; 
then, 
vel. before striking the ceiling = {u* — a^ x la) ; 

/. veLafter = €(w' — 2^ x la) , 

vel. before floor ={e*(u' — a^^ x la) +2ffX 12]*, 

vel. after = €{€*{u* — 2ffx la) + 2ffX la} . 

Hence the vel. on again reaching the ceiling 

= {d (w* — 25^ X 12) + 6* {2ff X 12) — 25^ X 12}* = o ; 

.*. €*u* = (d — e* + i) 25^ X 12 ; 

/. w= (i — 4 + 16)* *^24g = 2 V6 X 13^ = ioo'2i6ft. 

27. Let t?„ t^a, i?3 be the velocities of the ball just 

before striking the plane in the ist, 2(1, 3d descents; 

«i, «a> «3 the heights through which the ball succes- 
sively falls, e the elasticity = 2 = -g, 

5 

Then v, = {23 x 50)* ; s, = 50, 

Va" €* (2flr X ^o) 

' 2g 2g ^ ' 

&c. &c. &c. 

.•. whole space described 

= «! + 2^3 + ^*3+ to infinity 

== ioo(i + €* + €^ + ) -50 

"= 50 (^ - i) = 50 (tt?) = 50 X i^ = 50 X 2I = 106^ ft. 



UNIPOBMLY ACCELERATED MOTION AND GEAVITT. 826 

29. Let D and G be the 
points from which the balls are 
dropped; AD — a^ AC = b^ and 
the distance AB on the horizon- 
tal plane over which the bodies 
have moved before collision takes 
place = X. 

Since the velocity of each ball 
after impact equals its velocity 
before impact; 




- (f ) 



2,a\i X 



(««y)- 



== whole time = f L j ^ 



X 



{Zbgy 



1(35)* (3a)*J ^ ' ^ ^ ' 



• • X ^^ 






30. Let be the point where the bodies m, m' meet 

PQ = 2a, PO==x; 

•'. 0Q = 2a'-Xy 

h = height due to the velocity of m' on beinsr re- 
flected Sit Q. ^ 

Then velocity of m acquired down PO = Va^^a?, 

m' on reaching 0=^/2g {h — a,a + x). 

Now the velocities of t», m' after impact must be 
equal to their velocities before impact, and have op- 
posite signs; 

_ m^2ffX'-m' \f2g{k -za + x) -m' {^f^gx + Va^ (A- aa + a;)} 

.'. aw ^2gx = am' Va^' (A — aa + a;) ; 
.-. a? (w'* - w») = m'» (aa - A) ; 



.-. PO or a? = 



m' 



m'* — m' 



(aa — A). 



326 UNIFORMLY ACCELERATED MOTION AND GRAVITT. 

Obs. To satisfy the conditions of the problem, the mo- 
menta of m and m before or after impact must be equal to one 
another; hence, 

m '^%gx = m! '^^g (A — aa + a?), 
ftom which x is found as above. 



Ex, 8. 

1. Let 6 = the inclination of the plane, 

then «= -g^ sin^, 

% 

I • a 
or 9 = -5^ sm^; 

.-. Bmd = y = sm33»59'56"'4. 

5. The accelerating force down AC=g, 

AB—g sm.0. 




2AC\i 
9 )' 



Hence time down Av=\- 

/ 2AB \i_ (iAC\i, 
. sin' 0=-; .'. = sin- Q) = I9» a8' 1 6", 



ACmn0 _i 
AB 9 



UNIFOSMLT ACCELERATED HOTION AJSJ) QSATITT. 827 

6. In the figure to (5) let AP= x, PQ=>AC='h; 

AB=l-4ofL 

Then time down AQ^ \ ■ a \ 1 where d = 30", 

_f 20! \i 
" V.<7 sin e) ' 



AP 

^9 



AG 



/2h\i 



, A' T fX\i 

/. «+A = a?+ j + aAf-yj ; 
4JL 4 X 40 8 "> 



9. Let h be the height and I the length of the inclined 

plane; 

X be the space through which W has moved, when 
P ceases to act; 

t be the time during which W has moved, when P 
ceases to act; 

V be the velocity acquired hj TFwhen P ceases to 
act; 

/ be the accelerating force on the system. 

Then f ^llll.i^ZlMZ 
Then/ ^+ U^ ^ (p+ w)l ' 

Since the velocity v is just sufficient to take W to the top of 
the plane, it is equal to the velocity acquired by a body moving 
through the space l-^x down the plane, the accelerating force 

being /'=5rx-^. 



328 nHiFoniLT accslekated motion ahd eBAYirr, 

Hence vfie = v^ = s^f {I — x) ; 



"'f+f P \h + l)\ 

\f)~ P \{Pl-Wh){h + l)g] 



of the plane 
whose length 



11, Let a; = space described by the body 

projected from the bottom 

then ? — a? = space described by the body (i^. '> *^<1 ^" 

projected from the top j clination is a. 

t =s time of motion from the bottom, 

thenn + ^== top, 

/= accelerating force along the plane =^ sin a. 

Hence x = ct~j^ ...^ (i) 

Z -a? = a(w+ + ^/(^+0"; 

I — an — fn* 
whence t = ^ , which substituted in (i) gives a?, 

13. Let I be the length of the plane, and its inclination, 
t the time of a body's falling down the plane. 

Then t = f-^y= f .y .? = f -4%)*. 
\5r sm ^/ \y sm cos ^/ V^ sm 20/ 

Hence t is least, when sin 20 is greatest; 

/. sina^= i = sin90*'; 
.'. = 45**, and therefore the height = a. 



UNIFORMLT ACOELEBATED MOTION AND GRAYITT. 329 

15. Let h^kylhe th€ height, base, and hjpotheniise respec- 
tively. 

Then, the time dowp (A) =(— ) , and the vel. acquired = (a^A)* ; 
/ along (A;)= j, 

down «-/4Y=©^ 



vi>""'-^wh' ••■'=*+!*! 



• • 



/. A»4-AAj + -A;» = Z" = A' + *'; 

4 

17. Let X = side of the second triangle, 

/ the accelerating force = ^^ sin 6o^ the triangles being 
equilateral. 

The velocily acquired down (a) = Va/S, 

along the horizontal plane = V2/& x cos[6o**, 

after impact at the base of the second triangle 

= \^2J^ X cos' 60**, 

which being just suflScient to take the body 

to the vertex of the second A =: velocity 
acquired down its side (a?) = VaJ^ ; 

.'. x = a cos* 60** = -5: . 

16 

Again, the time down (a) = [-^ J , 

along {d) =-p= , 

^V2/axcos6o** 



np (-) = (7)*=(^)* 



42 



8d0 UNIFOBMLT ACCELERATED MOTION AND GRAVITT. 

iherefoie whole time 



19. Let ABP be a circle, AB 
a yertical diameter, FA anj chord 
drawn through -4, then velocity 
acquired in falling through AB is 

(2J7X^-B)* (I), 

vel. acquiied down AP 

«ad.AJP*=AM.AB; 
AP* 



.: AM= 



• 



A£' 
therefore vel. acquired down AP 




_ f ig . AP^ i 
AB J 



-( 



= AP 



2g\i 



m 



(3); 



therefore hj the question AP x (-^j — ^ ; 



.-. AP= -AB, 

' ^.^ AP AP I ^ , 

cos5^P=2g=^3p=-=cos6o«; 

.'. zBAP=6o'>. 



21. Let be the centre of the circle, GA and CB -the 
chords, such that z COA = 0, OOB = 26, then 



time down A 






I:. 



tnsriPOBMLY ACCELERATED MOTION AND GRAVITY. 331 



I ^ N* 



time down BC 



-r.n AX r 4-4 (7 COS -5 



{_2:BG_\ 
"t^sinOC^j 



^ sin 

I N* 

4^0 cos -5 J 



^ cos 



and by the qoaatioiij ^ ^^ 



zAC M 
ycoB- d 



+^^*^5^ . a^^ 



= «». 



fl' COS ^ I >, 

^ g COS -0 

/. 2 COS*- ^ = n" COS ^: . 

/. cos^ = n'cos(?- i; 
/. cos^(n*-i) = i; 
whence sec ^ = n' — i. 



23. The axis of the parabola, and the tangent at its vertex 
being taken as axes of x and y, and (A, k) being the co-ordi- 
nates of P, we have, I being latus rectum, 

Let the vertical through P meet the horizontal line at a dis- 
tajice I below A and M, then 

time down PM= f^Mlj* 

time down PA = |^|* = p^^t }* 



= |^J*±i)J*= (time down FM). 



332 UNIFOBHLT ACCELERATED MOTION AND QRAYITT. 

25. Let be the inclination of the required diameters to 
the axis of x^ and n seconds the given time, then the length of 

the diameters = 



/. n=- 



(e» cos" e-i)^' 
4b H 



i65^ 



(e* cos' 0-i)i , . 4_ 

Jsm? / ' " ^ "5r»(l-C0S»^) (6»C0S'tf-l) ' 

/. n^g* {e* + 1) cos" - n*^V cos^ 5 - n^g* = i6J* ; 

.-. COS^tf -^COS"e = -^ J-TJ-y 

which quadratic solved gives for the position of the diameters 

and when the time is a minimum, 

t- /,/ . TTj Tlr is a minimum : 
sm (e" cos" — i)^J * 

•'. sin (6" cos" ^ — i)* is a maximum ; 
.'. e"cosd. 



sm^ ^ (e"cos"^-i)* 

i-cos^.^^ ;— 3 — ^ = o: 

(e»cos"^-i)* smtf 

/• 6"sin"^-6"cos"5 + i = o, or ae'cos"^ = 6" + i; 

^e" + 1\* 



/. cos^ 



'<-^)- 



26. Let AB be the axis of the cycloid, AL the vertical 
through A meeting the horizontal plane in i, then 

^X = ^5sin6o^ = a^. 

Let -4Pbe either of the chords drawn from the vertex to one 
extremity of the base, then, by the property of the cycloid, 

AP = {AB' + PB')i = (o» + ^y = I (,r» + 4)* ; 
therefore time down AP= \ Tr\ = ■!— ^ — r-^c • 



TTNIFOBHLT ACCELERATED MOTION AKD aSATITT. 333 

27. Let ABss 2a, be bisected in 0, 
ON^ X, NP=y. 

Then time down AP= (^)*= g^^)* 

_/ay fa'+(a-g)' |t 



PB 



aPB'Xi 



~\ffNB) 




= M* fa' + (« + «)' ]§ 

If c be the constant, then 

y'+(o-a;)' . y'+ (g + a;)' _j9>.. . . 
a — a? a + aj ^ ' 

the equation, of which the locus is an ellipse or hyperbola, ac- 

29. The accelerating force / during the fall of 12 feet 

Q,g 2 
2 + 9 11^ 

Vj, vel. acquired = (2^)*= J2 (—ff) x ^^\ = 11*852 ft. 

tjy the time = f-^j = (-^ j = 2*025 sec. 

Let Vi , Va he the velocities acquired when the 2 lb. has fallen 
through 15 ft. and 20 ft. respectively; 

t the time of describing the first 15 feet, 
t' last 5 feet. 



834 UNIFOBHLT ACCELEBATED MOTION AKD aBAYITY. 

Then v^ = Ja (^) x i^^= 13-25 ft. 

»,» = ».» + a<7X 5 = 2/7 (fY+ 5) = ^/7; 



t = /'^ 
(^ 



.'. V2 = 2iZ'304 ft.. 



— \ = ( — ^J =ria64sec. 



Now Wa = V, + ^«' ; /. t' = -^ ' = '281 sec. 

if 

.'. «a =« + «' = r 545 sec. 

31. Let 9 + 0? and 9 — alb. be the weights of the extremi- 
ties of the chord. 

^, J. (9 + aj) - {9 - ar) x 
Then f—j^ — ; . ;^ sSl = -sr; 

•••3x13 = ^ (^S'j X I3», since a = - /i. ; 

6x0 

Hence the weights are 9*12904 and 8*87096 lb. 

33. Let /be the accelerating force on the bodies P and Qy 
and/' P— 2>and Q. 

V, the velocity acquired by P in descending a feet, 
therefore the space through which P — p will descend = -^, 

_ (p-<?)(<?+p-j>) ,. 

~(i'+<2)(<2-i*+i') ' 

after which P— p will ascend under the action of/'. 



UNIFOEMLT ACCELERATED MOTION AND GRAVITY. 835 

35. Let be the centre of the given circle -4CB, AB a 
vertical diameter, Pthe given point. JDraw PS' parallel to AB^ 
join P4, and produce AP to meet the circle in C\ join OG 
cutting PK in Q ; PC is the straight line required. 

The triangles CPQ, CA 0, will 
be similar, and since CO = 0-4, 
.•. CQ=^ QP; therefore a circle 
PCK described with centre Q, 
and radius QC, will pass through 
P, and since OQ joins the centres 
of the circles it will pass through 
the point of contact ; therefore, as 
the circle PCK touches the circle 
A CB in the point C, it can touch 
it in no other point. 

Draw PREj any other line, 
from P to the circumference ABC; 

then the time down PU > time down P8, 

> PC, 

hence PC is the line of quickest descent. 




37. Let 8 be the focus, SP the radius vector, and its in^ 
clination to the axis major, then bj the polar equation to the 

therefore time down 8P= \ ^ — ^[ = a minimum ; 

(y cos ^ (i — 6 cos u)) 

.*. cos d (i — 6 cos d) is a maximum. 

DiflTerentiating, — sin ^ (i — e cos^) + e sin ^ . cos 5 = o ; 

.*. ae cos 5 — I = o ; 



.*. cos ^ = — , or 5 = cos' 



■{£)■ 



} 



886 MonoH UPON a cubyEi akd the simple pendulum. 



MonON UPON A CuSYEi AND THE SiMPLE PENDULUM. 



Ex.4. 

1. Let a, hf ehe t he le ngths of the planes ; then velocity 
aoqniied thiongh a «= Vi^a, the resolyed portion of which in 

the direction of ^is s^-ga\ 

.*. vel. acquired through J = ]^^a + V3 gb\ j 
the resolved portion of which in the direction of C7 is 



^/h^^: 



.*. vel. through G 



.% velocity required = - 1^ (9a + 6 V3 i + 8c) [ . 



2. Let the chords -4 (7= a, GB=J, and CD^d; 




MOTION UPON A CUBVB, AND THE SIMPLE PENDULUM. 837 

vel. acqiiired down -4 C7= '^ig.MG^ V *^ ^ J > 
BC^^2g.NG^f^2g x ^. 



The "ball {m) which descends down the arc -40 impinges 
directly on an equal ball at rest at G; hence, after impact, the 
velocity of the latter is equal to that which it would acquire in 
falling down the arc BG: now the 

m{i+e)aA/^ ^ r- 

velocity after impact = = " (^ + ^) \/"T ' 



m'\-m %^ Vc? 




4. Let I be the length of the pendulum which vibrates 
4 times in a second ; 

L be the length of the pendulum which vibrates 
once in a second. 



and I ='w a /— /. Z = — X iy=ij'446jziii. 



i6 



5. Let g^ d measure the force of gravity at the earth's 
surface and at the distance of two of the earth's radii above its 
surface respectively ; and t^ i the times of a vibration at these 
two stations. 

Then t^ir f. — , <' = 7r./-r: also i=7r./— : 
N 9 y 9 . N 9 



•'• ** = 3 (3) = 4*289 sec. 



43 



838 MOTION UPON A CURVE, AND THE SIMPLE PENDULUM. 

7. Let t = time in which the particle falls through 8 1 feet, 
then 

^ /2 X 8i\i /i62\i , 

. = (-^j=:(— j seconds; 

/. the pendulum makes i vibration in - ( — ) sec. 
Let I be the length of the pendulum, then 

whence I = — x :^ = -i = I'SaayS feet. 

gg TT* ir^ ^' 



9. Let Z = i+ i'05 = 40'i893in., - 

X = number of seconds lost in 12 hours. 

Then t 




Now (43200 — a?) f = 43200 X I sec. 
.-. 4320o-a? = 4320of-yJ =4263i'92; 
.-. a? = 568*o8. 

11. Let X be the required height in miles; 

i be the time of a vibration at the earth^s sur&ce; 

g^ g be the &rce of gravity at the lower and upper 
stations respectively. 

Then t^tr . - 1 , . , 



y 9 



\gj 4000+ a? 



MOTION UPON A CURVE, Ai^D THE SIMPLE PENDULUM. 839 

But (6ox6o + 3)< = 6ox 6ox I sec. ; 

4000 t 3600 

/. a: = = cti miles. 

1200 ^3 



13. Let X be the required height in miles ; 

t be the time of a vibration at the upper station; 

ffy ff' be the force of gravity at the lower and upper 
stations respectively. 

V s 

^ffj 4000 



Then i 



t 



="\/fi 



\g J 4000 



Now (60 X 60 X 24 — 48'^ ^ = 60 X 60 X 24sec. ; 

. X , 86400 48*6 
Aooo 86q^i'4. 86^';i'4. 



t • X 5-^ 



4000 86351*4 
4000 X 48*6 _ J . 



8635 



-^ — = 2- miles, nearly. 



15. Let ff, g' be the force of gravity at the equator and 

pole respectively; 

I, ^ be the times of a vibration at the equator and 
pole respectively. 



Then i = 7r 




' -'-^^w)- 



Now (86400 + 5 X 60) < = 86400 sec. ; :. t- 



288 
289' 



.-. g \ g' ^t^ \ I = 288' : 288^ + 2x288 + 1 

I 



144 : 144+ I + 



2x288 



= 144 : 145 



576* 



340 MOTION UPON A CURVE, AND THE SIMPLE PENDUIiUM. 

18. Let g^ g* be the force of gravity at Greenwich and the 
other place respectively. 

Then * = '^y^ «^d i^-^y-; ••• ^=(|-')*- 
But (86400 — n) < = 86400 X I sec. ; 

g V86400 — n/ \ 86400/ 



'=' + 8^°'"^^- 



19. At B, let t = ir fj-, and .'. i = ir*/ 

••'"(ITT)' ••'-^-^' 

Now (86400 — 6ow) t = 86400 X 1 sec. ; 

I / n y n , 

.'. I — 7— = I =1 nearly; 

Im \ 1440/ 720 •'' 

.'. I = 2 — nearly. 






22. Let X be the depth of the mine in miles; 

g, g be the force of gravity at the surface of the 
earth and in the mine. 

Then i=-n-^-, and « = 7r^^; 

,.,= (£)^f_4000_y 
\g} \40oo-a?/ 

since in the interior of the earth, gravity varies directly as the 
distance from the earth's centre. 

Now (86400 — 10) « = 86400 X I sec, 



X 

I — 



. • X •— " 



4000 t^ \ 8640) ' 

8000 , - 
g^nearly=: '93 nearly. 



MOnON UPON A CUBVE, AND THE SIMPLE PENDULUM. 841 

24. Let h = the height of the mountain, a the force of 
gravity there, and %m the number of seconds lost in a given 
time, then 



^^^w^r 



Also 91, n' being the number of oscillations in the same time 
at the surface and up the mountain, we have 

n — fi 2m h h , 

/. = — = — —i=- nearly; 

n • n r + h r •' 



.'. h = 



7,mr 
n 



Let G be the force of gravity in the mine, d its depth, and 
n" the number of oscillations there in the given time, then 

G=ff.^ and j=y|=y-^; 

n!' n — m m Ir — d d , 

.-. — = = 1 = a/ =1-— nearly; 



J 2^^ T 

n 



26. Let V be the point from which the particle starts ; 
VH horizontal meeting the axis in H. On AH describe a circle. 




and let the ordinates PN*, QN' of two contiguous points meet 
this circle in p, q; join flp, Hq^ and Ap cutting S[q in J. 



342 MOTION UPON A CURVE, AND THE SIMPLE PENDULUM. 

Now arc AP^ 7.^1 AB. AN 



-^j^% 



AH 




AH' 



I AB 

Similarlj, wca AQ = zAq a^ -j^', 

.: PQ^2{Ap-Aq)J^- 

Again, since the particle starts from rest at F, the velocity 
at P = vel. acqiiired in falling freely through the vertical 
height HN . 

and flince PQ is very small, the velocity of the particle whilst 
describing PQ will be very nearly uniform and equal to its 
velocity at P, and the smaller PQ is taken, the more nearly will 
this supposition be true ; also on the same supposition Ap — Aq 
may be ultimately taken equal to hp. 

Hence time of describing PQ = — -, — -~ ultimately, 

vei« aw x^ 

Taking, therefore, the sum of successive small intervals start- 
ing from F, we get the time of describing F-4, and the sum of 

the corresponding small angles is = z VHA = - ; /.if AB = 2a, 
the time down any arc VA = — ( — | = tt (-] • 



PROJECTILES IN A NON-BESISTINa MEDIUM. 343 



Projectiles in a Non-resisting Medium. 

Ex. 6. 

F' . 60" 

1. Range = — . sin 2a = — sin 30° 

5- = 55*916 ft. 

H=^ — Sin' a = — siu* \^ - ^^ {% - a/^) 

120-57714 ^ - 

= Q = 37457 ft. 

32'i9o84 ^ '^^' 

jr= — sin a = sm i ^** 

= - — —o- = -0648 sec. 
32*19084 ^ ^ 



3. R^%h sin aa = 1000 \ 

-. %V \i required a and F. 

T = — sm a =15 sec. 
9 '^ J 

Now 2,V^ sin'a= '^^ , 

2 ' 

F* sin 2a = looqy, since F* = 2yA ; 
... tan a = ^g = tan 74** 33' 47", 

F= --i^ = 250-46 ft. 
2 sm a "^ 

A 1 TT 7 • * 1000 sin" a ^ . - - /v 

Also, -ff = A sm* a = = = 250 tan a = 005-35 ft. 

2 sin 2a .^ ^ ^^ 



844 



PBOJECTILES IN A NON-KESISTING MEDIUM. 



7. Let APB be the path of the projectile. 




-4-^= f mile = ^640 ft. 
^P= 100 ft., AB=36ooft. 



0^ 



Now since ^ = a; tan a j z- is the equation to APB; 

^ 4Acos*a ^ ' 

2640* ' 



.'. icx> =» 2640 tan a — 



4A cos' a 



and o = 3600 tan a — | r- 

^ 4Acos'a 



required a and F= V5^, 



tan a 



j\h cos' a 3600 

^ =tana[i ;—-) =-5_tana; 

^640 \ 3000/ 300 



100 



.-. tana = -^x^ = ^ = tan8^5'4", 
264 8 176 "^ ^ ' 



76 



P = gX2A= 36ooy ^36oog 
^ 1^ tan a cos' a sin 2&a 

/. F= 645*1 ft. 



9. Since a = 45, we have 






T= 



F' 
-« = — , 



and the time in which the sound of the explosion was heard, 
after the ball had completed the range, is 

Y* F* 

F' Va/2, 
.-. by question, ^^ + -^ = 3 J, 



PBOJEGTILES IN A K0N-BESI8TINa MBDIUM. 

and, completing the square of this qnadratlci 



345 



(j^^)"- 






. ^ ■ 35 Va _ 54'ggi7<>7 

whence 1= 54:H£76lz422i22M = ,.36.145 ; 

V* 
.'. B='—= (a-36ai45)» x^r => 179-616 ft. 

11. Let /9 be the foctiB, and let the axis ES produced meet 
the directrix in K. 




Then JEff=A 



^ 



JE4 = jr=ilsin*a; 

Hence, latus rectum = 4A cob" a = ^ ^ ■ x f - j 

« ii!;(2&2&ft. nearlj. 



44 



346 PBOJBCnLES ni a KOH-BEglSITHa MEDIUH. 

and Ja8ory, = A8in*a — AcoB*a=sAf2 — j = 56ii*aft. 

14. Let H^h sin' a, and ff = h sin' ria ; then since 
the area of a parabola equals two-thirds of the circumscribing 
parallelogram if ^, ^' be the areas of the parabolas correspond- 
mg to a, %a respectiyelj, 

A : A'^-jRH: -BH'^an^a : sin'2a=i : 4cos'a. 
3 3 ^ 

17, Since — = ^; /. ^"^ ^ =^; 

In 4 Asm^a 4 

18. Let ^^sthe height of the line; then 

a* a» 

«=:atana i t- =a 7 , 

4A cos' a 2h 

, , tx (a + hy I I 

^ ' aA ' izA a + ft' 

a' a5 



a4" J a + J* 

21. Let a' express the direction of projection of the second 

body; 

f , t^ be the times of describing the oblique range on 
the plane whose inclination to the horizon is /3; 
then 

g cos/9 ' g cosp 

t _ sin (g — )S) 
''•?'"sin(a'-/3)' 



PBOJECTILES IN A KON-B£ISISTINa MEDIUM. 847 

Now since the bodies strike the same point in the inclined 
plane, 

, sin(a — )8) cosa_ ^ sin (a' — fi) cos a' 

.% asin(a — /8)cosa=asin(a'-)8) coso', 
sin(2a-/9) — sin/8 = sin (m*"— /8) -^sin/S; 
/. sin (2a - iS) = sin (2a'-/9) =:sin {ir- (2a'-)8)} ; 

t _ sin (g —fi) 
< cos a 

25. Let the top of the tower be taken for the origin of co- 
ordinateSy 

a; = a = the distance of the part struck from the foot of the 
tower ; 

— y = the height of the tower ; 

then — y = ajtana — gti^, 

I . X « 

or y = - gn^ — a tan 30** 

27. Take the top of the tower for the origin of co-ordinates, 
and let a?, — y be the co-ordinates of the point on the ground 
against which the body strikes, then 

-y = a? tana - |p (i + tan'a) ; 

/. by the question, — ioo = a? ^3 ~ -^ > 

whence a?* — ^^ -^x^ ^^—^ 

'^9 ^9 



348 FioraonLCB n a vov-BKunmiQ medium. 

which equation solved gives a; — — iZi_35= 128 feet; 

also, if ^ — the time of flight, we have 
x^Vtooea, <» 25^ — 128; 

, ia8 
.% < = -— «5-iasec, 

30. Givw a = 75*, « =45*, ^ = 250, 

, n, 2V* 8iii(a — Ooosa 

and S^ . — ^ T^. ; 

g co8*f 

_aF* sin 30*^. cos 75° 

..-. 250-—. ^.^^ ; 

. ^_ / ia5xco8'45» \f_/ 5o<y \ * 
-\^Bm3o».cos75'/ W6— V»/ 



(i^)--H-«»5. 



^^2F sin 50^ ^ 124-685 X V^ ^ rurrSr: 

^r •cos45» flr ^^^^ ^ 



sec* 



81. Let the tower be denoted by -40= 60 ft., the width of 
the river by 0^=300 ft. The equation to the path of the 
projectile AP is 




jfsa^tana — 



•u' flu* 

-Y 7— =a?V3 — T-, since a= 60® (i). 

44cos'a ^^ A ' ^ ^ 



The equation to JSP, the slope of the hill rising from the 
opposite bank A being the origin, is 

y + 60 = (a? — 300) tan 30** = (a? — 300) -j- {%) ; 

V3 



?B0J£(mLE8 IN A NON-BSSIfiTINQ MEDIUM. 849 

To detennine P, the intersection of AP and BP^ whose 
co-ordinates are AM^x^^ MPj=y^^ suppose; subtract (i) 
from (a), 

then 60 = a?x (-75 - Vsj + X " ^^^"^3' ^'^^'^ * =" 500ft.; 

Jet h* h* 

.-. X* - a*aj, ^ + -T =* (^^ V3 + 60) A + — = 199935*8 ; 

3 3 3 

/. a:, = Y * ± 447*15 = 735'8. 

Also jNP= (y, + 60) = (x^ - 300) 2^ 

= 4358 X '5773 = a5i'587- 
Hence BP = .NPcosec 30® = 11 x a5i*587 = 503*2 ft. nearly. 

33. Take the top of the plane A G for the origin of co- 
ordinates, and let a?, — y, be the co-ordinates of J9, and V the 
velocity with which the body leaves (7; then 

-y = aj tana - 1^ (i + tan»a) (i), 

^G 9 ^ * 

or — -T- =aa?— 'yi^i ^1^^ « = 45^* 

Let V be the velocity with which the body is projected up 
the plane -4(7, then by the question, 

v^^2gy.AGj 

but F' = t;*-a5rsin45®x-4(7; 

.vF^ = a<7^C7(i-^) (a). 

Substituting in (i), we have 

a^-Q.AcU - -^) aj = ^(7» (V2 - 1), 

which equation solved gives 

aj = ^(7; 

.-. AD^AG^'AGy.-^-viojiy^AG. 



850 PBOJECnLES IN A NON-fiESlfiTINa MEDIUM. 

If < be the time up AG^ then 

.•. ^= — 5 — :;5=( — ) {» - (4 - a Viir L 




and T. 



X AC X 1^2 



the time of flight of the projectile ; 
therefore the whole time 

= {AC)^ X '487 seconds. 

36. Take the top of the tower for the origin of co-ordi- 
nates, and let x, — y , be the co-ordinates of the point struck, 
(y being also the height of the tower), any angle of projection 
such that the body may strike the ground at the point in ques- 
tion, and Fthe velocity of projection ; then 

-v = ajtand 7 r^ = a?tand r (i + tan*^; 

^ 4Acos'd 4^^ '' 

whence tan'd-4* tand+ i -^ = (i). 

Let a, )8 be the directions in which the bodies are projected 
to strike the same point ; then tan a and tan/3 are the two roots 
of (i) ; therefore, by the Theory of Equations, we have 

tana + tan)8 = — , 

X ' 
and tan a tan /3 = I - ^=: i -X (tan a + tan/9)* ; 



PBOJEOTILES IN A NQN-BESISTING MEDIUM. 351 

— A ^ "" ^^^ g ^^^"^ /^ _ 3, COS tt COS /8 COB (tt + j8) 

••• y-4^- (tana + taniS)^""'^ sin'fa + fl)- 



as4A. 



(tana + tan)8)* ^ 8in»(a + i8)^ 

costt C08j8 cot(tt + j8) 
8in(a + i8) 



37. Take the point from which the bodies start for the 
origin of co-ordinates, the direction of ^'s motion being the axis 
of jfy and let a;, y be the co-ordinates of A^ x^y y^ those of B at 
a given time t\ then we have 

aja = t?< cos 30*, y^^vtBmyf — g^. 

Now by the natore of the centre of gravity, if 5, y be its 
co-ordinates at the time tj we have 



- Ax^ + Bx 

X — 3 



+ Bx, _ 'jjBvt ... 
+ 5 ~%{A+B) ^^'' 

A(ui--ge)+Bt-vt--gA 

^ — 1Tb i+5 

Aia + -Bvt-{A+S).-fffi 

A + S ^— •(«)• - 

From (i) we have 

._ 2{A + S) - 

which substituted in (2) gives for the equation to the path of 
the centre of gravity 

39. Let a be the required angle of projection, then to find 
the value of a, we have {t being the time m which each body 
has moved before collision) 

y = a?tana- ^a^ , (i), 

^ av'cos'a ^ ' 



352 ntononLBS a a HOH-RBnBtura medidii. 

t ^ {2), 

veoBa 

y^irf-I^ (3); 

from (i) ttnd (3) we get 

flctana-— ^ — r- = «*"T^ (4)i 



and from {%) «nd (4) we have 



tvsina — ^ = iU — g^; 



whence 8ina = -. 

V 



40. Let a;, jf be the co-ordinates of any one of the bodies 
at the end of a given time t^ then we have, 6 being anj angle of 
projection, 

a?= Vtco&e (i), 

y= Vtmi6 — gf {%). 

Now to find the locos of the bodies at the end of the given 
time t^ we must eliminate the variable qnantity 6 from (i) and 
{%) ; to do this, add the squares of (i) and (9), 

aj« + A^ + 1 ^^y = FV (cos' + sin» 0) = P^, 
for the required locus, which is a circle. 

42. Let fi be the inclination of any one of the planes, as 
XPto the horizon, then by formula 5 we have 

jy , sin (g - ff) cos a 

and for this to be a maximum we must have 

% cosa sin (a — /3) a maximum, 
or sin (iza — )8) — sin /8 a maximum, 
and this will be the case when sin (2a — )8) » i, 

or aa — /8 = 9o% 





PBOJECriLEB IK A NON-RESISTING MEDIUM. 



353 



and then we have 



i? = 



2* 



2h 



I + sin /9 i + costf ' 




if be the angle of projection with respect to the axis of y. 
Now this equation is the equation to a parabola whose para- 
meter = 4A, therefore the locus of P, P', P", &c. is a parabola 
whose focus [3 the point of projection. 

43. Let AB be the plane, 
I its length, a its inclination; 
and let 1= z GBF, BO being 
the direction in which the body 
is reflected at P; E the elas- 
ticity. 

brawPP vertical. Then the 
velocity acquired in falling from 

^ to P = Va^fZ sin a^, and this 
being the velocity with which the body impinges against the 
plane BF at the angle of incidence DBA = 90** — a, and the 
angle of reflexion being 90® — ^, by hypothesis, we have 

vel. of r^exion = . , ^ — ^ . VagZ sin a = j^ . sfaql sin a ; 

sm(90**-^) ^ cos^ ^ * 

and this is the velocity with which the body is projected from 
P in the direction PC/ ;. 

- • cos' a 2ql sin a . ^ 

.'. the ranffe — — r-^ .-^ • sm %d 

° cos' g 

= 4Z cos' a sin a tan 0. 



But tan (90® ~ ^) = - tan (90°— a), or tan ^ = e tan a ; 
/. the range = 4?e cos' a sin a tan a = f ^ j eZ, since cos a = y- . 

45. Let AG he, the plane, DE vertical, and suppose the 
body to fall down DE; 4:aw GE perpendicular to the plane, 
and make the angle (r-EJF equal to the angle DEG\ EF will 
be the direction in which, since it is perfectly elastic, the ball 
will be reflected after impinging on the plane at E] and the 
velocity in direction EF== vel. acquired m falling down 



h = *J2gh. 



45 



354 



PBOJEGTILES IK A HOy-BE8I8TING MEDIUM. 




Suppose the ball after tiie rebotuid at ^ to strike the plane 
at H. Take E as the^ origin of co-ordinates, and let rt?, — y be 
the co-ordinates of projection, then since 

angle of elevation = FEX^ 30% F= V^A, 
and y = a? tan 30® 5* -T- , 



we have — t- = 



9^ 



^3 V3 4^;i^2 



, whence a = a Vs^ ; 



.-. EH'^ Va5» +y» = ViaA* + 4A* = 4A, 



and time = 






46. Let XAB=0, XAP^i. 




Since the elasticity is perfect, and the hodj is reflected by 
the plane AP in the direction PT, which is vertical ; 



w 



/. z JKPr= ai, and z JBJrX= - + 21. 

a, 



PROJECTILES IN A N0N-BESI8TIN0 MEDIUM. 355 

Now y^xtmO — 7 ra (i); 

/. tan5JJX=^=taii<?--r-5— ,. 

dx %h COB* 9 

Also y-retan^ (2). 

Subtracting (ij) from (i) we get 

o=s (tan d — tan ^) a? — 



4A cos' d ' 
2; 



/. o = a(tan^ — tanO — 7 tti; 

^ ' ih cos' ^ ' 

/• tan£JIZ'»tand-a(tand — tani)ss — tand + atan^. 
But taniZS2r=tanr- + ii^j= — cotij^; 

/. tan ^ = a tan ^ + cot at =- (3tan* + cot4). 

2i 



48. Let 6 and ^ be the angles which the tangents to the 
parabola at the points P and Q make with the horizon ; t, i 
the times from the point of projection to the same points ; v die 
Telocity of projection, and a the elevation. 

At the end of the time t^ the horizontal velocitj of the body 
is t? cos a, and its vertical velocity is t? sina — ^. <. Wherefore 
if t;' be the velocity of the body at the point P, we have 

t?' cos ^ = the horizontal vel. at P= t; cos a, 

t?'sin^ = ... vertical = i; sina— ^^; 

vcosa 



Similarly tan0 = —\ 



t;cos a 



.-. tand-tanA= -^ — . {i-t)\ 

^ t; cos a ^ 

and ••. «' — ^x tan^ — tan^. 



356 PBOJECIILES IN A NON-RESISTIHG MEDIUK, 

50. When the ball has reached its greatest height, all its 
vertical velocity will have been destroyed ; therefore, if F be the 
velocity of projection, a the elevation, and € the elasticity of the 
ball, it will strike the plane with vel. = Fcos a, and rebound in 
a horizontal direction with velocity —cFcos a. 

The equation to the curve first described is 

of 
^ 4A cos* a 

whence the co-ordinates of the point of impact are obtained, viz. 

y, ss A sin' a, and Xi=^h sin aa. 

If the origin of co-ordinates for the equation to the curve 
described after impact be at the point of unpact, then in 

of 

V = a? tan f-, ^-7= 

^ 4A cos' 

\. a Tf (eFcosa)' , , - , 

we have & = o, A =^ —^^he^cos^a, and y = — yi; 

*7 



a;" 



/. — A sin* a = =—5 r- ; .'. x^ = Ac sin 2a ; 

4A €* cos* a ' ' 

.•. the required distance =a?x ^^a = (i — €) A sin aa. 

Whole time = ^^^ + -^^2 ^^E^(i + x) 

Fcos a €Kcosa Fcos a ^ ' 



aF . /8A8in*a^* 

= — ^ sm a, or 



/ 8A sm* g ^f 



62. Let A be the summit of the mountain, and at the 
bottom; BA the height required; HK the directrix of the 
parabola described by the projectile. 

Then AH^ — , GK= ^^ ; 



PROJECTILSS IN A KON-EKSISTIKQ MEDIUM. 



857 




/. BA= (7jr-^^=3E! = 32ii6oo^/^ 



3i7 



'^ 



(f) 



3 X 1600^ _ 



^^ ft. =rii8i miles. 



64 



« , sin (a + 1) cos a 

cos' ^ ^^ ' 



B = 4h 



and F' = i6oo* (-^j; since a 13-inch shell weighs 196 lbs.; 



2, 1600* X 3/> sin (a + c) cos a _ 



196 



cos** 



450 cosec i ; 



. ^- 75>< i96y cosec 5° 30^ cos' ^i;o go^ „ 
• • ^ i6oo»sin30«»cosa4*'3o' "" '^^^' ^^z., 



^ 



_ gF8in(a+t) _ F 



^ cost 



sec 5° 30 =1^3 sec. 



858 



MOMENT OF INEBTIA. 



ROTATION OF BODIES. 



L Moment of Inebtia. 

Ex. e. 

1. Let ABhe the rod, and suppose it to revolye about an 
axis perpendicular to it at B. 

Let AB^ a, BP^r, PQ = rfr, » ^ "^ 

then in this case dM is proportional 
to dr; 

ir3 



I—l- 



and when r^Cf 



r r s 



3 



2. Let OA=^a, Om^r, then the 
circumference mn — 2,irrf and dM is pro- 
portional to airrdr, 

and T(^^ T,, 
M 



^fairr^dr _r* 



JZ'rrrdr 



and when rssQy 



2 



3. Let 04 = a, J^Q=r, AQ^s, 
then cUf = acfo, and l^M^ ^Jr^ds, 




;fc».axc P4p=f-^^ 

^C — ar {a^-^ r*)* + a* sin"' - , 




MOMENT OF IKEBTIA. 359 

and this beiiig taken between the limits r^o and r^PM^c 
suppose, we have 

le. % sin"' -= — . % sin"' c ia*-c?)^ ; 

• • KT -" "" """ ^^^i"^*^^^"^^ • 

asm ' — 
a 



7. Let a, ft be the semiaxes of the ellipse, and x^ y the 
co-ordinates of anj point of the enrve referred to a, j^ as axes of 
co-ordinates, then for the moment of inertia about the axis of a 
of a quadrant of the ellipse, we have 

and /y (J» - y-)*rfy = !/(&'- f)^dy {^^^ ^ ^ \ 

= i JV(*" - y")*d^ (same limits) 
4 * 

therefore for the moment of inertia of the whole ellipse we have 

4 
and /^^irdhy 

hence i* = - i*. 
4 

If we denote the radius of gyration about the axis h by h\ 
we shall have by similar reasoning 




860 MOVENT OF IKEBTIA. 

10. Let ABC be the triangle, A the angnlar point tiu-ougli 
which the line passes, PQ a veiy small 

Sirt of AB. Draw Pif, QN parallel to 
Cf and let AB bisect BG. 

if 8 the mass of the triangle, a, &, c the 
sides^a, ^^,^^, 

u 8 moment of inertia of the triangle 
AMP about A, 

m 

Au 3B moment of inertia of PMNQ about A. 

Now we may regard PMNQ as a very narrow parallelogram 
whose centre of gravity is g the middle point of MP. 

.'. Att = mass of P^x-i^ + its mom. of inert about g; 
and mom, of inert, of PN about ^=?mom. of inert, of gN 
+ mom. of inert, of ^r^ 

_ mass PN gM* mass PN gP^ 
a 3 » 3 

.-. A« = (mass of PN) [Aq* + - Mfj ; 

areaofPJV^ y.Aa^.sma 

but mass of P^= ilf ^ ^p^ = M — — 

areaof-axjO i t • <rr 

,,a;Aa/ 
^aJf-r;-; 



also Ag = -4I> r , and Mg = - — ; 

a;Aa; 



.% Aw = aJf. 






.'. <?M = ^(^Z)» + ^o»Wd»; 



UOMEMT OF INESTIA. 



361 



13. Divide the polygon into isosceles triangles hj lines 
drawn from the centre to the angular points; then A being 
the area of one of the triangles, and B, r the radius of the cir- 
cumscribed and inscribed circles, we have 

i? + 2r' 
mom. of inert, of triangle = ^ — . A ; 



6 
.'. krM = — nA = ^ if=-T7- ( 1 + acos' 



6 '~ 6 

and if a be a side, we have 



. M' ( 



3. 



a' 



24 



1 + acos*- 
n 



sm'- 
n 



12 



2 + cos — 
n 



1 — cos 



\ 



27r 
n 



M] 



a' 



12 



^ . 27r 
2 + cos — 

71 



^ 



I —COS 



27r 



16. Let this diameter be taken as axis of x, the centre 
being the origin of co-ordinates. Let x, x + Ax be the dist- 
ances of the circular faces of a thin circular slice of the sphere, 
at right angles to the diameter from the origin; then y being 
the radius of this section, its volume will = in/' Ax, 

and the mass of the slice = ^^ y'Ax, 



and its mom. of inert. = ^ • y'Ax . — ; 



^3M 

4a3 

therefore mom. of inert, of the sphere 

g^iJff "-=1 to x = + a, 



46 



I 



862 



MOMENT OF INERTIA. 



or 



k^M^^J{a^^a^ydx^^,{a^^^a^a^ + ^x^) 



8a^ 



8a« 



2a* 



M; 



5 



19. Let X be the distance of any thin circular slice of the 
cylinder from the middle point of its axis ; Ax the thickness of 
the slice; a the radius, and 2c the length of the cjlinder. 
Then the moment of inertia of the slice about any £ameter 
being 

= -Tra^Aaj; 
4 

mom. inertia about the axis of gyration := ira^Ax [ a^ + ~ a* j ; 

J \ 4 / ( to a; = + c, 



= (2,. + 1 a.) if, 



3 4 

21. Let CB be the base of the cone, AD the axis of gyra- 
tion, MP any section, 

AC=Cy CB=^a=^nc. 

It may easily be found that 
the mom. inert, of circle MP 
round AD 



= circle ix^ H j 

= 7m' f I H — \oc^; 



A 1> 




\ 


yC ^ 


^""^^ 




^^A P 




-A B 



MOMENT OF INEBTIA. 



863 



and, for the whole cone, 



irn^x^ 



(-3- 



Tc'M^ 



bnt Jf = 



«/.< 



imrc 



im'c* 



h^> 



c: 



23. Let AB be the rod, A the axis of gyration. 

c 

Take any point P in ABy and let 

AP = r, -4-0 = Z, p the density at P, ^ 
such that p^fin^ where fi is con- 
stant, 



then Ic'M 



r^'^^di 



Jo 

Also M^Jpdr = iijr*dr = — : 

whence 4* = — : — P 



n + 3 



(I). 



-B 



Similarly, if the axis of gyration be perpendicular to the 
rod at By we find 

*" = (n + aH« + 3)^ ^*^' 

therefore from (i) and {%) we get 

*' = -(w + i) (w + a)i'% 

and if A; = 64', we have 
w" + 3n + a « 72, 
or n* + 3n = 70 ; 
whence n = 7 or — 10. 



364 



CENTRE OF OBCILLATIOK. 



11. Centre of Oscillation. 



Ex. 7. 

1. Let ABO be the isosceles triangle, A being the point of 
suspension. Draw AD perpendicular to ^ 

Let AE-^ X, 

JiBAC^e, 

A^aieskofAFBA, 

and let I be the length of the perfect pen- 
dulum, then 







but A = a^ tan - : 

.'. dA = 2 tan - . xdx ; 



therefore, substituting in (i), we have 

ja^dx. 4 ! 
and for the whole triangle, if AD = a, we have 



Z=3 



a: 



therefore, if L be the length of the second's pendulum, the time 
of oscillation is 



&-m'- 



4. Generally, if h denote the radius of gyration, h the 
distance between the centre of gravity and point of suspen- 
sion, and I the length of the simple pendulum; then 



1 = 



T^ 



(1). 



OENTBE OF OSCILLATION. 



and by the rules for the centre of gravity, 



A=fr — 



rad. X chord 
arc 



r xr / fi\ 



Since the arc contains 6o^ 



3 



Also by Ex. 6, (5) (Mom. Inert.), we find 



P=ar-(i-|); 



therefore, substitating in (i), ve have 



ar» I 



;» 



_hi) 



= ^r. 



865 



Hence, L being the length of the second's pendulum, the time 
of an oscillation is 






and is independent of the length of the arc. 



6. Let ABO be a section of 
the pyramid through its vertex 
perpendicolax to its base, AL the 
given axis. 

Let AE==x, EF=nx; 

Now the moment of inertia of 
the square of which HF is a trans- 
verse section, round AL 



A L 


n/ 


\ V 


• / 


B V 



B^ 



square 






= «' (l + -) a*. 



366 



CEMTBE OF OSCILLATION, 



Therefore for the moment of inertia of the pjramid AEF 
round AL^ we have 



and for the whole pyramid, we have 

A:*Jf= (I + 

5 V 

and since Jf = - n*A* x h : therefore, 

3 



?)> 



*-^-('-i)-iK-iS=)= 



laA' + g' 



and the distance of the centre of gravily from the point of sus- 
pension is ^ A; therefore if Z be the length of the perfect pen- 
dulum, we have 



1 = 



y i2A* + a» 



4 



15A 



therefore tlie time of an oscillation ns 

« 

iaA* + a*\§ 






9. Let ADB be the cone, G its 
centre of gravity, BK the given axis 
which is a tangent to the circumfer* 
ence of the base of the cone at B. 

Draw A (7, OH parallel to BK, 
and therefore perpendicular to the 
axis of the cone. Let = alt. of the 
cone, and a the radius of its base. 





mom 



CENTRE OP OSCILLATION. 367 

Now, 
mom. inert, romid AG = mom. inert, romid OS+Aa* x Jlf, 

and, 

• « 

mom. inert, romid BK= mom. inert, round GH+ GB* x M; 

.*. mom. inert, round BK 

= mom. inert, round AC+ {BO^ - A G") M^ 

and by Ex. 6, ai, moment of inertia, we have 

. inert, round -4 (7 = - ( c' H — j 2f. 

5V 4/ 

Also, AG^ = ^', GB^^^^a^' 

i6 i6 ' 

.'. mom. inert, round BK^ — M. 

and if Z be the length of the simple pendulum, we have 

, _ mom. inert, round BK 
GBYsM 

_ 23a' + 2c' . (i6a'+c')^ 
~ 20 ' 4 ' 

Therefore the time of an oscillation is 



_ f 230' + 2c* U 



5(i6a" + c») 

13. Let m, om be the masses of the particles suspended at 
the distances of 30 and x inches respectively from a horizontal 
axis; and let i = length of the second's pendulum =39*1393 
inches. 

Then jr ^ »^ x 3Q' + a^no;' ^ a:» + 15 x 3Q . 
m X 30 + %mx aJ + 15 ' 

> 
.•. a?» — ia?= I5(i — 30). 

Hence £c = 4a|in. nearly. 



368 



CBNTKE OF OSCILLATION. 



15. Let < be the time of an oscillation 

about 89 

I be tbe length of the correspond- 
ing simple pendulum, 

and L be the length of the second's 
pendulum. 




Then t 



• V 9 



..3 



Now i= 



moment of inertia about 8 
moment of themassabout ^S^ 




"" 53 X 40 — 33 X I!i ' 

which reduced and substituted in the above 
expression for <, gives the time of an oscil- 
lation = 1*0589 sec 



17. Let m denote the mass of the bob of the perfect pendu- 
lum, and I its length ; w the mass of the given weight, and x 
the distance of its point of attachment from the centre of suspen- 
sion : d the distance between the centre of suspension and the 
centre of oscillation of the compound pendulum. Then we shall 
have, m and n being both of indefinitely small volume, 

^ ml^ + na^ 

a = — y . 

^ ml + nx 

Now the shorter the rod of a simple pendulum, the shorter 
will be the time of its oscillations ; hence we must have d a 
minimum ; therefore diflferentiating with respect to x we get 

2nx {ml + nx) — n {mP -f wa?*) __ 

ijnl+nxY ' 

hence wo?* + 2w& = wZ*, 

whence, x = - {(m* -f mny — m]. 

n ^ 



CENTRE OF OSCILLATION. 869 

19. Let K be the radius of gyration about the horizontal 

tangenti 

and K* be the radius of gyration about the horizontal 
axis at right aniles to the former. 

Then Z"' = - a» + c» = ^oV Ex. 6, 6, 

4 4 

jSr'» = -o» + a» = 2o'; Ex.6, 2: 
a a ' ' ' 



•'?-[7) -^-[6)' 



21. Let k^M= the mom. inert, of the sector about the given 
axis ; h the distance of its centre of gravity from the centre ; a 
its radius, and d the angle at the centre. Then 



h 



'"-i- 



y 



but if= — , and A ==-.3, 
« 3 ^ 

(vide Ex. 6, 8, Statics, c being the chord of the arc) ; therefore 
the length of the simple pendulum 

Tc" yi^e a0 . 
= -7- = - — = — by question ; 

.-. 0=^2 sin-^ (^ . 

26. Let a = radius of cylinder, ac = the length ; then if K 
be the radius of gyration about the axis through one extremity 
of the cylinder, we have (Ex. 6, 19), 

A:« = la*+-c* + c* = -a» + ^c». 
4 3 4 3 

£ ^« J. 4 c* 
XT r _ moment of inertia 4 3 

"" moment of mass "" c 

47 



870 



CEKTRE OF OSCILLATION. 



Let X be the distance from the upper extremity of the axis 
about which the cylinder would oscillate once in n seconds, and 
2 the length of the corresponding simple pendulum. 

Then n = T-y/i, i=Ty^|; •"•" = (2)* 
and /. l=n*(—+^. 



But Z = H 



C —X 



••— {--(^^f)} '=("-> (^f)' 



an equation which determines x. 
28. Let OA^a, 

bisect CA in Ey and CB in a 

F) join jEF, and let G be 

the centre of gravity of the lever. 

Then by Ex. B, 19, Statics, 




J 



Hence I 



^^ (a* + i^ + 2a»i» cos 0)^ 
(a + b) Ca 3 (^4 + J4 + 2a"J» cos ^)* ' 



30. The general value of the tension of the string, or, 
which is the same thing, of the reaction of the curve (instead of 
the tension) against a particle moving along the curve is given 
by the formula 



da da p' df^' 



(I), 



where JT, F represent the resolved parts of the accelerating force 



CENTRE OP OSCILLATION. 871 

acting on the particle, parallel to the axes of co-ordinates ; and 
p denotes the radius of curvature of the curve : the positive or 
negative sign is to be taken according as the particle is moving 
on the concave or the convex side of the curve. 

Now the body oscillates in the cycloid hj the action of 
gravity ; and the resolved force of gravity along the normal in a 

direction opposite to the reaction is %g.-^\ and therefore by 

(i), the particle moving on the concave side of the curve, 

^ dy 1 ds" 

^=^l+-p-Ti' (*)• 

Now dy = (^Jdx; ds = [^)*dx; 

dy _. f ia — Q^ 
" <fe ~ \ %a / ' 

and the radius of curvature at any point of the cycloid is 

p = a {2a {^%a — «)}*. 

Also if A be the altitude due to the initial velocity of the 
particle 

Therefore substituting in (2) we have, 



= 3^ 



%a + h — %x 



[2a {la - x)]^ 
But by the conditions of the problem 



■■■^-^^]' (3)- 



87S 



CBNTBE OF OSCILLATION. 



Now if tf be the angle which the string makes with the 
of 0?, in any position, we have 

dx /a?\* 



. ^ dx /a?\* 
^ dy /aa — a?\i 



.\ 5 ss 65r cos 5 



(4). 



When the bofly oscillates in a semicircle, we have 



dy 



_ {r — x)dx ^ , _ rdx 



{irx'-x^y 



dy __r —X 
' ' ds r 



and p = rf 
and since r is the altitude from which the body begins to fall, 

and if be the angle which the string makes with the axis of Xy 
we have 



B = 6ff CO&0 



(5)- 



Equations (4) and (5) sUbw the truth of the proposition. 



d'alembebt's pbikciple. 



373 



ni. D'Alembebt's Pbinciplb. 



Ex. 8. 

1. Draw AB vertically downwards, and let lPAB 
also let ds^ d8\ denote tne elements of the ^ 
circular paths described by P, P', in a small 
time dtj estimated in a direction corresponding 
to an increase of 0. Then the effective moving 
forces of the particles PP\ are 

-pd^s jy d^s 
^' IF' 

the moments of which about the point A are 



= <?; 




.d*8 



d's' 



Also the moments of the impressed foices are 

-PxAPy.^Bm0, -FxAFxg amB. 

Hence by D'Alemliert's Principle 

Fx AP^ + F X AT ^ + {Px AP+r X AF')g Bine =o. 

THow da = AP.dd, d8' = AP'.d0; 

.: {PxAP' + P'xAP'*)^+{PxAP+PxAP)gBmd = o. 

Hence it appears that the rod will oscillate isochrononslj 
with a perfect pendulmn, 



? = 



PxAP+FxAF ' 



2. Let m^ m' be the masses of the two particles; a, a' the 
inclinations of the planes to the horizon; and a;, x' the distances 
of the particles from the common summit of the planes at any 
time. Then the impressed accelerating forces on the particles 
w, wi' estimated down the two planes will be ^sina, ^rsino', 



374 d'alembert's pbinciple. 



respectively; and the effective accelerating forces, estimated in 
the same directions, will be g -j^ , g -^ . Hence, by D'Alem- 
bert*s Principle, we have 

msina-m sina)=m-^ "^ ~^ ('^" 

If a denote the length of the thread, then 

, d^x d^x 

hence, from (i), 

d*x 
(m + m')-^«5r(msina-wi'sina') (a), 

which determines the accelerating force on the two particles. 

Let T denote the tension of the thread, then for the equili- 
brium of the impressed moving forces T, mg sin a, exerted on 

d^x 
the particle m, and the effective moving force ^-^ applied in 

a direction opposite to its own, we shall have 

d^x 



r=m(5r sin a- ■^) (3), 

and from (a) and (3) we have 

tension = — — ^, (sm a + sm a ). 

4. Let a be the radius of the pulley, Ml^ its moment of iner- 
tia; m, vti the masses of the bodies, and x the effective accele- 
rating force on the circumference of the pulley -4, which is like- 
wise the accelerating force on P downwards, and on Q upwards. 
Let Tg be the tension of the string AP^ and Tg fliat of the 
string BQ. Therefore the force impressed on the circumference 
of the pulley is 

Tg-'Tg, 

and therefore 

^" V w- 



d^alembert's pbikciple. 



375 



But the accelerating force on P is 

m ' 

and the accelerating force on Q is 

T - m! 
m 

.'. mx={m—T) g, 'nix ^{T ''m')g\ 

.\ {m + m')x={r-'T)g+{m-m')g (2). 

And from equations (i) and (2) we get 

MJd'x+ {m + m') a^x = {m-m) go"] 

- ^ _ . ^ lm-'m)ga^ 

therefore accelerating force = x =7 — ; — ,. . , ,^,« ; 

therefore tension of AP=^m (a — a;) = 7 — ; — ,. ^ . it? a > 
and tension of BQ = m {x+g) = 7 — ;^ — ,v ^.jiria • 



6. Let the square be moveable 
about AB as an axis: the point 
required is the centre of percussion. 
Let EF bisect the opposite sides 
AB, CD; then with regard to i 
EFy the parts AF, FB are sym- 
metrical; and therefore the centre 
of percussion (0) will be in the 
line EF, and be determined by the 
equation 

„^ _ moment of inertia about AB 
" moment of mass ...•• * 



o . 



N 



» 



F 



Let EN=Xj Nn = hy AB = a^nh, Jf= mass of -4(7. 



Then P^^ssofPg ^ah ^ 
mass oi AG a" ' 



h 



B 



-9 



c 



therefore moment of inertia of Pa about AB = if x - a:* ; 



376 d'alembebt's pbinciple. 

therefore moment of inertia of AC 



= M— , when » is increased indefinitelj. 



Moment of mass == M- ; 



6. For the cylinder revolving about its own axis the 
moment of inertia is -ifr', Ex, 6, 18, 

where if measures the weight of the cylinder =: loo lb., 
and r is the radins of cylinder. 

If/ be the accelerating force on the weight P= 15 lb., 

then ^_, °^Q^Pgfo^<^ - P9 ^ ^59 ^3 
^ mass moved p.jr^ 15 + 50 ^3 



r* 



Space required = ^J^ = ^(TZff)s' = 92*856 ft.. 



where ^ = 32*19084 ft. 



8. The accelerating force 



^_ moving force _ Pg 
•^ "" mass moved "" ^ \^^ 






KOTATION OF BODIES. 



377 




since for the sphere Ml^ = - x (radius)* x M\ Ex. 6, 16, 

5 

^=(^)*-o (311)* = 33.488 sec. 

v = V^= 10 (^) = ^-986 ft. 

10. Let w = the weight of the pulley; then, considering 

a" 
the pulley as a small cylinder, its moment of inertia = —«(?. 

Suppose P to draw Q over the pulley, then the effective 
accelerating force on P downwards, or on Q upwards is (as found 
in Ex. 4), 



P+ Q + -W 

2, 



therefore time = (-j^) = ( — Z~j ~ ^'79 ®^^' 

12. Let P and Q be the weights, P drawing up Q. Let 
a, h be the radii of the cylinders, and Mk* the moment of inertia 
of the machine about its axis. We shall then have impressed 
forces. Pa at distance a, — Qg at distance b ; of which the 
moment is Pga — Qffby hence we have 

accelerating force on P = ;^^|^ , 
accelerating force on ^ = „;. ^ 7)a. i ^z.. ' 



Fa' + Qb* + Mk' 



48 



378 EOTATION OF BODIES. 

and from the data P=s=ioo, ^ = 500, J = -, A; = 3, Jf = 80, 
*aa5, and « = io; 

/. /== the force with which Q ascends 

= C5oa-ia5),^ , and (since. = Iy^'), 

i X (50a - 125) X 5- 

we have 1 , . a « 10, 

1 00a' + 845 

or a" — 20a + 58 =3 o, wheuce a » i6'4 ft. 

Now if if J Jf" be the masses of the wheel and axle re- 
spectively, then 



/^M'a* + ^M'V\ 



is the ezpiession f<» the distance of the centre of gyration of the j 
machine trom the axis ; 

II I ' 

i if V + i if "i* i34-48if ' + 1 if" 

¥ So 3. 

and af' + if" = 8o, 

hence if' = 5- lb. nearly. 

Let r, T* he the tensions of the ropes by which P, Q 
respectively act, then 

_, P {Qb* + Qab + MJe') ^ _ Q {Pa* + Pah -\- Mk^) 
Pa'+Qb* + Mk* ' ~ Pa'+Qb* + Mle' ' 

Pressure upon the axis = T+ T + M 

_ PQ{a + hy+{P^Q)Mk' , ,- 
Pa*+Qb* + Mk' "•" 

_ 50000 (i6'9)*+ 600 X 80 (3*) _ 
~ 100 (i6'4)* + 500 X '25 + 720 

= 6io*4 lb. nearly. 



PKESSUBE ON SV&FACES. 



S79 



HYDROSTATICS. 



Prebsube on Surfaces. 



' Ex, 1. 

1. Let ABG be the equilateral triangle, A being in the 



snrface, and AB vertical ; at', g'% a*" the centres of gravity of 
A G, ABy BG respectively. Througn g'" draw g'"a perpendicular 
to the surface of the fluid ; g"'a will pass through g. 

Now the sides pressed being equal in length, the pressures 
will be proportional to the depths of their centres of gravity; 

.•. press, on AG I press, on AB : press, on BG=ag' : Ag" : qg"'j 

but Ag" = na^y ag'" = ^ag' ; 
/. press, on -40 : press, on AB : press, on J5<7= 1:^13. 

2. The surfaces pressed being equal, the pressures will be 
to each other as the depths of tne centres of gravity of the 
triangles below the surface of the fluid. Now if a be the per- 
pendicular from the vertex on the base, the depth of the centre 

of gravity in the first position is - tis ^^^ ^ ^^^ second - a ; 

3 3 

.*. ratio of pressures -« - a : - a = 2 : 1, 
^ 3 3 



7. Let ABGD be the rectangle immersed 
in fluid, BE the line dividing it as required, 
g the centre of gravity of the triangle BED. 

Let ED^x; draw gK parallel to BD; 

then gK^-BD; therefore the depth of g 
3 

below the surface = - BB, and the area of the 
triangle = yx ; 

.•. pressure on the triangle = 70? x — xp, 

3 



B 



/l 


f 



E 



K 



380 PBESSUBE OK SUBFACES, 

and press, on remaining portion = 9 x 14 x 7p - ^ ^ ; 

3 

/. by hypothesis p(^9i47-^— ^ — J : ^—^ — p = 5 : 3, 

or : — = 553; 

^3 3 

••• 9 ' Y = 8 : 3; 

2^_27 

••• 3 - 8 ' 
_8i_ , 

9. The areas of the triangles pressed being eqnal, the pres- 
sures on them will be proportional to the depths of their centres 
of gravity below the surface of the fluid. 

Now depth of centre of gravity of A BOA = ^ BC, 

is^BOC^'BG, 

a, 



^DOG=^^BC; 

o 



.. L I M : N^^BO : - BG : ^BG 

6 a 6 

= 1:3:5- 

12. Let a = the axis, and h the double ordinate of the 
parabola; then the area of the parabola is - (the area of the 

circumscribing rectangle) = - a5 and the depth of its centre of 

gravity =2 a; 

/. pressure on the parabola = - a^Jp, 

5 



PBESSURE ON SUBFACES. 381 

similaxly we find the pressure on the rectangle 

/. pressure on parabola : press, on rectangle 

= - a^hp : - a^hp = 4 : 5, 

13. Let h be the altitude of the given parabola, and K that 
of the parabola cut off; also let c be the double ordinate of the 

original parabola, and that of the parabola cut off will be c [ -7 ) . 
The pressures on the two parabolas will be 

- Ac . — A . p = - A*cp, 
3 5^5^ 



, 2 ,, fh'\i 3 ,, % A'* 



5 A* 

therefore the pressure on the lower portion of the proposed para- 
bola will be 

hen6e, by the hypothesis, 



A'' ,, A" 
A* A* 

.'. A'' : A^ = m : m + n, 



whence A 






17. Let the edge of the cubical vessel = aa, 

> 

p, <r be the densities of water and mercury respectively. 
Then pressure on upper half of a side =s^%a.a.-p^ a^p. 



382 PBBSBtJfiE OK StJBFACES. 

To find tKe pressure on the lower half we mnst first find the 
thickness {z) ot a lamina of mercnij, the weight of which is 
equal to the- weight of water : then 



ea- 



p 



Pressure on lower half ^ aa . a (a ^ + - j <r ; 

therefore pressure on a side = a^ (p + ap + cr), 
pressure on base = 4a* {ap + aa). 
Hence, pressure on sides : pressure on base = <r + 3p : <r + p« 

18. The area of each of the equal sides of the pyramid is 
22*561 X 5 ; and the area of the base is loo. 

The depth of the centre of gravity of each of the sides 
• 44 

therefore pressure on the bottom = 100 x 22 x p, 

pressure on the side = ii2'8o5 x ^ x p, 

the weight of the water =- x 100 x 22 x p= p; 

therefore press, on base : press, on each side : weight of water 

, 44 2200 
= 100 X 22 : 22*561 X 5 X -^ : 

s= 3 : 2*256 : I. 

21. If p be the density of water, and r the radius of the 
sphere; then 

pressure on the sphere = ^irr^p^ 

weight of metcury =« - tit* x i3'568p ; 

3 

therefore pressure : weight = i : -2-5 — 

3 

= I : 4'5^3- 



PBSSSUBE ON SUBFAGEfl* 383 

24, Let A be the height of the upper segment cut off by the 
dividing line, r the radius of the sphere, and p the density gf the 
fluid; then, 

the surface pressed = a^rAr, 
and the depth of the centre of gravity = - A 

(vide Ex. 6, 35, of Statics) ; 

therefore pressure on upper segment = irrh^p^ 

but pressure on whole sphere = ^irr^p ; 

therefore pressure on the lower segment = irrp {^ — A*), 
and, by the hypothesis, 

4r* - A* = A» ; 

/. A « r X a*. 

25. Let 2?„ i?a, z^ be the depths of the dividing planes 
below the surface of the fluid j p the density. 

Surface pressed in the upper portion =5 ri7rr«, ; 
depth of centre of gravity = - i?, 

therefore pressure on uppermost portion = trrz^p (i). 

Surface pressed in next portion = %irT {z^ — Zj)y 

and depth of centre of gravity = - («?a + «i) > 
therefore pressure on next portion = tit («,* — z^^) p ^{2), 

Similarly the pressures on the remaining portions are found 
to be wr {z.* — z/) p, and ttt (r* — z^^) p, 

and by the hypothesiS| these pressures are equal to each other, 

therefore »,* = z^* — «i* = z^* ^ af^* = r* — z^^^ 

whence 0j* = -£f * = -^,^=-r^ 

^3 4 

whence », = - r, z^^-rxo^^, ip, = - r x q*. 



884 



PBE88URE ON SUBFACEfi. 



28. Let ABO represent the hemisphere, with its axis BD 
making an angle a with the vertical. 




Let g be the centre of gravity of the surface of the hemi- 
sphere. From g draw gb perpendicular to the surfiBbce of the 
noid, and gl to meet the vertical through D in I; then 

I T 

gb=^r^W^r — r cosa = - (ij — cosa), 

the sorfieu^e of the hemisphere = airr' ; 

.'. pressure required = wr^p {2, — cos a). 

83. Let Zif e„ e.f z. be the depths of the sections, and h 
the height of the cylinaer. 

The surface pressed in the first annulus = i^irz^^ 
and the depth of the centre of gravity = -«i ; 

.*• pressure on first annulus = ^jirpz^^ (i). 

Surface pressed in the second annulus = I4'7r {z^'— z^^ 

and the depth of the centre of gravity = -(«, + «i) ; 

the second annulus = *jirp {z^ — z^^) {p). 

the third annulus = ^irp {z^ — z/)^ 

fourth =7^p(V""V)> 

and fifth =7Tp(A"-0> 

and, by the hypothesis, these pressures are all equal to one 
another, and to the pressure on the base; 



/• pressure on 
Similarly, pressture on 



PBESSUBE ON SUBFACES. 885 

••. A* = 5«,« = 35A; /. 4 = 35. 
Breadth of third anntdns = ^f^ — «, 

= «x (V3 - Vii) = 7 V5 (V3 - \^») = 4*9749 i»- 

36. Using the same notation as in Exs. 25 and 33, the 
pressures are irrpz^^ trrp (z^ — «,*), ^rrp (^3" — «,*), Trrp (ao* — «j*), 
and bj hypothesis, these pressures are in Geom. i?rogression, 
haying 2 for a common ratio; 

/. %z^ = «/ - «,•, «,• = 3«x% 

a«a« - %z^ = «3» - «,•, «,• = 7«,*, 

2o« - «3» = a«3* - a«a*, ijo* - 7«x* = 8ifx" ; 

•••«. = ^ (15 X3)*> «^d 05 = |(i5X7)* 

38. Let A = altitude of cone, Z = its slant height, and 
r = radius of base. 

Convex surface of cone = tttZ, 

depth of its centre of gravity = - A ; 

3 

.*. pressure on convex surface = - irrhlp. 

Also, pressure on base = irr* xrxp] 

.'. whole pressure = - irkr {q,1 + 3^) p. 

3 

40. Let A = altitude of cone, I = slant height, and a = radius 
of base, z = depth of dividing plane. 

Then, A — 5? = altitude of cone below the dividing plane, 

z + = - (A + a«) =s depth of its centre of gravity, 

3 3 

7 

surface of this small cone = tt t (A — «) x t (A — «) ; 

49 



1 



886 PSESBUIUS ON SUBFACES. 

/. prewnxe on « -^ (A — «)" (A + 2z) p. 

Bat pressure on given cone = wal x - hp; 

3 

...^(A-.)-(A + a.)p = |(^xAp); 

.-. (A-«)»(A+a«)=-A*, 

or 45* — 6hz* + A3 s= o ; 

the real root of wKicK equation determines the depth of the 
dividing pbine. 

42. Let A =s axis of the cone, r =s radius of its base, and 
p ss density of the fluid. Then the pressure on the concave 
surfiu^ of the cone 

= i awrhp (r* + A")*, 
3 

and the pressure on the base of the cone 

therefore, by the hypothesis, 

?(r« + A«)»:r = 4: 3, 

whence t = -7- = teii ^o* ; 
* V3 • ^ 

.'. the vertical angle = 6o*. 

The ratio of the pressures = - (r* + A*)* : r 

3 

A* / A*\i 

Since -; is always positive, (i + -ij is greater than i; 

therefore the ratio of the pressures cannot for any cone be less 

than - : I or than 2, : ^. 
3 

44. Let A be the axis of the vessel measured vertically 
upwards ; a;, y the co-^ordinates of any point in the curve, then 



PBESSUBE ON SUBFAGES. 



387 



Q.y is the diameter of any horizontal section ; Try' its area, and 
^ (A — a?) y* the volume of the superincumbent fluid, which is 
therefore proportional to the pressure. Hence, 

yj^ (A — 0?) = c, a constant quantity, 

c 



y*= 



A-aj' 



is the equation to the curve. 



Ex. 2. 

1. Let a = radius of circle ; 



jLPGQ^e 




area of Pj= PQ x Nn ultimately; let Nn — dx\ 

then, pressure on Pj = aa?yda? x p ; 

.*. pressure on segment PAQ =i2pfxydx, between the limits 

x= o and x = AN, 

B 

Now a; = a + a cos - , and y = a sin - , 



7 (1,0 

aa? = — sm 



- • dOi /. xydx^ (sin* - + sin" - cos - ) d0. 

Hence, 



== C—a,pa^ ( 



5 — sin^ . I . 



+ - sm* 
3 



-D- 



388 



PRESSUBE ON SUBFACES. 



This integnl YanuheSi or the pres8iiie=o, when ^ zv; 

.*. C^pahr; 



/. preasuie on PA ^ = a* -jw — 



e-mnO 2 . .0 



— sin* 
a 3 



ii'- 



If PC be the required dividing line, this pressure = 



-pressure 



on the circle = — p; 







/. w (tf — Sin ^ — sin* - = — : 

2^ -3 2 2* 


/. 4 sin* — 3sin^ + 35 = 3w. 

2 



2. 



Let AB^r, 
aCAP^0, 
AG^a, 
Pp = dr, 

zP^e = (?5, 

p = density of fluid. 
Elementaxy area 
fQ = PQ X j^ = Td0 . c?r ultimatelj, 

pressure on 

fQ^ {rd0dr) xr cosO p, 

whole pressure on loop ==p\lr* cob 0drd0 




^pjco&0 d0(~r^+c\ from r = o to 

r = a(cos5t^)* 

=:/) - [cos 5(i -asin" ^*rf^ 

= p - X a* j * coBd (c»-Bin» ^*<?^, 

3 «/ — — 



if c» = -. 

3 



PSESSUBE ON SURFACES. 

Let Bm0=sz^ then cos 6 dB^dz^ and 

J * cos (c» - sin» 0* de=r\<^ - «»)*efo. 



389 



Now 



J(c» - i8»)* dz^z{i^- «»)^ + 3 J^f* (c» - «») cfe 

= -3[(c»-^»)*^ + 3<5'f(c'-«')*^ 

4 4 -^ 

Hence, whole pressure = p x -^ — = ^-^ . 

4. Let be the inclmatlon of the hemisphere's axis to the 
vertical, r the radius, p the density of fluid. 




Then, pressure on concave surface = %icr^ x HG . p, 
plane =7rr" x KG .p. 

Now HG^FG cos ^= (G^(7+ CF) cos d = ^^r + r tan tf) cos ^, 

and KG^r sin^; 
.*. whole pressure = irfrr^ (cos 5 + j:^ sin 5 + sin ^. 



390 PRESSURE ON SURFACES. 

Bat COS ^ + 3 sin ^ is a maximtim, when differentiating 

— sin tf + 3 cos ^ = o. 
Hence tan 5 = 3, or tf = tan""3. 



7. Let X = the depth of the chord below the surface, p the 
density at any point in the chord, and therefore = lafj k being 
constant ; 

f* f* k 

pressure at depth x = / pdx = A; I a^dx = - ic^ ; 

Jo Jo 3 

therefore pressure on chord = ij (a" — x*y x - oj* a maximum ; 

3 

differentiating (a' — a^) a^ a maximum ; 
6a*x^ — Soj? =s o ; 



/. X = (^j X radius, 



10. Let-45=ar, -4iV^=a?, NM^dx, NP^y, 

m 

A 



X 




B 


N 


1 


F' 


f 


\ 




C 


ff 


i 


^ 


/f 


\ 


M 




/ 



p = density at depth r. 
Then, area of Pj » ayda?, ultimately. 

Pressure on Pj »= weight of fluid superincumbent on the 
particles in contact with P^. 



PRESSURE ON SURFACES. 391 

To find the weight of a colomn of particles AN^ 
let AU=^z, EF^ &, p = density at E\ 
then p : p ^ z : r; 

,\ weight of the column AN= Ipdz = I 2 zdz = ^- ; 

J Jo' V* 

/• pressure on Pq = 2ydx x ^— = ^ {x^ydx) ; 
.•. whole pressure on circle = - / o^ (sjra? — a:*) (fo 

^r 8 

= 1 7rr3 X density at the centre of circle, 

15. Let a denote the axis of the cone, h the radius of its 
base, y the radius of any horizontal section, and x its depth ; then 

Now let p be the density of the fluid at depth x ; and there- 
fore = fee", k being constant. 

f* f * k 

Pressure at depth x = I pdx =k j a^dx^^-x^. 

Jo Jo 3 

k 
Pressure on horizontal section at depth x = iry^ x - a?*. 

3 

irb* k 

Hence — r- (a — a?)' x -aj^ or (a — a?)' a?* is a maximum. 
or ^ 3 

Differentiating, we have 

3aj* (a — xY — 2^ {a — x) = o, 

— 3 

• • a? ■— - "" t*. 



17. Let a; = the depth of the horizontal section below the 
centre of the spheroid. 



392 



PBEaSSUBB OR BVBFACES. 



The aection will be an ellipse, the semi-axes of which ate 

(36 -«»)*, and i (36- SB»)*; 



IT 



/. area of section = - (36 — a:*). 

Let p = density of the fluid at depth z, and therefore equal 
to Tcz where h is constant or the density at depth unity. Then 

pdr = - z^. 

IT I 

Hence, pressure on section = — (36 — a*) x - (6 + a?)*, 

3 ^ 

and this is to be a maximum ; 

.-. 2(36 — 0*) {6'\-x) — 3a?(6 + a;)* = o; 

.'. 40^+ laaj — 73 = 0, 

whence a; = 3 ; and the depth of section = 9 ; 

.'. pressure = -^f- (36 — 9) (6 + 3)* = - w* x 9^. 

% 



1. Let u45: 
AM. 



Centre op Pressure. 

Ex. 8. 
= a, AD=^h, /.ABC 
= x, MN=hf 
Density of fluid = p. 



a; 



=h; 




Then area of Mn = a^ sin a, 
pressure on 3fn = a^ sin a x a? sin a . p ultimately. 



OENTKE OF PBESSURE. 



393 



Moment of this pressure about AB=^ aA sin a x a; sin a.p x a; sin a 

= ahp sin^ a x a?*. 

For X write A, aA, 3A, nk successively; then 

sum of all such moments 

= ahp sin^ a {A" + (aA)» + (3A)" + + (nA)»} 



3 \ w/ \ aw 



:) 



= - aJ3 sin^ a . p, 
3 



since nh = 5, and n is increased indefinitely, A being diminished 
indefinitely. 

Again, area of ABGD = ah sin a, 

pressure on ABGD == ai sin a x - J sin a . /). 

This pressure may be considered to act at P the centre of 
pressure ; 

/. moment of this pressure about AB = - oJ' sin' a. px BP. 
Hence BP x - aJ* sin* a.p = -ah^ sin^ a.p; 

.'. BP^-b sin a, 
sma 3 



2. Let ABGD be the rectangular 
plank, irE being a straight line joining 
the middle points of AB, GD, and pro- 
duced to meet the surface of the fluid in 
0. Take Ox, Oy as axes of co-ordi- 
nates. Let PP,pp' be two lines parallel 
to AB cutting iiF in two points M, m, 
very near to each other, then 

element ij)' == 2,ydx, 
and its depth = x ; 



o 



A 


E 


B 


y 


P 




M 


p* 








F' 


P 


tn 




( 


P 


.r 


D 



60 



SM 



or, soioe jf is 



laLmiK OF FBBBBUBE. 

TuiaUe, 

x I xdx^j a^dx; 



• • SC "^ "" • 



-*^); 



3't"-*'* 



4. Let the 
in tlie point D, 




(70 prodnoed meet tiie sm&oe £2>L 
let OD^ki and let OJl, Of piodaced 




indefinitely be the axes of x and y. Let a;, y be the co-ordi- 
nates of a point P of the square, and x + c2r, y + ^y , of a point 
p indefinitely near to it; then the depth of the elementary 
square of which i^ is a diagonal, below the surface of the 
fluid, being 

x + y 



A + 



V^ ' 



and its area being dxdy^ we have, a denoting a side of the 
square AB^ 



'fJS 



h+ 



X 



^)^-£r(*^^)«^*- 



v« 



The coefficient of x in this equation is equal to 



I 



CENTRE OF PKESSUBE. 395 

and the right-hand side of the equation is equal to 

Jo\^ %kI% 3V^/ ^ 4V^ 3V» i^Vii 

(6*V« + 7o). 

TT — a ya + 6 V^ . A 
Hence a? = — . -, — 5— . 

(Walton's HydrostcUical Problems). 

Otherwise, bj the Integral Calculus, 

I sc^ydx I xy*dx 
^ = -7^ ' ^"I'T^ ' 

for the hypothenuse, ^ + t- — i ; 
therefore, substituting for y 

Jo ^ li^ 

/ft I 

= — a*5* ; (between limits) 
I , - I 

5. Let C7^=a, 05=*, CM=^x, MN=h, 
P the centre of pressure, and JSP= z. 

Now pressure on Mn = ^^ 7"^^ A.ajp ultimately. 



396 



CBHTBE OP FBEBSURB. 



M 



r I 


9 ▲ 

/ 




A 




P / 



B 



Moment of this pressure about CA 

For X write h, 2,h, 3A, nh sucoessiTdy; then 

Stun of all such moments 

^^\h{h' + (3A)»+ (3*)* + ...+ («*)»} - {*» + (a*)»+ (3**) +... 
= ?*e f JA». »(«+')(«« + ') _ AS. «' (» + i)* l 

= 06*/) f J , when w is infinitely great and nA = &, 

Again, 



pressure on ABC ^-cib.-bp^ 



ah' 



moment of this pressure about CA =-^p.z ; 

ab* oJ* 

/• « = - ft. 
In like manner we maj find Cff^ - a. 



CENTRE OF PBES8UBE. 



397 




6. Let ABGD Be the 
trapezoid, draw DE parallel 
to GB and CJf, DN perpen- 
diculars on AB. 

Let ulB = a, DO^by 
DN^GM^hy 

P the centre of pressure of trapezoid. 

• 

Then pressure on ABGD = press, on NDGM+ press, on 

A'aADN.RGM, 

pressure on NDGM= bh x - p, which may be considered 

to act at its centre of pressure, (Vide Ex. 3) ; 

.*. moment of this pressure about AB ^ - bh^p x - & = - bh^p, 

A's ADN, BGM= area of ABE. 

moment of press, on ABE about -4-8= (a— J)- x- px- , 

(Vide Ex. 85), 

press, on trapezoid » (a + J) - x [ , ] - p ; 



\a + %bl 2, 



7. The intersection of the plane of the circle with the 
surface of the fluid being taken as axis of y, and the diameter 
perpendicular to the surface as axis of Xy the formula to be 
used is 



I afydx 
I xydx 



and since y" == {2,rx — of) ; 



OF 






5^ ^ 
(belweto lnnibi) =« ^^— = ~ r. 



9. Lei the oeotre of die ciide be taken as erigin of oo- 
oidmatea; tlieii, aiiioe 

y=(r»-a:»)* 
Ca^if^^x^^dx ^ ^ fx{f*^x')dx 

tlien <£e»roo6 0<29, (r*-a:0*»r cos 0; 
Now Jsin* ^ COB* d d» = - jrin' atf <W = I J (i - cog 4^ «W 

/•§» I 

and Ja?(r«-a:')*&j«--(r«-a:»)*+C7, 

between limits ^-r*. 

3 

also I aj(r' — a^)di» = -r*; 



CENTBE OF PBES8UBE. 



399 



m • OS ^ ' 



y = 



3 



.J. 
i6 



TTTi 



3r 



10. Let ABC be the sector, 
the radius BA being perpendiculAr 
to the surface of the fluid. 

Let-45=r, BAC^a, AF^r^, 
anj radius-vector inclined at an 
angle 6 to AB^ and let PQ be the 
element described bj (frx, the radius 
vector r, + dr, having revolved 
through an angle dO. 

Then PQ = dr, x dd, 
and ^ = p .PjSr=pr, cos ^; 

.•• pressure on PQ ^pfr^dr^ cos tf dtf. 

Taking the moments about AB and Ay^ we have if S, j^ be 
the co-ordmates of the centre of pressure, and since 

NP^ r, cos ^, -4-^= r, sin tf, 
jj r^'coaffdr.de^^j j r^^ cob* 0dr^d0y 




X 



r,' COS d dr, dd =: / I r,3 cos tf sin 5 dr, d^. 

5 J O J O 

Now jjr^^cos*0dr^d0^ j^r^^coa^0d6+C; 
/. 11 r,3cos»ddr,d»= -/-{i +cos2^} dd 

= 5{<? + ^sinatf}+0; 

na Hi 

r,^ cos" tf dr, dtf =: -g- {a + - sin i»a}. 



400 



or HLUBUKB. 



Sin a. 



Similailj, f Tr^' cob 0dr^de = - a 

Jo* O 3 

Also f J'r,*ci»0m0dr,d0=j^.^aa20de+ C 






— o 2 



« = 



— sma 
3 



Mi 



sina 



+ C06 a 



}■ 



y 



sHaiii* a 

8 3 • 

I , . o 

3 



11. The equation to the parabola, leferred to the directrix 
and its azisy as axes of co-ordinates is 

y*=4a(a:-a) (i), 

where 4a is the Utas rectum of the parabohi. Let h be the 
length of the axis of the parabola. 




The immersed area being symmetrical with respect to the 
axis of a?, the formida to be used is 

- __ 11^ dx dy 
^^ JJxdxdi/ ' 

Integrating with respect to y we have 

_ (a^dx.y 
fxdx.y 



CENTRE OF PBESSUBE. 401 

Therefore fix>m (i), 

I %<^Qi? {x — a)^dx j a:^{x— a)* dx 

"^ rk "" rk 

I 2€r x{x-'a)^dx j x{x — a)dx 

■ Jo J a 

{x — a)* {15a?* + I2ax + 8a*} 



_io5 



^ (a? - a)* (3a; + aa) 



I 15a?' 4- I2ga? + 8a' 
7 ' ^x + (ia 



and between the limits a? = a + ^, a? = a, 

7 V3* + 5«/ 



13. From Ex. 1 it appears that the depth of the Centre of 
Pressure of any rectangle is equal to two-thirds of its height. 
Hence if we conceive the cylindrical tub to be composed of a 
great number of similar staves, each of which differs insensiblv 
from rectangular ones, they present a plane surface to the fluid, 
and the tub when foil of fluid will be kept together by a 
single hoop passing through the centre of pressure of all the 
staves. Hence the hoop must be at a distance equal to two- 
thirds the height of the tub from the top. 

15. Let r be the radius of the sphere ; and the centre of 
the sphere being taken as origin of co-ordinates, let a?, y, z be 
the co-ordinates of any point P on the surface of the sphere, 
and the plane yz horizontal. 

The pressure on an element at P, resolved in a direction 
perpendicular to the plane xz is equal to the weight of the 
column {x + r)dxdy. Similarly the pressure resolved perpen- 
dicular to yz = the weight of the column {x + r) dy dz. 

Let X and y be the distances of the resultants of these paral- 
lel forces from the planes ay, and a»5 respectively, then 



^_ JKx + r)xdxdz - SRx + r) y dy dz 
JJ{x + r)dxdz ' ^ IJ{x + r)dydz 



61 



! 



408 CENTBB OF PRESSURE. 

Now //(«» + »«)<&<&= I (iaj>+^r«»+ c\dz 

[between the limits « =- (r* - «»)* and a: = + (r* - «»)*], 

= - wr* ; between the limits « = — r, z =+r; 
4 

ff{x + r)dxdz = l (-a:^+rx+ Ojdg, 

= j2r{r^'-z*ydzy limits as before, 

= ar •^-^— ^ + T-sm -^ + C, 

(a ij rj ' 

= wr*, from « = — r, to ^ = + r. 

Hence 5 = r=-r (i). 

Again, 

limits y = o to y = (»^ — «*)*, 
« V 3/ 31 4 

= h -3- from « = — r to « = + r. 

3 8 



C£MTBB OF PfiBSSCRE. 403 

Also SS(x + r) rfy efo = //{r + (r» - y» - «•)*} dy dz 

limits aa before, 

s-i-— from « = — r to «« + r. 
o 

If a; be taken negativelj, we obtain by repeating the in« 
tegrations, 

.*. the resultant pressure on the plane oiyz = ^-= -=- = -wr^; 

3 
The resultant = |(^ ^) V (^) j* - ^lll^ . 



17. Let AD = a, AB^h; 

AM^ a?, MN^ Bxj 
p the pressure at the depth x. 

Then p^a^i and =caj* suppose, 

f* I 

•' 3 

.*. pressure on Mn = a.&» x - ca?*, 

moment of this pressure about AD = a.ixx-cix^\ 

/. snm of all such moments ^--aci oc^dx 

3 ^o 

I I,. 
s= - ac X - J*. 

3 5 

/•ft I 

Again, pressure on AG = I cm&j x - <?»> 

= - oc X - &♦. 



D 



at 



404 



CBNTBB OF PBB8SUBE. 



If now P be the centre of prawnie, we hare 

V3 4/3 5 

5 

19. Let the tangent at the vertex and the axis of the para- 
bola be taken aa co-ordinate axes. Let i^ be any elemental 
area of which the sides are c£r, dify and whose depth is x ; and 
let p be the unit of pressure of the fluid on I^ ; then, o denoting 
a constant quantity, 

j>s| pdx^ CI xdx^-cgc^f 

Jo Jo ^ 

and, dxdy being the area of i^, 

- ffsc^xdxdy 
jja^dxdy ' 

and, integrating with respect to y, since y = %€^x^y 

I 7^dxy\ I %(?x^dx 



x^ 



/■* f^ I 1 

I x^dxy I 2a*x*dx 



?A^ 



7 



Ex. 4. 

1. Let AB be the heavier 
fluid, and CB the lighter. 

Suppose the radius = i. 

Let z^OjB= z50(7= a. 

B7 hypothesis nJSW=: EM. 

Let BD^e, then 

JEN^rs 0J&- OJV^= cos tf - cos (a - ^), 

jEAf- 0^- OM^ cos ^ - cos (a + ^; 




CENTRE OF PRESSUBE. 



405 



.•. n{cosd — COS (a— ^1 =cosd— cos (a + ^; 

.'. n {i — cos a — sin a tan d} = i — cos a + sin a tan ^, 

or (n + i) sin a tan d = (n — i) (i — cos a) ; 

, A n— II — cos a n— i^ i 
.•. tan^ = . — ; = tan -a. 

w+i sin a n+i 2 

3. Let ABhe the heavier fluid, and CB the lighter; also 
let PJS^=a?„ P£!=-x„ PD^z, 




then the length of arc AB = length of arc -8(7= (ara)* ; 

.'. 7, 42rx^ + a Viira?a = a Vara, 

or Vi»i + V^J. = V« (i). 

Now p.BE^p'.BF, or (a;, -«). p= («,-«) p; 

• • » -^ * • 

Also, since BC^BA^ 

a Vor^— 2 '^zrz^ % V2ra?,4- 2 Var«, 

or Vi»a-V«i = %'^Z^%Kl ^^^^f («)• 

From equations (i) and (2) we obtain 

''"i6U+pV' ^'~i6U + p';- 



a;, 



Specific Gravity. 

Ex. B. 

1, Let <r, p, a be the specific gravities of the cube, water 
and air respectively, v the volume of cube = 4^ inches. 



406 SPECIFIC QSAYITT. 

Then vo- — va « 16047 S^- ^^ weight in air, 

and vc-'Vp^ 95S^ water; 

/. vp — t?a= 1615^1; 

16152 770 
= 252*703 gr., the weight of one cabic inch of water. 

2. Let Mj M\ represent the Yolumes of the bodies, «, s\ 
their specific gravities, and TT, W\ their weight; then 

Jlf : Jf' = 3^ : 53 
« : «' = 3 : 4 

.-. W \ W =M8 : Jlf V = 81 : 500. 

4. 960 lb. = 15360 oz. ; 

.'. ^ o = — = no. of oz. in a cubic inch of the second metal: 
1720 9 

•'. if tf, «' be the specific gravities of the metal^, 

, 80 

^ 9 
= 1-1655 : I. 

7. Let if = the volume of the globe, and a- its specific 
gravity ; then its absolute weight = Mc and its wei^t in 

air is 

Tr= if<r^lfa = if (<r-a), 



and in water, 



w = if (<r — /)); 

W a-a 

• • — > 

w a—p 

, Wp — woi 

whence, <t = -^ ; 

W — w 

also, if = = . 

cr — a /o — a 



SPECIFIC GKAVITY. 407 

Hence if d be the diameter of the sphere, we have, 

\ir p-a / 

9. Let V, V* be the Yolumes of gold and diamond, 

pj p sp. grav 

Then vp + vp = 69! gr., where p = 16*5, 

and vp + vp —{v + v) x i = 64^ gr., and p' = y^ ; 

.*. v + v' = 5, 

vp + i?'p = 5 X 16-5 = 8a"5, 

^' (p - p1 = 13 ; 

10. Let V be the volume of gold in the crown, 

. v' silver 

V crown, 

p, p', a the sp. gr. of the gold, silver, and crown. 

Now F<r-Fxi=i5F(r; 
.'. o- = i4. 

Similarly, p = — and p' = — . 

But vp -f v'p' = (r -f v') o- ; 
V / /v \ 

vp p — a- p' 77 2 X lix 9 

— 14 



408 SPECIFIC ORAYITT. 

18« Let IT a the weight of the hydrometer; then 

1F'+6o» water displaced, 

and Tr» spirits displaced, 

and the volume is the same in each case, since the instrument is 
sunk to the same point, and since the specific gravities of two 
fluids are in the ratio of the weights oi equal volumes of the 
fluids, we have 

I : -866 = W+ 60 : TF, 

whence TF= ^ " 

•134 

= 38776 gr. 

16. Let a be the specific gravity of the mixture. 
Then a8<r = ai x 1*84 + 8x1= 46*64 ; 
.'. a = i'6657. 

20. Let Fbe the volume of the proposed substance, and p 
its specific gravity. 

Then Vp-V\^P\ ^ 

j^ __ TT — 7)1 "^^ which equations V and /> aie to 
yp yi^— ss> \^ eliminated; 

and F(i/-/i) = Q-R] 

.•. Q(ji-\+V'-fi)-R(ji-\)='F{v-fi); 
.-. Piv- fi) + Q {\- v) + R (ji-\) ^o. 

21. Let a be the specific gravity of the mixture, whose 

volume = (v -f v') { i — ) . 

Then {v + v) [i - - j cr=: t;/> + vp ; 



n-^ 1 \ V + V J 



SPECIFIC GRAVITY, 409 

m 

23. Since the surface of a spherical segment is propor- 
tional to its height ; we have 

^ X lo in. = height of segment immersed in water ; 

.'. g X loin. = naphtha. 

Let <r be the sp. gr. of the sphere ; then 

^ X io^<r = ^J3 ^ ^^""^ (i^ ^°) ' (i^ ^^) ^' 

+ ^js X lo -- 2 ^1 X ioW| X loj X 708, 

Vide Coll. of. Ex. p. 97 ; 

••• " == (3 " ¥) (i)'-^ (3 - S (i)"x -708 = -9076. 

2/i • Let Aj Bj (7 be the globes whose sp. grs. are 3, 4, 6 respec- 
tively, V the volume of each \ Xj y^ z the distances of their 



-e — e — e- 

centres from the fulcrum ; then, since thej balance one another 
in vacuo, and in water, 

3 F. 0? + 4F.y + 6F. « =s o, 
and (3-1) F.a?+(4-i)F.y+(6-i) F.«»o. 
Hence 305 + 4^ + 6«ssso, 
and iW5 + 3y + 5« = o; 
.-. «+ y+ « = o; 

.*. 305+ 2y = o, whence aj= — y, 

3 

I 

y + 3«=0, 2f = --y. 

The globes A and (7 are therefore placed at distances % and 
I respectively on one side of the iiilcrum, and B at the distance 
3 on the other side. 

52 



410 SPECIFIC «RAVITT. 

29. Let A be tlie aiea of the plane of floatation, TP] tbe 
weight of the ship, V the volnme of water displaced bj it, 
then 

W^ 1-026 X r, 

and the volume of water displaced on entering the river 

.% w^{r+3,A)xj, 

and the volume of water displaced after i2cx>olbs. of cargo have 
been discharged is 

= V+A; 

/. W—i20oo^V+A; 
therefore eliminating Ay F, we obtain 

^(^ ^ 1^ " ^^^^ ^ a = 24000, 

, ^jy. 24000 X 1*026 ^ „ 

whence W= --^ — - — ^ = 947076*923 lbs. 

= 422*802 tons. 

32. Let s be the specific gravity of the cylinder, and s, s" 
that of the fluids, then 

8 : « ' = i : I whence s' = ^8, 
5 4 

and 5:5' = -: I whence «'=-*; 
3 ^ 

/. "o" * = specific gravity of the mixture. 

Let X = the portion of the axis immersed in it, then 

^ II _ 

S • *Q* 8 -^ 01/ I I J 
g 

.'. a? = — X the axis of the cylinder. 



SPECIFIC GRAVITY. 411 

35. Let X = internal radios in inches. 



Then weight of iron = 4![ J f «. ^ j - aj»|- x 7*2, 



alcohol = — aj3 X '8, 
3 

water displaced = — [ojH j xi; 



.'. ](i»+ — j -ic'h X 7'a+aj3x '8 = (x+—\ X i; 



.*. (x H J X 6'2 = a^ X 6'4 ; 

whence a? = 9*4 in. 

36. Let r be the radius of the hemisphere, 

.... p sp. gr. of the fluid, 

.,.. X additional weight required. 

The solid content of a spherical segment, h being the 
height, is 

^ {6r - ah) h* or IT fr A j A*. 

W+X _2S 

•"•• TT ~ 8 • 
Dividendo Tfr= "o" ; •'• ^ *= ^'5 >< W^* 



412 



EQUILIBRIUM OF FLOATING BODIES. 



88. Let d be the diameter of the sphere, a its sp. gr. 
Then ^ <^<^ =* ^ (s^t— v^d) (•65^)* x i ; 

.-. cr = 17 (-65)* = 7181*5. 

When the air has been admitted, let x be the height of the 
segment immersed in water ; then 



TT 



TT 



TT 



g (3^—00:) 35* X I + J? {3^— a(rf— a?)}(rf— a:)*x"ooiM5 = ^ci5»o-; 

/. 3£fe*— 20*+ (rf^ — 3diB" + ax*) X -001225 = rf3x 71825; 

/. (2«» - $dixf + d^) (i - •001225) := rf^ (i — 71825) ; 

.*. a^ - r^da^ + "35895^^ = o. 



Equiubbiuh of Floating Bodies. 

Ex. 6. 

1. Let ABC and FQ be the sections of the prism and of 
the plane of floatation ; J7the centre of gravity of the flaid dis- 
placed, O the centre of gravitj of the body, and if the meta- 
centre. 

Let h be the length of the axis 
of the prism ; then 

The moment of inertia of the 
plane of floatation about an axis 
through its centre of gravity perpen- 
dicular to ABC 



3 4 

and the volume of the fluid dis« 
placed 



= -A.P(7. (7(?.sinO=Axareaof AP(?(7; 




B 



.'. hxAFQGxHM^—PQixh: 

12 



EQUILIBRIUM OF FLOATING BODIES. 



413 



.-. J3af = — PQ^-r- (area of A PQC) 



mersed. 



= — (line of floatation)^ -r- area of triangular part im- 

l2» 



4. Let ABCD be the square, floating with one angle C 
immersed in water. Let PQ be the line of floatation ; GP^x, 
CQ^y. Join AG and bisect it in O. 

Bisect PQ in F, join CF, 

and take HF^ i GE. 

3 

Then G and 5" will be the 
centre of gravity of ABGB, 
and GPQ respectively. Join 
OH and draw FE parallel to 
SO. Then, in order that the 
lamina may be at rest, O and 
M must be in the same verti- 
cal line; hence EF must be 
vertical and therefore perpen- 
dicular to PQ; therefore the 
lines joining PE, QE will be 
equal. 

Let a be the length of a side of the square. 

Then, the weight of the displaced fluid being equal to that 
of the square, we have 




-ay=:pa*y or xy^2pa*, 



(I)- 



Again, PE^ = ir» + GE^ -t.x.GE cos 45^ 
QE^^y^ + GE* - ay . GEcob 45^ ; 
.•• since P^=^^, 

{x—y) {x + y-2,GEcoB 45*) == o 

but GE: GO^GF: GH; 
/. GE^^GG; 



414 EQUIUBRIUM OF FLOATING BODIES. 

4 ^ 4 ^ 

hence («— y)(« + y — a)s=o, 

which equation gires 

either x — y^o (2)9 

or a? + y-|a = o (3). 

Taking (1) and (2) we have 

which gives one position of eqoilibrinm if /> be less than - • 
Taking (i) and (3) we have 



a* 



4 
or aj-y = ±j(9-32p)*» 

and a; + y = -a; 

4 

y = 7{3 + (9-3«p)*}- 



Thus there will be two other positions of equilibrium if x 
and y be both possible and positive quantities less than a ; that 
is. if 



P^Ti^ and(9-3ap)*<i, 

that is, if p lie between — and — . 
^ 3^ 3^ 



There are therefore three positions of equilibrium when one 
angle is immersed, and therefore twelve for the whole square. 

^2) The second case may be deduced from the first by 
writmg (i — p) for p. 



) 



EQUILIBRIUM OF FLOATING BODIES. 



415 



7. The nature of the equilibrium evidently depends on the 
position of JIf the metacentre. 

Let CL = a?, PL = v be the height 
and radius of base of the cone im- 
mersed ; and CD = a, AD = J, of the 
"whole cone. Let H be the centre of 
gravity of the fluid displaced, and Q 
that of the whole cone ; then the mo- 
ment of inertia of the plane of floatation 
{JPQ) round a horizontal axis through- 
out its centre of gravity L^ is 




also, OH:=^{a — x)y 
4 

and the volume of the cone immersed is 



hence HM=^ «^; 

4 X ' 

But the part immersed bears a constant ratio to the whole 
cone ; and the cones being similar solids are to each other as the 
cubes of their heights or of the radii of their bases, hence 

a^ = -^a^ and y^= ^b^; 



8 



^_i b\ 
X 2 a 



\ OM=^b 
4 



% a \ 2/) 8 [a ) 



416 



EQUILIBRIUM OF FLOATING BODIES. 



Now Binee the cone in right-angled at the yertez we have 

and therefore 

flfJf=o, 

or, Jbf coincides with (7, and the equilibrium is nentraL 

9. Let ABG be the cone, O its centre of gravity ; g the 
centre of gravity of the portion in the lighter fluid, h that of 
the part immersed in the neavier, FQ the plane of floatation. 

Let as axis of the cone equal a 
rad. of base, then 

Oh^la, Cs^^a, CO^ia; 

7 Q 

Let p s density of the lighter fluid, ^the centre of gravity 
of the fluid displaced, and let gS^ x, then we have 




X 



^ii^<^'p''ri'"''''lpQ ''-'')' 



whence a? = ^5"= 



6a 
"9 



rro 15 ^ 1617 



2(8 



119 333* 



and therefore HQ == a = ^^r? «• 

333ij 1666 

Now the mom. inert, of the plane of floatation is 



= ^7raS 



8 



and the volume of the fluid displaced = ^ ira^ ; 

4 



.-. GM^ 



81 



TTO* 



81 



ira^ 



^ 441 ^^^^ 441 ^ 
1666 2 1666 



106 • • • 

= P^- a, a positive quantity, 

or M is above G, therefore the equilibrium is stable. 



EQUILIBRIUM OP FLOATING BODIES. 417 

11. Beferriug to the figure in Example 7, we have, if / be 
the mom. inert, of the plane of floatation, and V the volume of 
the fluid displaced^ 

..MM y^'^y 

and bj the conditions of the question 

4 a? 4 

^ 3 

whence sLs=_-3tan30**; 
X v3 

therefore the vertical angle of the cone is 6o^. 

15. Let Xfffhe the height and radius of the base of the 
paraboloid immersed, and a, b, of the whole paraboloid, then 
volume of the fluid displaced is 

and the volume of the whole paraboloid is 
And from the condition of equilibrium we get 

whence v'aj = - J*a (i), 

3 

but y' = Zaj and J* — Za; 
therefore, substituting in (i), we get 

aj ss f - j X a, the depth required. 

53 



418 EQUILIBRIUM OF FLOATING BODIES. 

18. Using the same notation as in the preceding example, 
we have 

moment of inertia of plane of floatation 



I . 
4 ^ 



Yolmne of the fluid displaced 



I . 2, X' 

"iry^x 

n 

and HO = - (a — a?) ; 

Q, X 3 ^ . 

Let 2 =: the latus rectum of the generating parabola, then. 



since x^^a, 



♦ GM^-l — [a a), 



3 

and for the position of indifferent equilibrium we have 

-I — a = o; 
« 3 

whence -y^^- 



Elastic Fluids. 
Ex. 7. 



i 



1. Let p be the density of the air at any point P above the 
earth's surface, c the compressive force of tiie air at that point, 
or the weight W of the superincumbent air, h = the altitude of 
an homogeneous atmosphere at P, upon supposition that the I 
density is uniform, then j 

Wqc pxh; j 



ELASTIC FLUIDS. 



419 



, W c 

.*. hoc — oc - , 
P P 

but c oc p ; 
hence h is constant. 



2. Let BP be the space occupied by the air in the tube 
at first; and when the cylinder is immersed in the 
water, let the water stand at Q ; then the air which 
"before occupied the space BP^ now fills the space 
BQ, And since the elasticity of the air is propor- 
tional to the density, and the density is inversely as 
the space occupied by the same quantity, we haye 

elasticity of air in BP : elasticity in BQ^ BQ : BP. 




Now BE^ aoft., J5P= loft., and the elasticity of 
the air when it occupies the space BP would support 
a column of water 3 j ft. ; also the elasticity of the 
air in BQ + the weight of the water in QE, are 
balanced by the pressure of the atmosphere, and 
therefore would support a column of water 33 ft.; 
hence, if QE=x, the elasticity of the air in BQ would sub- 
port a column of water = 33-.:; hence 

33 • 33 - ^ = 20 - a? : 10, 

or a?» - 53aj =:= - 330, 

whence x = 7*206, 

the less root alone satisfying the conditions of the problem. 

" 6. Let AB represent the surface of the water; CD the 
cylindrical tube; DP the height to which 

the water rises in the tube. Let V be 

the content of the whole tube, F' the con- 
tent of the portion aboye MPN. When the 
air in the tube occupies the space CP, the 

Sressure of the water is proportional to the ' 

epth -4P= a? + 1, and the pressure of the 
atmosphere is proportional to 32 ft. ; there- ^ 

fore the whole pressure upon the air is equal 
to the weight of a column of water whose height is 

32 + a? + I = 33 -f a?. 
But the elastic force of the air is as its density, or inyersely 



480 



ELASTIC FLUIDS. 



as the space occupied when the qoantitj remains the same; 
therefore 

4:1:: V: F' :: 33+aj : 3a; 

.'. 33 + a^ = i^8, 

and a? = -4(7= 95ft. 



9. LetAB = A, 
JEF«Ar, 
jEB=a, 
a » density of atmosphere, 

press, at D = 33 + A + a — i, 

press, of air in BDH _ vol. BCK 
press, of air in BGK" vol. BDH' 

YolBCK^^.GE'.BE^'^.L.BE'; 



•-- 


7-- 


r 


B 


I 


T> 


° 










P 




C 


, 




- K 



L being the latos rectum ; 



TT 



YolBDE^^.L.BF'; 



press, of air in BDH _ BE* = f ^ Y 
•'•press, of air in 5(7^ jRF^^U-Ay* 

■R f press, of air in BCK _ p ^ 
press, of atmospheie a ' 

.'. press, of air in BGK^ - x 33. 
Hence 33 + A + a-A;=^^3^) a^335 



•••P = (i + 



33 



10. Let ^ represent the unit of pressure at every point of 
the air before, and ^ after, the insertion of the piston. 

Then, the pressure on the base of the piston being tt x 2 xp\ 
we have 



7rx^xy = 5 + 7rx2x^, 



(I). 



ELASTIC FLUIDS. 421 

Let X be the distance of the piston from the top of the 
cylinder, when it has ceased to descend ; then we shall have 

p* :^ = a4: 24 — Xj 

or p {%4 — a;) = 24^, 

and therefore from (i) we have 

24 X 80 

So + 'irx^ X o 
4 ^ 

bntjp^ ^^^y^8 ^35'5; 

• • a?= -7—^^ — s = I'l in. nearly. 
1745-048 -^ 

12. Here A = 30-151, ^ = 59*9, 2^=59-9, A' = 26-474, 
*' = 49-i, and 2" = 50-88; 

.-. ^ - 60345 f I + i^^^^) log-^— r^525i ^ 

^^^ \ 900 / ^^ 26-474 (i + 9-02 X -0001} 

= 63362-25 xlog^l^g 

= 63362-25 X -056091 = 3554-052 feet. 

14. Let. W be the weight of the balloon with all its append- 
ages ; and let x denote the specific gravity, or density, and z 
the height of that stratum of the atmosphere at which the 
balloon will cease to ascend. 

Then since the weight of the air displaced by the gas 

^-irr^xx. 
3 

and that of the balloon and its appendages and gas is 



TF+-^rHxa ; 
3 



therefore, by the question, 



3 3 

ctW 
whence x = -^ — r + a : 

4irr* ' 



422 



ELASTIC FLUIDS. 



therefore, if the temperatures at the height z, and at the soi&oe 
of the earth be considered the same, we nave 



z = 60345 log -j^ 



3 
47rr* 



+ a 



1 6. Let X s radius of balloon. 

Then a x 5280 = J.log |^. 

h A. 
7x528o = -^log=; 

1 ^ 

2 ^^1 H a 

.'. -T = 167. 
■D i Pi. * 1*2 

= -i-a!» X 'la oz., 
3 



P B 

Now weight of gaa 

appendages =8ooxi6oz., 



air displaced = — a^ x p^ ; 



Air , 1*2 Air ,0 ^ 

.\^— a^x — ; =-^aj3x -la + Scx) x i6; 

/. iC*. TT ( -— •l6j = 800 X 16; 



• • M? ^~ 



8 ^ 

800 X 167 

^ — T" 

7r(i'6 ); 

V 100/'^ 

1 6*44 ft^ 



ELASTIC FLUIDS. 423 






Now, weight of copper = ^ iUo + j^) - S^\ 8788 oz., 

gas s=:^x5o»x'iaa5oz., 

3 



a"«'^pla««l = ?(50 + 7^) p. 

car, &c. =5000x1602. 



whence the value oip^ is found. 



^ , 60000 
+ 5o3x'ia25 + 



/. «« 60345 log ^^^^ 

Pn 

= 141^^3 ft* or 14623 ft, 

according as small terms or none are omitted in computing the 
value of f>n. 



Ikstbuments and Machines. 

Ex.8. 

1. When the fluid in the tubQ falls i inch, that in tK^ 
cistern rises 

w X I" I . 

r X I = - m. : 

w X 3» 9 ' 

therefore the true variation 

. I 10. 
.« I + -« — m. 

9 9 



424 



IKSTRUMKMTS AND MACHINES. 



3, -AB»33iiL, AC^z^iiLy AP=:Xy 
pressme of air in BC^ 30 — 29 = i in. 

Agmin, 

junasare o{ nJT ia BP' _BC _ 4 
BG' ""5? " 33 - a; ' 



B 



I' 




.*. pressure in BP" 



4x1 
33-^ 




Hence — h« = 25; 

33 -« 

.-. (33-«)(a5-a')=4. 
a* — 580? + 89* = 841 - 825 + 4 = ao, 

a; = 29 ±4-47213 

= 24*52787 in. 



5. Let X, y be the length of tabe, which the air left in the 
barometers, of the mean densi^ of the atmosphere, would oc- 
capj; A, k, the heights at which the perfect mstrument would 
stand on the two days ; then we have the following equations : 

(i) {h-a){l-a)^hx; (3) {h-a'){l-a')=hy; 

(2) {k-b){l-b)=kx; '(4) {k-b'){l-h')=hy; 

whence, eliminating h and k, we get 

{ a'{l + a-a') b'{l+b-b') \ f a'(Z-o') h'{l-b') ] 
'-\ a b i • \a{l-a) i(i-ft)j' 

{ a{l + a'^a) b(l + b'-b) \ . f a(?-a) b{l-b) \ 
3''"| ^' b' )'\a'{l-a') b'il-b')}- 



7. 



Generally, p^=U~j p; 



1 



IKSTBUMENTS AND MACHINES. 425 

Therefore, by the question, we have 

f B y 1 



R I 

whence jB : 6 = i : 4* — i. 

9. -Let A be the air-pump with barrel of i foot capacity, 

B a feet 

p^ be the density of air in A after 5 turns, 
ft B 3 

(P) quantity of air exhausted from A 

= quantity of air at first — quantity after 5 turns, 
= io/> - io/>^. 

Similarly, {Q) the quantity exhausted from B 

= io/> - lop^. 

Hence, 

/loy 

P_ lO/g— lOpj _ \II/ IQ? ^^*""^o5_^ 1 

11. Let X be the number of turns ; then after x turns the 
quantity of air in the pump 

therefore by the question we have 



(7? \* 



54 



426 



INSTRnMENTS AND MACHINES. 






14. Let^F=Z, EA = a, EC^c, FP=^x; 

h the standard altitade, or height of mer- 
curial column in the true barometer. 

^p, pressure of air in FA __ FP __ x 

pressure of air in FF FA l—a^ 

.*. pressure of air in FA = j—- ; 



hx 



+ a = pressure at E=h (i). 



If p be the density of atmospheric air in the receiver 
at first, 

and p^ be the density of this air after n turns, 



Now if A measure the pressure of the air of density />, 
and A' p^ 



we 



have •r = — ; •'• h' = gh. 
h p 



hx 



. , pressure of air in FG _ FP ^ x 
pressure of air in FP^ FG " l—c' 



. . , + c = pressure at .Bunder the receiver = qh (2) 

Equations (1) and (a) give the values of A and x; thus 

{qh — c) (?— c) = Aa; = (A — a) (Z — a) ; 

.'. A {q (?— c) — (Z--a)} = c(Z— c) — a(Z— a); 
, /c l-a\ (/ m Y l-a] , . •. / w \* 



INSTRUMENTS AND MACHINES. 427 

16. Let B be the capacity of the receiver of the condenser, 
and also the capacity of the receiver of the air-pump, b the 
capacity of the barrel of the air-pump, and p the density of the 
common air in the receiver and barrel at first. Then tne mass 
of air in the receiver of the condenser at first is JRp. The mass 
of air in the receiver of the air-pump after n turns is 



(sf.)"*- 



therefore the mass of air discharged into the receiver of the con- 
denser after n turns is 



^''-(sTi) ^P' 



and therefore the mass of air in the receiver of the condenser 
after n turns of the air-pump is 

Therefore the density of the air in the receiver after n 
turns is 



{ 



M+b 




18. Let p be the density of the air in the receiver at first, 
and consequently of the air in the gauge. After iiij turns the 
density of the air in the receiver is 

30S + 12S 7 
306 P 5^' 

Now the density of the air in the receiver increases in arith- 
metical progression, and as the density is inversely as the space 
it occupies, we shall have, if x be the distance of the globule of 
mercury from the receiver after IQ, turns, 

7 

/. 100= 140 — 70?; 
.•. 7a; = 40, 

ic = 5|- inches. 



428 



1N8TBUMENT8 AND MACHINES. 



19. 



Let AB''=^ 2om.y 
AQ=^x„ 



AR = X 



3> • • • 



A, p^y p„ the densities of the air 

in the receiver at first, and after one, 
two, &c. descents of the piston. 



Then p. 



E + b 



M 



II 




13 



13 



Ps^-^^p, &c. 



lO 



T^T pressure of air in BP BA 20 
pressure of air in BA BP ao — a?, ' 

/• • • T>Ti 20 X QO since the barometer stands 

.'. pressure of air in -fiP= ^- , . ^^ • . 

*^ 20— a?j' at30in.; 

.*. press, at ^ = *^ + a?.. 

^ 20 - ajx ' 

T» , . , pressure of air in the receiver _ Pi _ 1 1 

pressure of external air "" p ~ "" ^ 



10 



II 



.". pressure at .4 = — x 30 = 33. 



10 

TT 20X30 , 

Hence — h a?i = ^^, 

20 — a;, ^"^' 

(20 - x^) (33 - x^) = 20 X 30 ; .' 

Similarly, (20 — x^ (36 — x^ = 20 x 30 ; .' 

(20 - aj^) (39 - 0:3) = 20 X 30; .' 
i&c. = &c. 



x^ = 2-2318, 

x^ = 3'2ii74> 
&c. 



21. In the figure, let Pbe the position of the piston 
when the valve in it opens after r descents, 

AB = a, -4P= ajy. 

When the piston is at A after r descents, the density 
of the air in AB is expressed by 



=■ [rTi} p- 



A 

P 

I ■■ I ^ 



B 



INSTBUMENTS AND MACHINES. 429 

When the piston has descended to P, the density of the 
air in PB has increased till it equals the density of the ex- 
ternal air, after which the valve in the piston opens ; 



'• p'^ AB" a ' 



^^=^{'-"(ra)]- 



22. Let A, K be the altitudes of mercury at 
;8^ and P respectively. 

8G^x. 

rn^ pressure of air in AP AB K 
^"•^ AB =:iF=J' 

But pressure at P 

.'. a; = (A' — A) <r — ^ ; 



J? has therefore descended through 

h 



a 



X 







23. Let Fbe the content of the bell, V that of the part 
of the bell which is clear of water, h (= 53 ft.) the height of a 



430 



INSTRUMENTS AND MACHINES. 



column of water whose pressure is equal to that of the atmo- 
sphere ; and let x be the depth to which the top of the bell is 
sunk ; (a) the radius of the hemisphere. 

Then the pressure at any point of the surface of the water in 

the bell is the weight of a column A + a? 4- - , being as the depth 

below the surface of the water ; and the elastic force of the air 
being inversely as the space occupied, we have 



A: h + x + ^ = y:^=V' : V. 



(I). 



Now F=-iro», and V = '^ (6a-a)- = ^iraK 
Therefore, substituting in (i), we get 



33 5 33 + 






H 3 



, a K II 
.-. a?+- = =^ = ^-^ = 7^-6 ft. 



25- 



Let AB =2o{t., 

AC = 17 {ty 

AD = 16 ft. 



(i) When the range of piston is J9Z>. 

Let Pbe the greatest weight, and AP= a?, 

pressure of air in DQ ^ BP _ 20 — a; 
B^^RP^'16^^' 

But pressure in 5^ 4- a? = 33 ; 
.'. pressure mI>Q = ^ ^^^ ^ = ^^y 

^o X 33-53^ + ^' = 16 X 33 - 33a;, 

a?' — 20a? + 10' = — 4 X 33 + 100 = — 3a ; 

.'. a?= 10 ± V— 32, 




INSTRUMENTS AND MACHINES. 



431 



Hence, there is in this case no greatest height, and the 
water flows out. 

{7,) When the range of piston is BC. 

Similarly. (?^=^M^ = 33; 

.-. aj»-53aj + aox33 = i7x 33 -330?; 

.'. x^ — 20X + 10* = 100 — 99 = I ; 

.•. a:= io± I = II or 9. 

Hence -4P=iift. or 9 ft., according as the piston is at 
B or G respectively; the valve at A being open. 



26. Let -4(7= c, AEWiq range of the piston 



Let-4P=«, P being the surface of the water 
in the pump, after any stroke. 

Now, pressure of air in AP-\- weight of water 
column GP = pressure at (7 = pressure of atmo- 
sphere ; 

.*. pressure of air in -4P= a — (c — «). 
On the piston descending to E, 

pressure of air in EP __ AP _ z 
pressure of air in AP^ EP ~~ z — x' 

.'. pressure of air in EP=—^ . 

^ z — x 

Now this pressure must be greater than the 
pressure of the atmosphere, or the valve in the 
piston at E would not open, and the action of the 
pump has ceased. — 

Hence — ^ must be greater than a : 

« — a; ° 

.'. az — cz-\-z*>aZ'-'ax] 
.'. (ix> cz — «*. 
But cz—z* is a maximum, when 

c — !Z« = o, and .'. ;» = - c : 

2 



4Si INSTBinCENlS AND MACHINES. 

.•. 4MX must be > — c* ; 

4 

therefore the pump cannot work unless the length of the dtroke 
be greater thaii €^'7-40. 



27. The same distance is diyided in the three Thermometers 
into i8oPy locfy and So** respectivelj. Hence indicating the 
respecdTe Thermometers bj F^ (7, By we haye 

^ ' ^ ' ^ ' 180 100 80 

III ^ 

9 5 4 ^ ^o 

30. Let F be the volume of the bulb or part below the 
fireezing point. 

Then 180 x 'oooi F= expansion from freezing to boiling point 

.*. K= 5 = -^ — = ^'054 cub. m. 



EFFLUX OF FLUIDS FROM VESSELS. 



433 



HYDRODYNAMICS. 



Efflux of Fluids from Vessels. 



Ex. 9. 

1. Let V be the velocity of the fluid at the aperture ; 

area of aperture = - ('125)* = — ^ , 

volume of cylinder = -x5*xiiZ = J^ir. 

4 

Hence vxgox — -? = y^w, since the time of filling = 90 sec.; 



90 '^ 



3. 



Let AF= Xj, AB= 3, AN= 5. 




The issuing fluid describes a parabola 



y = x tan a — 



aj^ 



2^ > 



4A cos* a 



56 



434 



EFFLOX OF FLUIDS FBOM VESSELS. 

in which a «» o, PB or * = «, — 3, 
and when y = — 3, « = 5; 



•• - 3 = - 



5* 



4 («. - 3) ' 



.-. a.. = 3 + ^ = 5nft- 
Similarlj, a?,, x^ may be found. 



6. Let ABCD be the cylinder, Ml of water, and inclined 
at an angle BAN^ 6d^ to the horizon, 



HC 




L N 



AB^iift., AD=:BC = 5ft., the orifice at the middle 
of ^IB. 

The initial velocity of the issuing stream is .equal to that 
acquired by a particle falling down HO = OE+BK; 

.'. the impetus A = 9 sin 60® + 5 sin 30**. 
The equation to the parabolic curve described is 

y = x tan a r — , 



EFFLUX OF FLUIDS FROM VESSELS. 



435 



where a = 



P m, .^ 



-30^ y= 



V3 



Oi = — 9sin6o% LN=^x; 
I »• 



r- a — 



V3 



3(9=?+ 



S' 



.-.»» + i (27 + 5 V3) a^ = ^ (7^9 + 135 V3)> 

whence x is found =: 8*978, and then 
AN- AL + LN^ 9 cos 60^ + a; = 13*478 ft. 

9. Let -452) be the cone, ADAB-^d"^ AC =^7 ft. 

the orifice at a vertical depth AQ = z. 

Now the pressure of the ^ 

fluid at any point on the 
concave surface of cone beinff 
in the direction of the normal 
at that point, on making an 
orifice the fluid will spout in 
the direction OE perpendicular 
to AB. 

The ecjuation to the curve 
of the issumg stream is 

y = a?tana-^j^^^^, inwhicha = 450, h^z; 

and in order that the curve mav pass through B we must have 
2,r= OJV=-(7-0), mix=^NB=NO^y-z; 



/ Q 






7 


\\ 


^- " 




— — ^ 


^•* C 


;*■ ^^— -^ 



.-. -(7-«) = 7-«-^^*; 



I 

4»x- 



.'. 7 — « = 4j5; .'. « = i*4; 
.'. -40=-4Q sec45® = i'4X V^. 



12. Let AB be the cylinder standing on the top of the 
plane BOD : and let E be the point at wnich the orince must 
be made. 



} 



436 



EFFLUX OF FLUIDS FROM YESSEI^. 




Let .4^=0?, CZ> = y; 

v3 
Now CD'^j^AExEG, 

or y» = 4ajfio-ir + -^j; 

(2X\* 4 , 



.-. y = ;4{aJ±(30«-2«*)*}; 
.•. J52) =-{x± (30a! — aaj*) } = a maximum ; 

/. I± ^ ^^--^=0, 

7. (300? — aa?*)' 
whence 6aj* — 90a; = — %2^, 
which gives aj = 3'i7; 
the less value alone being admissible. 



15. Let ky k' denote the area of the orifice in the vessels ; 
and Xy y the altitudes ; then K being the area of the base in 
each case, we have 

time of emptjring first vessel 




and time of emptying second vessel 



^/^ 



-/f4- 



Now bj the question the times are equal, and 

k \ k' =^2 I ly or k^ %k' , 



EFFLUX OF FLUIDS FROM VESSELS. 437 

Therefore, substituting, we get 

/. a? : y = 4 : I. 

19. Let X be the depth of the orifice ; then the latus rec- 
tum of the parabola first described by the spouting fluid is 4x ; 
and the time of emptying a depth x is 

but jK" : A;= la : I, or K— i7,k; 

.'. time = 24 a/— : 
therefore the time of one vibration of the pendulum 




^9 
therefore, if ? be the length of the pendulum, we have 



\ g V 2g' 



whence Z = '^^^^ , 
hence, the ratio required 



7r» 



= 4X : 3— - = 7r* : 8. 

22. Let A; the orifice = 7ra% AC=^b, BE^'xB, 

AP^p, AQ^q, AB^h, 
GM=:m, GN^n. 

rr.1 ^ 

Then y = a; tan a 7 t" > 

^ 4A cos'a 

the equation to curve described, where a = o. 
Hence — J = , 

and — J = . 

Aa 



EFFLUX OF FLUIDS FSOH TESSEU. 




Now : 
and i 



Kf>/3A~ 



■ffzh- 



'•/oJi —^a\ 



■■■ t/lb. 



'^^'VZ' 



m n a — r 



,^f m , m — n 

'Jih = ~r= + — 7=- X 
V2i V2* 



I fma — nr\ 



VaAV 



EFFLUX OF FLUIDS FEOM VESSELS. 439 



^ ^ - 1 [ms — nrV 
/. AB or A = T I • 



S ^ T 






25. The general formula is 

and a cone being a solid of revolution, we have 

also, from the geometry of the figure, we get 

a? : y-5 :: a4 : 4 :: 6 : i, 

whence y' = -^^ ^-^ ; 









= - aJ* + 4oa;*-f iSooa;*; 
5 



therefore, between limits we have 



* V^ V24 ( 5 3 i 



1246 IT 



440 EFFLUX OF FLUIDS FROM VESSELS. 



28. 



= — ? JL [ 






'** ITX — Ql^ 



- ^. ^ r — ix 



w 



f4^|.2^J^ 



k^/2g\i 5 

= — 7=-.--^r* 
kW2ff 15 

_ 77rr' /2r\f 

31. Let a== the height of the segment; also let t, i be the 
times of emptying through the orifice in the vertex and base ; 
then 



k^zg] a,* k^2g} A ^ 



w 



h^%g 



Ki&raj* - a?*) da; (i) ; 



•^^-^v^/ '^'-lr'""'"' '^ 



'^~ib7^i^^^^^"^)^"^"'^^^"^)^*"'^^J^-'-(^)' 



therefore by the question, we have 



j ^{2ra^ - a^) dx :j {a {2r -- a) x"^ - 2 {r - a) x^ - a^} dx = 2, : 3, 

or -a»-4^a^ . ^ a* + 4 (r-a)t - i^a(2r-a)a* = i& : 3, 

whence, reducing, we get 

3a — lor : 8a — 2,or = 2 : ^, 



EFFLUX OF FLUIDS FROM VESSELS. 441 

whence 9a — ^or = i6a — 40^ ; 

10 
/. a = — r ; 

7 
therefore volume of segment : volume of sphere 



=?(«'-?) 



loor' ^ 4777^ 
49 T' 



2iJ X loor^ 

34. Let y = Ax^ be assumed as the equation to the required 
parabola. Also let a = the common altitude of the paraboloid 
and cylinder; then 

_ TT [y'dx 



k{a,g)^J x^ 



I' 



k^2a J 






^%g J x^ ks/ig ^^^l 

2 



airA' 



4«+i 



between a? = o and a; = a = 7 r->— r — .a * . 

{4n +i)k sj^g 

For the cylinder, we have 

K-=^TrA\ar, 

and i = -y— 7 X a*; 

.'. ^ : e' = I : 4n + I = I : 9 by the question, 

hence n = i^, 
and the equation to the parabola becomes 

y = Ax\ 

36. Let the equation to the ellipsoid be 

x^ y* z^ __ 
a' 6* c' 



56 



442 EFFLUX OF FLUIDS FBOM VESSELS. 

the axis of z being vertical. Then, for the velocity of the 
descending surface, we have 

di ~K ' 

putting the equation to the ellipsoid under the form 

^ I y I 



"■•(-D »•(-$) ' 



we have K= irab (i — jj ; 



«» 



, irah c* , 

.-. dt = r.— — — ?«« 



Jc{2fff{c--Z)^ 



TTOb 



kc" [ig) 

irab 
h^ {2g) 

irab 



(c + «) (c - «)* dz. 



Hence ^= yI {2c— (c — «)} (c — «)*cfe 



AjC«(25r) 

TTOJ /4 3 3 3 S _S\ 

T -c X a» X ca — X aa X c* 



_ i&irai /c\i 
~ 15k Kg) ' 



(ixal-ix.f) 



37. The solid being one of revolution, we have 



k^2g} rjfi 



h{2gY k{2gY 



EFFLUX OF FLUIDS FROM VESSELS. 443 

fl/x^dy = fy{2a — x) dx = yifia — xY + - / (ija — a?) » rfy, 

3 3 

^^ ^ -^ J x^ 3 5 

TT f , i 8 , \"^ 8 /o - X 8 i 2 At 

h{2gf^ i 3 3. 5 '^ 

,« fa; = otoa? = 2a, 
and irom i 

(y = o to y=^ira, 

we obtain, for the whole efflux, 



7ra2 



1c {gy 



h-|}=?^©'(-l)- 



38. Let Try' be any circular section of the vessel whose 
altitude is x ; and k the area of the orifice ; then the velocity at 
the orifice is 

and that of the descending surface is 

= — ^ aJx =• constant quantity ; 

•*• y' = V^ >< constant = ^/Ax suppose ; 

.'. y^ = Ax, 
the equation to a parabola of the fourth order. 

40. The velocity of efflux, being that due to an altitude of 
27690 feet, will 

= {2g X 27690)* 
= 1335*17 ft- nearly. 

42. Let p = the density of the external air, and therefore 
the density of the air in the receiver at first; and suppose j&to 
be the elastic force of the external air ; then the density of the 
air in the receiver after 15 descents of the piston is 

and its corresponding elastic force is 



444 



BESISTANCES. 



Now if v be the initial velocity, we have 
fj s {2g '^ X 27090}* 






= \y - (^) } X 64-3817 X 27690 1 

= {'877105 X 64*3817 X 27690}* 
= 1350*45 ft. nearly. 



Resistances. 

Ex. 10. 

1. Let C be the middle point of the rec- 
tilinear boundary of either face of the lamina, 
P any point in the curvilinear circumference 
of this face, and CA its axis. 

Let AG = a, zAGP^O, arc AP=s, t = 
the thickness of the lamina. Let P, Q denote 
the resultant pressures of the fluid on the la- 
mina, accordingly as -4 or (7 meets the stream. 

Then, t? cos ^ being the normal component of the velocity 
(t?) with which the fluid impinges upon the lamina at P, the 
component of the pressure upon an elemental rectangle rds of 
the edge of the lamina, parallel to A (7, will be equal to 

I I 

- p . rds . {v cos Oy. cos ^ = - prav^ cos^ d0. 

Hence, P being equal to twice the component of the pressure 
upon the quadrant AJB, parallel to A (7, we have 

P= pTav^ I cos3 0d0== prav^ I (i - sin'^) d sm0 




= prav^ f I j = -prav'. 



The pressure Q upon the base of the lamina will be equal to 



,3 



- p . Zra . V' = prav'. 
Hence P : Q = 2 : ^, 



RESISTANOES. 



445 




3. Let a be the length of the 
semuaxis major A (7, oh the length of 
the axis minor BB\ of either face of 
the lamina. 'Let C-4, CB, produced 
be taken as co-ordinate axes of x and 
y respectively ; let CM = a?, FM = y , 
AP=^8, Pp = ds^ p being any point 
of the curve AB indefinitely near to P. 
Then, t denoting the thickness of the 
lamina, and p the density of the fluid, 
the normal resistance on the portion 
I^ of the lamina, the vertex A being 
supposed to move foremost, will be equal to 

and, the component of this, parallel to A (7, will be equal to 

Hence, if P be the resultant resistance on the lamina parallel 
to A (7, we have 



dy' 



But from the equation to the curve 



we may get 



dx^ a^ 



dy^ h^'y-y^' 



y 



hence, integrating between the limits — y and +y, we have 






I , a5» f a' 



i**n la 

by * 



^-y}, 



446 BE8ISTAKCES. 

or, b being the value of y in the limit. 

But, if be the resistance when the motion of the lamina 
takes place with its base foremost, 

6. Let r be the radius of the sphere ; the centre being the 
origin of co-ordinates ; then the resistance 

Nowy' = »--a;», and| = (n-^)* 



=(-s)*=E' ! 









which, between the limits y = o, y = r, 

.*. resistance on the sphere = - TrpvV*, 
and the resistance on the circular plate is 

therefore resistance on sphere : resistance on plate = i : a. 



RESISTANCES. 447 

8. The sections of a cylinder made by planes parallel to 
the base are circles. Hence if R denote the resistance upon the 
diameter {d) of one of these sections, and I the length of the 
cylinder; the resistance upon the cylinder when moving in a 
direction perpendicular to its axis is 

3 

also the resistance npon the base of the cylinder when it moves 
in the direction of its axis is 

d' 

^4 d 

Id 4 

On comparing the latter resistance with the former, we obtain 
the ratio 

4 3 ■ SI 

11. Let 2a, a J be the axes of the generating ellipse ; v the 
velocity of the spheroid, and p the density of the fluid ; then 

dx _^a —y 

doc^ _ h^ — & V + g'y' _ b^ + aVy* 

.-. resistance = irpv^b^ l ^.T^if / 

_ irpd^y f (£ Ve' - aVy' •¥¥-}/) ydy 
~ aV j 6^ + aVy' 

_ irpv^b^ ( f {b^a^e^ + b^) ydy _ ) 

and between the limits y = o, y = &, 

resistance =^ |(^ + i) log (i + ^ - i} 






448 RESISTANCES. 

12. B7 the nature of the cycloid, we have 

dx sine , e fdxy ^6 

•'• 3" = — : 2 = tan - ; .'. i + ( 3- ) = sec* - . 

ay I+COS0 2 \dyj 2 

n 

Also ydy^a' {d + Brnff) (i +cos ^ d0=!ia^ {O + am 6) co8»-d9; 

.-. resistance = %irpv^a^ J{0 + Wi0) cos* - d0. 

I 
Now cos* - = 3 {cos a^ + 4 cos 5 + 3} ; 
a o 

.-./cos*- erfe = | |- (aesinae+cosae) + 4(5sin5+cos^ + 3^}-. 

Also fsind. cos*- <?e = --4 fcos^ -<Z cos - = — cos*-; 
^ 2 ^-^ a 2 3 a' 

•*. resistance 

. , /tfsina^ cosi^ flsin^ . cosfl ^^ 2 a0\ ^ 
'^ \ 16 32 2 2 16 3 2/ ' 

and taking the integral from = o, to d = 7r, 

• resistance = aw/w^a* J3_^ — I ^ 

and when the solid moves base foremost, 

resistance = - pv V x if'a^ ; 
therefore the resistances are as 

27r/wVr? — ^K—) : -/ot;'7r3a* = 97r"- 16 : I27r". 



449 



APPENDIX. 



Ex. 16 [20], p. 79. Companion, p. 91. 

Here the functions are, 

X =a^+ 24XS+ i2$x* - 376x3 - 1 726a5« + 5592a; - 4080, 
-STi = 3x5 +60054 + 2 50x3 -564a!>- 1726X + 2796, 
Xa=5aj4 +68053+ 5^x2 - 908X+ 1018*4, 
X3=529x3 + 3736x«- 1437074X+ 13934*87, 
X4= 84285 75xa - 29399I58-695X+ 25M304'35* 
-X's = 3295i383i7097o«-574925395<59293, 

X6=+, 

which shew that aU the roots are reaJ, three positive and three negative. The 
limits are 2 and - 1 + 

X Xi X.2 Xj JL^ X.^ JL(5 X JLj X.% JLj Jv| X^X6 

If x=o, -++ + + -+If x=o, - + + + +-+ three changes, 

x=i, - + + + +- + x=-i, " + + + + - + 

X =2, + + ++ + + + X=-2, - + + + +-+ 

«=-3» -+ + + + - + 

x=-4, + + + _ + 

x=-B, + + + - + 

x= -6, +- + + + -+ four changes, 
x=-7, + + + -+ 

x=-8, + + + - + 

«=-9» ++-- + -+ 

x= -10, - + --+-+ five changes, 

x=-ii, - + -_ + -+ 

X=-I2, -++-+- + 

a5=-i3> + - + - + 

x= -14, +- +-+ -+six changes ; 

.'. the positions of the roots are expressed hy 

{-14,-13}; {-10,-9}; {-6,-5}; {1,2}; {1,2}; {1,2}. 
In fact, the roots, worked out hy Homer's Method, are 

- 137459<5 ; -97445<5 ; - 5*74165 ; 174165 ; 

I 74456 ; I 74596 



THE END. 



67 



dnnlnlrgf: 



PRIVTBD BT O. J. OL^T. X.A. 
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Principles will be carefully explained, clearness 
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Now Heady. 

MENSURATION, An Elementary Trea- 
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Pembroke College, Professor of Mathematics, 
Boyal Staff College, Sandhurst. With nu- 
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" On the whole we are of opinion that Mr. 
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more, for Judging by the specimen J>efore us, we 
believe they are likely to prove valuable additions 
to the very few really aood School books which we 
have on mathematical «w/^Vet«."— Educational. 
Times, Jan. 9. 

ELEMENTARY HYDROSTATICS. By 
W. H. Besant, M,A., late Fellow of St. John^s 
College. 4«. 

ELEMENTARY TRIGONOMETRY. 

With a Collection of Examples. By T. P. 
Hudson, MA., Fellow of Trinity College. 
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ARITHMETIC. For the Use of Colleges 
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Harvey Goodwin, D.D., Dean of Ely. 
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ELEMENTARY DYNAMICS. By the 
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ARITHMETIC AND ALGEBRA. 



Arithmetic. For the Use of Schools and 

Colleges. By the Bey. A. WBIGLEY, M.A. 3«. %d. 

m 

Principles and Practice of Arithmetic. Com- 
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Dy Artificers, Gragers, and Land Surveyors. Designed for the Use of Students 
By the Bey. J. HIND, M. A., late Fellow and Tutor of Sidney Sussex College 
Ifinih Edition, with Questions. 49. 6<^. 

A Second Appendix of Miscellaneous Questions, (many of which haye been 
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*«* KET, with Questions for Examination. Second Edition, 6s. 



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With Answers. By the Rev. JAMES WATSON, M.A., of Corpus Christi 
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Fuhlished ly Detghtan, Bell, and Co., Cambridge. 



TRIGONOMETRY. 

Elementary Trigonometry. With a Collection 

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College. 3«. 6^. 

Elements of Plane and Spherical Trigonometry. 

With the Nature and Properties of Logarithms and the Construction and XJse 
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Solutions of the Trigonometrical Problems pro- 
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MECHANICS AND SYBRO STATIC 8. 

Elementary Hydrostatics. By W. H. Besant, m.a. 

Fcap. 8vo. 4*. 

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Treatise on Hydrostatics and Hydrodynamics. 

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WALTON, M.A; 8vo. 10*. Qd. 

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Examples of the Principles of the Differential 

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CONIC SECTIONS AND ANALYTICAL GEOMETRY. 

Trilinear Co-ordinates, and other methods of 

Mode rn A nalytical Geometry of Two Dimensions. By the Rev. "W. ALLEN 
WHITWORTH, M.A., Professor of Mathematics in Queen's College, Liver- 
pool, and late Scholar of St. John's College, Cambridge. 

Elementary Geometrical Conic Sections. By 

W. H. BESANT, M.A. [In the Press. 

Conic Sections. Their principal Properties proved 

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Elements of the Conic Sections. With the 

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Solutions of the Geometrical Problems, consisting 

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"With an Appendix containing several General Properties of Curves of the 
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ASTRONOMY, ETC, 

Elementary Treatise on Astronomy. For the use 

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Examination in the Senate-house. By P. T. MAIN, B.A., Fellow of St. 
John's College. 8vo. 7». 6rf. 

Practical and Spherical Astronomy. For the Use 

chiefly of Students in the Universities. By the Rev. R. MAIN, M.A. 
RadcHffe Observer at Oxford. 8vo. 14». 

Brunnow's Spherical Astronomy. Part I. In- 
cluding the Chapters on Parallax, Refraction, Aberration, Precession, and 
Nutation. Translated by the Rev. R. MAIN, M.A., F.R.S., Radcliflfe 
Observer at Oxford. 8vo. 8». 6<f. 

Elementary Chapters on Astronomy from the 

" Astronomic Physique" of Biot. By the Very Rev. HARVEY GOODWIN, 
D.D., Dean of Ely. 8vo. 3*. M. 



FuhUshed hy Deighton, Bell, and Co., Cambridge. 7 

Exercises on Euclid and in Modern Geometry, 

containing Application s of the Principles and Processes of Modem Pure 
Geometry. By J. McDOWELL, B.A., Pembroke College. Crown 8vo. 8«. 6rf. 

Elementary Course of Mathematics. Designed 

principally for Students of the University of Cambridge. By the Very Rev. 
HAEVEY GOODWIN, D.D., Dean of Ely. Fifth Editum. 8vo. 15*. 

Problems and Examples, adapted to the 

<< Elementary Course of Mathematics.'* With an Appendix, containing 
the Questions proposed during the first three days of Uie Senate House 
Examination. By T. G. VYVYAN, M.A. Third Editum, Svo. 5». 

Solutions of Goodwin's Collection of Problems 

and Examples. By W. W. HUTT, M. A., late Fellow of Gonville and Caius 
College. Third Edition, revised and enlarged. By the Eev. T. G. VYVYAN, 
M.A. Svo. 9». 

Newton's Principia. First Three Sections, with 

Appendix, and the Ninth and Eleventh Sections. By the Rev. J. H. EVANS, 
M.A. Fourth Edition. Svo. 6«. 

Examples in Arithmetic, Algebra, Geometry, 

Logarithms, Trigonometry, Conic Sections, Mechanics, &c., with Answers 
and Occasional Hints. By the Eev. A. WRIGLEY, M.A., Professor 
of Mathematics in the late Boyal Military College, Addiscombe. Sixth 
Edition, corrected. Svo. 8«. 6<f. 

A Companion to Wrigley's Collection of Ex- 
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Methods of Solution. By J. PLATTS, Esq., Head Master of the Government 
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Figures illustrative of Geometrical Optics. From 

SCHELLBACH. By the Rev. W. B. HOPKINS. Flatee. Folio. 10a. ^. 

A Treatise on Crystallography. By W. H. 

MILLER, M.A. Svo. Is.M. 



A Tract on Crystallography, designed for Stu- 
dents in the University. By W. H. MILLER, M.A., Professor of Mine- 
ralogy in the University of Camhridge. Svo. 6«. 

Physical Optics. Part 11. The Corpuscular 

Theory of Light discussed Mathematically. By RICHARD POTTER, M. A., 
late Fellow of Queens' College, Camhridge, Professor of Natural Philosophy 
and Astronomy in University College, London. 7». 6<?. 



CLASSICAL. 

^schylus. Translated into English Prose, by 

F. A. PALEY, M.A., Editor of the Greek Text. 8vo. 7». 6rf. 

Aristophanes. ComoediaB Undecim cum Notis 

et onomastioo. By the Bey. H. A. Holdbk, LL.B., Head Master of 
Ipswich Grammar School, late Fellow and Assistant Tutor of Trinity CoU^e, 
Cambridge. Second Edition, 8yo. \b». 

The Plays separately, 1«., Is. &f. and 28. each. Notes, 4«. 

Demosthenes. The Oration against the Law 

of Leptines. With English Notes and a Translation of Wolfs Prolegomena. 
By W. B. BEATSON, M. A., Fellow of Pembroke College. Small 8vo. 6». 

Demosthenes de Falsa Legatione. Second Edit 

carefully revised. By B. SHILLETO, M.A. Svo. S*. 6<f. 

Demosthenes, Select Private Orations of. After 

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With English Notes. For the use of Schools. By C. T. PENROSE, A.M. 
Second Edition, 12mo. 49. 

Euripides. FabnlaB Quatuor. Scilicet, Hippo- 

lytus Coronifer, Alcestis, Iphigenia in Aulide, Iphigenia in Tauris. Ad fidem 
Manuscriptorum ac yeterum Editionum emendayit et Annotationibus instnudt 
J. H. MONK, S.T.P. Editio Nova, Svo. 12«. 

Separately — HnppoLYTUs. Svo. cloth, 6». Alcestis. Svo. sewed, 4*. 6<?. 

Titi Lucreti Cari de Rerum Natura Libri Sex. 

With a Translation and Notes. By H. A. J. MUNRO, M.A., Fellow of 
Trinity College, Cambridge. 2 vols. Svo. 21s. 

Plato's Gorgias, literally translated, with an 

Introductory Essay containing a Summary of the Argument. By E. M. 
COPE, M.A., Fellow of Trinity College. Svo. Is. 

Passages in Prose and Verse from English 

Authors for Translation into Greek and Latin; together with selected 
Passages &om Greek and Latin Authors for Translation into English : forming 
aregular course of Exercises in Classical Composition. By H. ALFORD, M.A., 
late Fellow of Trinity College, Cambridge. Svo. 6*. 

Plautus. Aulularia. With Notes, Critical and 

Exegetical, and an Introduction on the Plautian Metres and Prosody. By 
Dr. WM. WAGNER. [In the Ftess. 

Plautus. Menaechmei. 

Ad fidem Codicum qui in Bibliotheca Musei Britannici exstant aliorumque 
nonnullorum recensuit, Notisque et Glossario locuplete instruxit J. HILD- 
YARD, A.M. Editio Altera. Is. ed. 

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Sophocles. (Edipus Coloneus. With Notes, 

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opposed to conjectural emendation. B7 the Rev. 0. E. PALMER, M.A. 9«. 

Tacitus. Opera, ad Codices antiquissimos exacta 

et emendata, Commentario critico et ezegetico illustrata. 4 yols. 8yo. 
Edidit F. EITTEB, Prof. Bonnensis. PubUshed at 1^. 8s., reduced to Us, 

Theocritus, recensuit, brevi commentario in- 

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**It is the best as well as the briefest of Commentaries on the Syracusan bard; and 
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unattempted, and few unsolved." — Saturday Review. 

Arundines Cami. Sive Musarum Cantabrigien- 

sium Luflufl Canon. Collegit atque edidit HENRICUS DRURY, A.M., 
Archidiaconus Wiltonensis Collegii Gaiani in Graecis ac Latinis Literis quon- 
dam Prsslector. Equitare in arundine longa. Editio Sexta. Curavit 
HENRICUS JOHANNES HODGSON, A.M., Collegii SS. Trinitatis quon- 
dam Socius. Crown Syo. 7«. 6^. 

Foliorum Silvula. Part I. Being Passages for 

Translation into Latin Elegiac and Heroic Verse. "With Notes. Edited 
by H. A. HOLDEN, LL.D., late Fellow of Trinity College, Head Master of 
Queen Elizabeth's School, Ipswich. Third Edition. Post 8vo. 7«. 6rf. 

Foliorum Silvula. Part II. Being Select Passages 

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Passages for Translation into Greek Verse. Edited with Notes by H. A. 
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Foliorum Centuriae. Selections for Translation 

into Latin and Greek Prose, chiefly from the University and College Examina- 
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Progressive Exercises in Greek Tragic Senarii, 

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prefaced by a short Account of the Iambic Metre and Style of Greek Tragedy. 
For the use of Schools and Private Students. Edited by B. H. KENNEDY, 
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Platonis Protagoras. The Protagoras of Plato. 

The Greek Text revised, with an Analysis and English Notes. By W. 
WAYTE, M.A., Fellow of King's College, Cambridge, and Assistant Master 
at Eton. 8vo. 69. Od. 

Dissertations on the Eumenides of r^schylus, 

from the German of C. 0. MULLER. "With Critical Remarks and an Appendix. 
Translated from the German. Second Edition. 8vo. 6*. 6rf. 



10 Published hy Beighton^ Bell, and Co,, Cambridge, 

WORKS BY THE LATE J. W. DONALDSON, D.D. 

A Complete Latin Grammar, Second Edition. 

Very mach enlarged, and adapted for the use of TTniyersity Students. 
8to. 14«. 

Th0 mlarged Edition of the Latin Grammar has been prepared with the same 
object at the eorreeponding work on the Greek Language, It is, however, 
eapeeiaUff deeigned to serve as a convenient handbook for those students who 
wish to acquire the habit of writing Latin; and with this view is furnished 
with an Antibarbarus, with a full discussion of the most important synonyms, 
and with a variety of information not generally contained in works of this 
description, 

A Complete Greek Gram^mar. Third Edition. 

Yearj much enlarged, and adapted for the use of Uniyersity Students. 
8yo. les. 

This enlarged Edition has been prepared with the intention ofplaciny within 
the reach of Students at the Universities, and in the highest classes at Schools, 
a Manual of Instruction and Reference, which, without exceeding the limits 
of the most popular works of the kind, would exhibit a more exact and 
philosophical arrangement of the materials than any similar book ; would 
connect itself more immediately with the researches of comparative Fhilologers; 
and would contain the sort of information which the author's long experience 
as a teacher and examiner has indicated to him as most likely to meet the 
actual wants of those who are engaged in the critical study of the best Greek 
authors. 

Without being formally based on any German work, it has been written with 
constant reference to the latest artd most esteemed of Greek Chrammars used on 
the Continent 

Index of Paoages of Greek Anthers qnoted or referred to in Dr. Donaldson's 

Greek Grammar, prioe 6<2. 

Varronianus. A Critical and Historical Intro- 
duction to the Ethnography of Ancient Italy and to the Philological Study 
of the Latin Language. Third Edition, revised and cor*siderably enlarged. 
8vo. 16«. 

Ind^tendenily of the original matter which will be found in almost every page, 
it is believed that this book presents a collection ^ known facts respecting the 
old languages of Italy which will be found in no single work, whether British 
or Foreign, and which must be gleaned from a considerable number of rare 
and expensive publications ; and while the lists of Oscan and Etruscan glosses, 
and the reprint of fragments and inscriptions, may render the treatise an 
indispensable addition to the dictionary, and a convenient manual for the 
professed student of Latin, it is hoped that the classical traveller in Italy will 
find the information amassed and arranged in these pages, sufficient to spare 
him the trouble of carrying with him a volitminous library of reference in 
regard to the sul^'ects of which it treats. 

The Theatre of the Greeks. 

A Treatise on the History and Exhibition of the Grreek Drama : with yarious 
Supplements. Seventh Edition, revised, enlarged, and in part remodelled; 
with numerous iUustrationB from the best ancient authorities. Syo. 14«. 

Classical Scholarship and Classical Learning con- 
sidered with especial reference to Competitive Tests and University Teaching. 
A Practical Essay on Liberal Education. Crown Svo. 5s. 



Published ly Deighton, Bell, and Co,, Cambridge, 11 

The Greek Testament : with a Critically revised 

Text; a Digest of Tarious Beadings; Marginal References to Verbal and 
Idiomatic Usage ; Prolegomena ; and a Critical and Ezegetical Commentary. 
For the Use of Theological Students and Ministers. By HENBY ALFORD. 
D.D., Dean of Canterbury. 8vo. 

Vol. I. FIFTH fiDinoN, Containing the Four Gk)8pels. 1^. 8«. 

Vol. II. FOUBTH EDITION, Containing the Acts of the Apostles, Epistles 
to the Bomans and Corinthians. 11. 4«. 

Vol. III. FOu&TH EDITION, containing the Epistles to the Gklatians, Ephesians, 
Fhilippians, Colossians, Thessalonians, — ^to Timotheus, Titus, and Phile- 
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Vol. IV. Part I. THIRD EDITION. The Epistle to the Hebrews: The 
Catholic Epistles of St. James and St. Peter. 188. 

Vol. rV. Part II. SECOND edition. The Epistles of St. John and St. 
Jude, and the Bevelation. 14«. 

Companion to the New Testament. Designed 

for the use of Theological Students and the Upper Forms in Schools. By 
A. C. BABBETT, M.A. Fcp. 8vo. 5*. 

An Historical and Explanatory Treatise on the 

Book of Common Prayer. By WILLIAM GILSON HUMPHBY, B.D., late 
Fellow of Trinitv College, Cambridge ; Vicar of St. Martin's in the Fields, 
Westminster. Third Edition, revised and enlarged. Fcp. 870. 4«. 6d. 

Annotations on the Acts of the Apostles. De- 
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Papers. By the Bey. T. B. MASKEW. Second Mition, enlarged. 12mo. 5«. 

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