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and x =

yo

A
9a sin.
A =

2
2a sin.
2

2
which determine the position of the parabola.

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149. Let SP = p, and the traced by it = 0, then by the question

0 oC p = apa and we hence obtain the area S pw+1 dp =

P Р
2

2 (m + 2)
0
mt? mte

OC 0 m 2 (m + 2) Let A = 26, 47, 67 ...... 2nt successively, then the corresponding areas, described by 1, 2, 3, ..... n revolutions of p are evidently as +g m+2

m+2 1,2,1

3

ma

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m-+2

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....... n

150. The equation to the conchoid referred to the centre C(Fig. 65,) of revolution of its generating line p, is

6

ta

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62

A

al tan. 0 + ab . l. tan. + + + C 2

2

2 See Appendix to new edition of Simpson, p. 335, or Lacroix.

Let 6 = 0. Then

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A BCQ = * BQ = V6–a) •– bo

62

tan. 0,

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Hence
ABNP = ACP - BCQ + PQN

a? a
= ab I. tan. + + 0+ sin, 2 0.

This may also be expressed in terms of PN = y, or BN = d.

For y = a.cos. 8
:. sin. 0 = Na? - yo

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cos.-ay+ y.way?

:. ABNP = ab I. a + va 9% +

y a result which is the same as that deduced by Simpson in his Fluxions, p. 147, from the equation

xoy2 = (b + y) x (a- y2.)

151.

Let d, d', be the diagonals of the trapezium, & their inclination, m, m', the segments of d'. Then, since the trapezium

= sum of two A, whose common base is d =

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m'd sin. 6

2

m+m
2

d sin. 6

dd'

sin. 0. 2

152.

m2

m

(as it

Let x, y, be the co-ordinates of the hyperbola Vv, (Fig. 33,) originating in the centre A, and measured along the asymptotes, then by property of the curve

XY = const. = m, the equation. Let BC touch the curve in P, then the subtangent MC may easily be proved,) =

ydx

Х dy

yo = x= AM

Hence PB = PC, and A ACB = 2 AMPN = 2xy X sin. A = 2m sin. A = a const. Q. E. D.

A geometrical demonstration may be seen in most books on conic sections.

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153. Let the radius of the circle CA (Fig. 66,) = 1, AN = x, PN = y; then by similar A

Dn y= (1 – x)

(1 - x) tan. vers.-1x).
Сп

X = 2z,
then x = vers. 2z = 1 - cos. 2z = 2 sin. 2,

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Let vers.

-1

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2x— ?

the equation to the curve. Hence fydx = $dx 2x – 2+ +

(1-r) dir

2.0 x2
= area CDĄ + 2x - x? vers. - 17.
Let x = CA = 1. Then
Sydx = ACQ + 1 - ABQ R CA

= CA ACQ.

154. Let A, B, C, be angles of the A, of which make B = 0, A = * 30, the side AB = a, and BC = p; then pia :: sin. 30 : sin. (*

30 + 0) a sin. 30

(1) sin. 28 the polar equation of the locus of C.

pido a ? do sin.' 30 .. Area =

2

2

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cos."

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156. Let a, b, C, denote the two sides of the A plane or spherical, and C the included Z; then in the former case, the

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Now in the latter, A, B, being the other two L, it is well known that the area

= A + B + C But by Napier's Analogies

- 6

COS.
A + B
tan.

x

2 2

cos.

= cot.

a+b

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COS.

2

COS.

:. the spherical A, T is given by the equation

- 6

с
T=C+ 2 tan.-- {cot.

2
a+b

2
Another expression for T is

b
cot. cot.

+ cos. C
T
Cot.

2
2

sin. A which the reader will have no difficulty in investigating.

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2

157. Let ABC (Fig. 67,) be either end of the oblique triangular prism A'C, and (supposing the prism to be produced,) let Cbc be a section made by a plane I its axis, and :: to its sides ; then NCD being the intersection of the planes ACB, bcC, and D that of the lines bc, ND, if through Ab, Bc we make planes to pass IND, the LINA, CMB between the intersections of these planes with those of ACB, 6Cc, will each be the inclination of the plane ABC to the plane 6Cc, i.e., the complement of the inclination of ABC to the axis, (since it is I, by construction, to 6Cc). Let this L=

2

0.
2

CD X AN
Now A ACD =

2

and ABCD =

CD 'X BM

VOL II.

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