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a

2

Hence the side (s) of the octagon being the chord of a, is expressed by S = 2 sin.

= 2 sin.

= 72

4 =R xv 2, if instead of being = unity the radius of the sphere is = R. Hence the area of each face is

R XN2

R? 2R'.

x

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R2

3, and the distance of the face from the centre, or the 2 altitude of the equal prisms whose bases are the equal faces of the octaëdron and common vertex at the centre, is

2 R?

R2 = R X
Hence each of these prisms

R?
J 3X RX

R3,

and the 3

6 whole octaëdron is therefore expressed by

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X

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2

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174. Let 6 be the radius of the base common to the hemispheroid (H) and paraboloid (P), and A their height, then H : P :: 4 : 3.

6
For H=". Sy’dx = -S (2ax – x2) de

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175. Let PQ (Fig. 69,) the transverse axis of the ellipse = 2a, and its conjugate (to be determined) = 2b, and put A ABC = T, A APQ = T, cone ABC = C, and segment APQ = C' then C:C:: T}: TH.

For since PQ is the axis of the ellipse, its plane is I plane ABC which passes through the axis of the cone.

Hence pm being the intersection of the o B'C' with the ellipse PQ, and B'C' that with the plane ABC, it is easily shewn that

pm – Bʻm x mc'.
But Pm : B’m :: sin. B : sin. P ?
and mQ : m :: sin. B : sin. QS

sin. P x sin. Q
:: Bộm x mC' = Pm x mQ X

sin.' B

T = Pm x mQ X

T

= pm', p and po

pa being the perpendiculars let fall from A upon BC and PQ respectively.

Let mQ = Pm = 1.26 = 6.
Then

pe

T TT
18 = ax

p>
T

p.
and the area of the ellipse is :: expressed by

TT.

р Now it is evident that may be decomposed into triangular prisms of the same altitude p, and the limit of the sum of whose bases = the base of C', hence

TT =

1.t

T! T..... (1) р

Зр

a

mab =

C =

Х

VTT

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Let T = T, then C = C, and we have

(2) Зр ::C: 0 :: T: The Again, to find the equation to the surface of the cone, we will suppose the rectangular axes of x, y, 2, to originate in its vertex, and the axis of z to be that of the cone itself, as in Fig. 69. Then, L being any point in the surface, let OLA be that position of the generating 4 in which it passes through L, and draw LN I AZ, LL' I plane of (x, y,) and LM I AX. Let, therefore, AN = 2 , AM = x, and L'M = y, and we have

A
Z:NL = AL = V x2 + y® :: OA : 00 :: 1: tan.
A being the angle at the vertex of the cone.
.. Z x tan. A =

= V x2 + y2 the equation of the surface of the cone, whose rectangular co-ordinates originate in its vertex, and that of z coincides with its axis, which, in practice, will be found the most commodious way of conșidering the question.

2

THE RECTIFICATION OF CURVES.

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176. This is a particular case of the Theorem of Fagnani, which we will first demonstrate generally.

Let 1, and b, be the semi-axes of the ellipse, and e its eccentricity. Then the equation of the curve referred to the centre by rectangular co-ordinates x, y, is

y= b 1-X. Hence, denoting by E, that part of the elliptic arc measured from the extremity of b, whose abscissa is x, we have

dy

e x2 1+

= 5 dx2

x2 and putting

- U2

XVI-X u N1-u? e?u?

NI - eRx? NI

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or

we get

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e%22 d E t du.

=ed. u 11 - 22 u

„1 - elus .. E, + E. =

e u VI

+ C.

N1 - eu3
But u = 1 when * = 0.:: C=E,
and E, + E,-E,=
eau 1 - 2

.. (1)

vi- eu? which is the theorem of Fagnani.

1

(as we readily find by substituting 7176

u? -e?u?

Let u ==

Then equation (1) becomes

e ? 2E E, =

1+6 ito which solves the problem.

1

= 1 - b,

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US

1

177. Let 2, 26, and e, be the transverse axis, conjugate axis, and eccentricity of the ellipse, then its equation, referred to the centre, is expressed by

y = b 1-2 and the length of any arc by

dya

ex dx

1Let

or que N17 x

1+ Then

1
1.- doa =

1 + bor?
s exca
1+ u'

1+u
du
and dx =

(1 + u^)
Hence

du . N I + b*u*

(1 + ?) which, by putting

+ b2u2 부

1+ and making the requisite substitutions transforms to

6*u'du
d Ez =dP

(3)
(1 + u) (1 + b*u*)
Again, make

1 + ua

1 + bu?
Then
b'v? - 1+11+2. (2 6°) v* + bo*,

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