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360°, the elementary triangles yy © contain those already computed.

It would be easy to supply this defect by calculating the elementary trapezia comprised between two adjacent spires. The same inconvenience will be found to occur in the common formula fuč, if several ordinates correspond to one abscissa. Ex. VI. In the hyperbolic spiral a:-:::y: Mrs. Yk;

A therefore COMC = is

-YYX

a

a

Now cy=ab, therefore Yy4=by,

K

B:

a

and the space comprised between the curve and two ordinates = by+C.

ON THE RECTIFICATION OF

CURVES.

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111. If we conceive the point m to be taken indefinitely near to M, Mm will be the fiuxion of the arc AM, and considering this as a right line, we shall have flux AM = v(x?+ya). Consequently AM=fvi?+y)+C; a formula which equally applies, whether the ordinates are parallel to one another, or whether they proceed from a fixed point.

112. Example I. In the circle y = v(aa---2); and y =

n

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and 2

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ax

Di

کی

A

B

MB=y+*+

+ &c.

1

1.3 QM-S

=X+

+
(aa-xx)

2 Jan. 2.4
1.3.5 x?
+

+&c.; therefore the arc 5 a4 ' 2.4.6 7 ao

1 y3

1.3 ys

2 3aa' 2.5 5a4 Example II. In the parabola AM = Sy (1+*99)=824V (pr+yy). Now we

=PP) PP р

2 have shewn (71) thal. Si N rx + aa =

Q C+ x^(2x +aa) + aal (r+ Nxr+aa) Therefore AM=C+?n (yy +$ 11') +

P 1 pl {9+v(99+£rp)}. Make y=0,

A

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m

PP

and we shall find C=- pl } p. Consequently AM = y

P (94+ pp)+4 p2 (4+ Vyy + PP.

p 113. It may be remarked, if from A as a centre, and with the semi-transverse axis BA=} P, we describe an equilateral hyperbola BN, the space ABNQ will be

(yy + pp). Therefore AMX) p = ABN'Q; from which

PP it follows that the rectification of the parabola depends upon the quadrature of the hyperbola, and reciprocally.

Ex. III. In the ellipse, if we suppose the semi-transverse axis =l, we shall have y=6V (1-xx); and making (1-6), or half the dis

N' tance between the foci =c, we obtain

A

n

D

M

B

BM=s&V(1—ccx.x)

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B

a

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1.3

an integral which (1-2x)

A

A cannot be obtained by the preceding rules. We must therefore reduce the expression into a series ; but for the pure pose of simplification, we shall only re

N'

D duce (i--ccxx); we shall then find

1 BMEA

1.3
(1-} cc XC-
c4 x4

1.3.5

c8x8-&c.) V (1-xx)

2.4 2.4.6 2.4.6.8
xx à
1
It

25
-cos
c4 S

co/

-&c (1-rr) (1-xx) 2.4 (1-xx) 2.4.6

N (1-xx)

1

(1By (61) it appears that S

--XX ✓(1-2X)

2i (2i –1) r-5 (1--rx

"

(21-1) (-3)...3 -&c.

x (1-4X ,
2i (22—2)

2i (22-2)...2
1.3.5...(21-1)
+

And therefore
2.4.6...i

VI-XX) 304

c BM =(1

3.5 сб. 32.52.7 CS

- &c.) S 2.2 2.4? 29.4*.62 22.44.6.8

(1-**)

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(ا-2

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.

+c*: (1==3){ +
+c* *° (1 –xx)} {+&c.}

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+ .

3 c2 32.5 c 22 22.42 22.42.62

+&c.} 1 3.5 c 3.52.7 c+

+

2.42' 2.42.6% 2.49.64.82 + &c. But the arc DN=S

.

consequently we know all the

(1-IX quantities that enter into this series, of which the law is easily perceived. Let x=1, and we shall have c 3 c4 32.5 c6

-&c.) AND. Therefore the pe2*

22.42.6* riphery of the ellipse is to that of the circumscribed circle as 1 c 1.1% SCA 1.1%.32 5 c

-&c. : 1 (callingthe semi

at 24.4*.62 as transverse axis a). This series will be very convergent when the

1 foci are near to one another. For example, if c= = a, the circum

10 ference of the ellipse will be to that of the circumscribed circle as 0.997495 292 861 261 : 1

The rectification of the 'hyperbola may be found in nearly the

AMB = (1

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.

rame manner.

Consequently S V(*+*") =sýv(1 + 1 (1++c

27

Ex. IV. The equation of the second cubical parabola is y'=as'.

8 Sy

a 4a

8 Making y=0, we have C=

a,

and any arc of this curve come

27 puted from the vertex=8

=.
8

1

27 E

998–1

a

{(1+-1} Ex. V. In the cycloid j=iviams). Therefore vír*+j)=id

and taking

;

the fluent, we obtain AM=2 ax=2 AN by (37).

Ex. VI . In the logarithmic curve ya=ay, and wid+ja) = 3

(+*}

(yy+aa). Let (yy+aa)=2, and we shall have yy=zz-, and y

ze

and again (ca + y) = x+. of which the

aaz

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22-aa

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ord (aa+yy)—al {a+v(qa-+ xy)} +C;

2

integral is z+

yy
1
z ta

y the expression for any arc of the logarithmic curve, in which C is easily determined.

Ex. VII. In the spiral of Archimedes, if we make

B AGFBN=x, CM = y, we we shall have Mr =

yx

D

M

कार

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P

(99 +

+ If we describe a parabola

2a* CN, whose parameter

ot by making CQ=CM, and drawing the ordinate QN, we shall have CNESS veri+yy). Therefore CN'=COM. From which we may conclude, that there exists some analogy between the Archimedean spiral and the parabola.

Ex. VIII. In the hyperbolic spiral, the arc COM= 2(6b+yy).

y Therefore if we describe a logarithmic curve NK, whose subtangent =b= to that of the spiral, we shall have MOC= the infinite arc NK, by taking the ordinate NR=CQ= CM. But if we desire an expression for any arc of the spiral, or of the logarithmic curve contained between two ordinates y, y, we shall find it to be ✓ (6b+yy)-V (66+44')+614 (6+Vb6+yy

y (6+156+y'y

=s ,V66

A

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