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Hence the value of the curve, from the fluent of each of these, gives the four following forms, in series, viz. putting d = 2r the diameter, the curve is

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3.5.73 %= (1 + +

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ť

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Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the square of it, is known to be some small simple number. Now, the arc of 45 degrees, it is known, has its tangent equal to the radius; and therefore, taking the radius r = 1, and consequently the tangent of 45°, or t, = I also, in this case the art of 45° to the radius 1, or the arc of the quadrant to the diameter 1, will be equal to the infinite series 1 Št}+ &c.

But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc of 30 degrees, the tangent of which is = v's, or its square t = 4: which being substituted in the series, the length of the arc of 30° comes out

1 1 1 1 (1.

+ &c) / . Hence, to com3.3

5.32 7.33 9.34 pute these terms in decimal numbers, after the first, the sueceeding terms will be found by dividing, always by %, and

quotients again by the absolute numbers 3, 5, 7, 9, &c; and lastly, adding every other term together, into two sums, the one the sum of the positive terms, and the other the sum of the negative ones; then lastly, the one sum taken from the other, leaves the length of the arc of 30 degrees; which being the 12th part of the whole circumference when the radius is 1, or the 6th part when the diameter is 1, consequently 6 times that arc will be the length of the whole circumference to the diameter 1. Therefore, multiplying the first term by 6, the product is ~ 12 = 3.4641016, and hence the operation will be conveniently made as follows:

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So that at last 3•1415926 is the whole circum

ference to the dia

meter 1. glass, Exam. 2. To find the length of a parabola.

EXAM. 3. To find the length of the semicubical parabola, whose equation is ax* = y3.

Exam. 4. To find the length of an elliptical curve:
Exam. 5. To find the length of an hyperbolic curve.

OF QUADRATURES; OR, FINDING THE AREAS OF

CURVES

62. The Quadrature of Curves, is the measuring their areas, or finding a square, or other right-lined space, equal to à proposed curvilineal one.

By art. 9 it appears, that any flowing quantity being drawn into the Auxion of the line along which it flows, or in the direction of its motion, there is produced

al the fluxion of the quantity generated by the flowing. That is, dd X de or ya is the fluxion of the area ADE. Hence this rule.

RULE.

Y 2

Hence the value of the curve, from the fluent of each of these, gives the four following forms, in series, viz. putting d = 2r the diameter, the curve is

3.82 3.5.x3
(1 +
2.3d
+ +

+ &c) v dr,
2.4.502 2.4.6.7413
guz 3g*
(1 to

3.57%
+

+ 2.3r2 2.4.574

th t6 7 (1 +

+
3r2

- &c) t,
584 76
93, –

+
2.353

+ &c)r. 2.4.555

2.4.6.7.20+ &c)y,

998

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Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the

square

of it, is known to be some small simple number. Now, the arc of 45 degrees, it is known, has its tangent equal to the radius; and therefore, taking the radius r= 1, and consequently the tangent of 45', or 1, = I also, in this case the art of 45° to the radius 1, or the arc of the quadrant to the diameter 1, will be equal to the infinite series 1 - +}-}+;-&c.

But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc of 30 degrees, the tangent of which is = v's, or its square t = 4: which being substituted in the series, the length of the arc of 30° comes out 1 1

1 (1

+ &c) V. Hence, to com 3.3 5.32

7.33 9.34 pute these terms in decimal numbers, after the first, the sue ceeding terms will be found by dividing, always by 3, and these quotients again by the absolute numbers 3, 5, 7, 9, &c and lastly, adding every other term together, into two sum the one the sum of the positive terms, and the other the sur of the negative ones; then lastly, the one sum taken from the other, leaves the length of the arc of 30 degrees; whic. being the 12th part of the whole circumference when ti. radius is 1, or the 6th part when the diameter is 1, consquently 6 times that arc will be the length of the whole ci cumference to the diameter 1. Therefore, multiplying t. first term v } by 6, the product is ✓ 12 = 3•4641016; a hence the operation will be conveniently made as follows

+ Ter

1

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of a circle

DE the ordiCH.. TINTA LIDE

e diameter. BE,

SO, A = = = 21626. Tekil, Sajuki, O Oute TE-to spect, eget the given equation

1 HE Erasuring times is the fluxion of

ound, and substituted .uent of the expression, of the solid required.

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e of a sphere, or of any seg

RULE.

63. From the given equation of the curve, find the value either of ic or of y; which value substitute instead of it in the expression gt; then the fluent of that expression, being taken, will be the area of the curve sought.

EXAMPLES.

EXAM. 1. To find the area of the common parabola.

The equation of the parabola being ax = y*; where a is the parameter, x the absciss AD, or part of the axis, and y the ordinate DE.

From the equation of the curve is found y=vax. This substituted in the general fluxion of the area yi gives */ ax or aixíš the fluxion of the parabolic area; and the fluent of this, or a*m= kit Vax = xy, is the area of the parabola Ade, and which is therefore equal to of its circumscribing rectangle.

Exam. 2. To square the circle; or find its area. The equation of the circle being gé = ax — **, or y = Wax - x?, where a is the diameter; by substitution, the general Auxion of the area yx, becomes xv ax - 3?, for the fiuxion of the circular area. But as the fluent of this cannot be found in finite terms, the quantity Vax - 72 is thrown into a series, by extracting the root, and then the fluxion of the area becomes

1.3.713 1.3,5x4 * Vax (1

&c); 2a 2.44% 2.4.6a3 2.4.6.894 and then the fluent of every term being taken, it gives

2 1.x luz? 1.3.23 1.3.5.24 tvar x 15 ๔

&c); 5a

4.7a? 4.6.903 4.6.8.1 lat for the, general expression of the semisegment ADE..

And when the point D arrives at the extremity of the diameter, then the space becomes a semicircle, and x = a; and then the series above becomes barely 1 1.3 1.3.5.

&c) 5

4.7 4.6.9 4.6.8.11 for the area of the semicircle whose diameter is die

ExAM. 3.

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