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OF RECTIFICATIONS; OR, TO FIND THE LENGTHS OF CURVE LINES.

60. RECTIFICATION, is the finding the length of a curve line, or finding a right line equal to a proposed curve.

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By art. 10 it appears, that the elementary triangle Eae, formed by the increments of the absciss, ordinate, and curve, is a right-angled triangle, of which the increment of the curve is the hypothenuse; and therefore the square of the latter is equal to the sum of the squares of the two former; that is, Ee Ea2+ae2. Or, substituting, for the increments, their proportional fluxions, it is +ÿÿ, or =+; where z denotes any = 2 zż curve line AE, its absciss AD, and y it ordinate DE. Hence this rule.

RULE.

61. From the given equation of the curve put into fluxions, find the value of or y', which value substitute instead of it in the equation =+ j; then the fluents, being taken, will give the value of z, or the length of the curve, in terms of the absciss or ordinate.

EXAMPLES.

EXAM. 1. To find the length of the arc of a circle, in terms both of the sine, versed sine, tangent, and secant.

The equation of the circle may be expressed in terms of the radius, and either the sine, or the versed sine, or tangent, or secant, &c, of an arc. Let therefore the radius of the circle be CA or CE=r, the versed sine AD (of the arc AE), the right sine DE=y, the tangent TE =t, and the secant CT = s; then, by the nature of the circle, there arise these equations, viz.

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Then, by means of the fluxions of these equations, with the general fluxional equation +2, are obtained the following fluxional forms, for the fluxion of the curve; the fluent of any one of which will be the curve itself; viz.

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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 d2r the diameter, the curve is z

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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 r1, and consequently the tangent of 45°, or t, = 1 also, in this case the arc 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.

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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 , or its square which being substituted in the series, the length of the arc of 30° comes out

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1

+
7.33 9.34

&c). Hence, to com

pute these terms in decimal numbers, after the first, the succeeding 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 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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+ Terms.

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EXAM. 2. To find the length of a parabola,

EXAM. 3. To find the length of the semicubical parabola,

whose equation is ax2= 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 a proposed curvilineal one,

By art. 9 it appears, that any flowing quantity being drawn into the fluxion of the line along which it flows, or in the direction of its motion, there is produced the fluxion of the quantity generated by the flowing. That is, Dd X DE or yż is the fluxion of the area ADE. Hence this rule.

RULE,

RULE.

63. From the given equation of the curve, find the value either of x or of y; which value substitute instead of it in the expression yx; 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.

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The equation of the parabola being ar y2; where a is the parameter, the absciss AD, or part of the axis, and y the ordinate DE.

From the equation of the curve is found yax. This substituted in the general fluxion of the area yx, gives

I I
Σ

I

ar

or axx the fluxion of the parabolic area; and the fluent of this, or a2x2=x/ax=3ry, is the area of the parabola ADE, and which is therefore equal to of its circumscribing rectangle.

EXAM. 2. Το square the circle, or find its area.

The equation of the circle being yar x2, or y = Jaxx2, where a is the diameter; by substitution, the general fluxion of the area yx, becomes axx, for the fluxion of the circular area. But as the fluent of this cannot

be found in finite terms, the quantity axx is thrown into a series, by extracting the root, and then the fluxion of the area becomes

xaxx (1

X x2 1.3.x3 1.3.5.x+
2a 2.4a2 2.4.6a3 2.4.6.8a

4

&c);

and then the fluent of every term being taken, it gives 2 1.2 1.r2 1.3.x3

xax ×

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1.3.5.x+

&c);

3 5a 4.7a2 4.6.9a3 4.6.8.11a4

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

a2

2 1
3

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5 4.7 4.6.9 4.6.8.11

for the area of the semicircle whose diameter is a.

EXAM. 3.

EXAM. 3. To find the area of any parabola, whose equation is a""y"+",

=

EXAM. 4. To find the area of an ellipse.

EXAM. 5. To find the area of an hyperbola.

EXAM. 6. To find the area between the curve and asymptote of an hyperbola.

EXAM. 7. To find the like area in any other hyperbola whose general equation is "y" = am+".

TO FIND THE SURFACES OF SOLIDS.

64. In the solid formed by the rotation of any curve about its axis, the surface may be considered as generated by the circumference of an expanding circle, moving perpendicularly along the axis, but the expanding circumference moving along the arc or curve of the solid. Therefore, as the fluxion

B

of any generated quantity, is produced by drawing the generating quantity into the fluxion of the line or direction in which it moves, the fluxion of the surface will be found by drawing the circumference of the generating circle into the fluxion of the curve. That is, the fluxion of the surface BAE, is equal to AE drawn into the circumference BCEF, whose radius is the ordinate DE.

65. But, if c be 31416, the circumference of a circle. whose diameter is 1, AD the absciss, y = DE the ordinate, and z = AE the curve; then 2y the diameter BE, and 2cy = the circumference BCEF; also, AE =¿ = + therefore 2cyż or 2cyx2+2 is the fluxion of the surface. And consequently if, from the given equation of the curve, the value of x or y be found, and substituted in this expression 2cy, the fluent of the expression, being then taken, will be the surface of the solid required.

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EXAMPLES.

EXAM. 1. To find the surface of a sphere, or of any seg

ment.

In

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