Example 3.-From a given point P, in the Hence MN2CM2 + CN-2 CM CE= transverse axis of an ellipse, to draw PB, the shortest line to the curve:

and from this all the other quantities may be
determined.

ON THE METHOD OF DRAWING TANGENTS
TO CURVES.

By this method the tangent and subtangent to a curve are determined when its equation are given, and vice versa.

E

Example 1.-It is required to find the length of the arc of a circle, in terms of its sine, versed sine, tangent, and secant.

Let C in the preceding figure represent the centre of the circle; call the radius AC, r; the versed sine AD, r; the sine DE, y; the tangent TE, t; and the secant TC, s. Then by the property of the circle 9.2 12 $2 g.2 y2 = 2 rs = r2 = ~ +$

and by taking the fluxions of these equations, and making proper substitutions in the general fluxional equation, i = √2+j2, we obtain

If A E be a curve, let it be required to draw a tangent TE at any point E. Draw the ordinate DE, and another da e indefinitely near to it, meeting the curve, or the tangent produced in e, and draw E a parallel to the axis A D. Then the triangles Ea e, and TED are similar, and therefore e a: a E:: ED: DT. Orj:::: y=DT, the subtangent; r being the absciss

AD, and y the ordinate DE.

Any of these quantities may be expanded in a series, and the fluent of each term being taken, a general value of z will be obtained. We shall = take as an example i This form of the series, however, is one of very slow convergency, so that a great many terms of it must be collected before a result of sufficient practical accuracy can be obtained.

But it may easily be transformed into series of almost any required degree of convergency. The following are amongst the most useful forms that have yet been discovered; A representing the

Example 1.-To draw a tangent to a parabole, circumference to radius unity, and a, ß, y, &c., whose equation is a x = y2. the preceding terms in each series:

2 ax

Here a i= 2yj; whence

y i

2y

=

2r; or the subtangent is double the corres

ponding absciss.

Example 2.-Draw a tangent to the cissoid of 1st A =

Diocles, whose equation is y2 =

Here 2 y j =

quently y

I

3 ar i 2 r3 i

y

3 ar

and conse

1+

+

3

· 10

TO DETERMINE THE LENGTHS OF CURVES
WHOSE EQUATIONS ARE GIVEN.

43. In the annexed figure E a, e a, and E e, are
simultaneous in-
crements of x, y,
and %, or of the
absciss AD, the
ordinate D E, and
the curve AE;
and the triangle
Eae is (see arti-
cle 42) similar to
TED;
therefore be considered as a right-angled tri-
angle. Hence, — ï' + j', or i = √ i 1 + j 2.
Therefore substituting for y its value in terms of
, and taking the fluent, the value of z is
obtained.

42-3

4 n 8 a

27

- 1

n

ข

ny

2n-2

2 n

1

2 a

4 n

3

.

on- 5

&c.

6 n

4

But when 2 n-2 is either unity, or an aliquot part of it, this series will always terminate, and consequently the length of the arc will be accurately obtained from it.

a i

√ *2 + y2 = i = 21, and 2 c y ŝ = a ci
ï2
2 y'
whose fluent is a cr, the value of the spherical
surface. But a c is the circumference of the
generating circle; hence the surface of any seg-
ment is equal to the circumference of a great
circle, multiplied by the versed sine; or height of
the segment; and, when this versed sine is the
whole diameter, the expression is c a2, or four
times the area of a great circle of the sphere.
Example 2.-Let the proposed curve surface

TO FIND THE AREAS OF CURVES, WHOSE EQUA- be that of a parabolic conoid.

TO FIND THE SOLID CONTENTS OF BODIES. 46. As a curve surface may be conceived to be generated by the expanding circumference of a plane moving forward, as the solid itself which the surface bounds may be conceived to be generated by the plane itself. Hence, if r, y, and c, represent the same things as they did in art. 44, we have cy'i for the fluxion of the solid, and the fluent of this quantity will be the required solid.

Example 1.-Required the solid content of a cone, whose altitude is a, and base b.

45. A surface may be conceived to be gene- 62 rated by the circumference of a plane moving forward, and expanding at the same instant; therefore the fluxion of the surface is equal to the fluxion of the curve, in which the expanding circumference moves forward at any instant, multiplied by the periphery of the variable circumference at the same instant; and the fluent of this fluxion is the value of the generated surface.

If c the circumference of a circle, whose diameter is 1, a the abscissor, ythe ordinate, and z the curve in which the expanding circumference moves forward; then 2 c y = the circumference, and 2 c y i — 2 c y √ i2+j2 the fluxion of the surface S, and consequently by taking the fluent, S is obtained.

xa is

с аз
6

Hence a sphere, or a spheroid, is two-thirds of its circumscribing cylinder, for the solidity of the cylinder whose base diameter is b, and altitude a, caba ca b2 is of which is evidently two-thirds. 6

4

TO FIND THE POINTS OF CONTRARY FLEXURE OF
CURVES FROM THEIR EQUATIONS.

47. It is evident when a curve is concave towards its area, that the fluxion of the ordinate decreases with respect to the fluxion of the absciss; and the contrary when the curve is convex towards its axis; hence, at the point of

contrary fluxion
Or is a constant quantity, and
consequently its fluxion is 0. Therefore, if
from the given equation the value of or

y

be

301

as of DE; hence the equation may be put in
this form GC-ii GEy-ÿý, or GC
2:2 + j;2
the terms of this equation be respectively multi-
3; and if each of

GE· y

found, the fluxion of that value will give an plied by the equivalent expressions GE equation, from which the relation of x and y at the point of contrary flexure may be found.

=

Example 1.-Required the point of inflexion in a curve, whose equation is a r2 a2y + xry. This equation in fluxions is 2 axi a2 j + 2 x y z + x2 j, whence fluxion of which made the = (a2 + x2) · (u x − x a2 + x2 gives a2

a2 + x2
0, gives 2 x i (a x —xy)
2ax- 2 x y
ỷ — x y), and this again

a

-; which being equay

gives

a2 + x2 a2

ted with the former value of

= a.

and consequently y

Example.-Required the radius of curvature at the joints of an ellipse, at the point corre

; whence ra√, sponding to the absciss and ordinater and y, the
equation of the curve being a2 y2 — c2 a x —
By taking the first and second fluxions of the
- x2?
and 2 a 2 + 2 a2 y ÿ — — 2 c2, considering
given equation we have 2 a2 y j= c2 ¿ · a — 2 x,
c2ia 2 r, and
2 a2 y

Example 2.-Required the point of contrary flexure in a curve whose equation is a y2= a2 x + x3 ?

Here ya x x

y=

a3 j 2 + c2 ¿ •
ay

a

and - ÿ=

a (√3—1)

cia — 2x

fluction of this put = 0, and reduced, gives r= 2 √ a √ a2 x + x3

4√ 12

ought simply, the ellipse in that case degenerating into a circle.

TO FIND THE INVOLUTES AND EVOLUTES OF

CURVES.

49. If a thread wrapped close round a curve were fastened at one end, and unwound from the other in the plane of the curve, the thread being always kept stretched, the end of the thread= Put the involute, that from which the thread is r

G; draw Ed, and ed parallel to ED and AD, to winding off will describe a curve which is called

represent the fluxions of AD and DE.
AD=r, ED=y, and AE 2.
similar triangles GC: GE :: ý: i, or GC -
Then by

GEj. Whence by fluxing GC ï +
GC GE+GEy. But GĊ
BG, therefore, GC

x

BGGE

y

GE. Now is the fluxion of BG, as well as of AD, and y is the fluxion G E as well