164. As an example of deducing the equation to a curve from its geometrical definition, which is exactly the converse of tracing a curve from its equation, we will investigate the equation to the cycloid from the nature of its generation.

Let C be the centre of a circle in contact at A with the straight line HK. Let O be the extremity of the diameter through A. Suppose this circle to roll, without sliding, along HK; the point 0 of the circumference will then trace out a curve OPK, which is called the Cycloid. Let OAx be taken as the axis of x, and Oy, at right angles to OA, as the axis of y. Suppose that, when O has arrived at a point P of the cycloid, the circle has revolved about its centre through an angle ; then its centre must have advanced, parallel to

HK, through a space a0, a being the radius of the circle: for, since the circle rolls without sliding, it follows that the velocity of its point of contact, parallel to KH, due to its rotation about C, must be equal to the velocity of its point of contact, parallel to HK, due to the translation of C.

From C draw CQ, making OCQ=0: draw MQ at right angles to 40, and produce MQ to P, making QP equal to al. Then Q will be the position into which O would be carried by the rotation alone, QP being its additional progress due to the translation of C. Let OM = x, PM = y.

Then

The curve will evidently be symmetrical on both sides of the axis of x for if we put

0 for 0, we see that y retains the

same magnitude with an opposite sign, and x remains entirely unchanged.

If Ꮎ

=

= π, Y АК = πα = АН. Also arc 0Q = a0 = PQ. If for we write 2λ + 0, λ being any integer whatever, the expression a (1 cos 0) remains unchanged, while a (0+ sin 0) receives an increment 2λra. This shews that the two equations between 0, x, y, represent a series of similar, equal, and similarly situated cycloids, with their vertices arranged along the axis of y, both in the positive and negative directions, at intervals of 2πα,

Tangent and Normal to the Cycloid.

165. The general equation to the tangent of a curve is

x'dy - y'dx = xdy – ydx :

this equation becomes, for the cycloid,

x' (1 + cos 0)-y' sin 0 = (1 − cos 0) (1 + cos 0) (0 + sin 0) sin 0

If be the inclination of the tangent to the axis of x, then

a result which shews that the tangent at P is parallel to the chord OQ, and that consequently the normal at P is parallel to the radius CQ.

Arc of the Cycloid.

166. Differentiating the formulæ for x and y, we get ds2 = dx2 + dy2 = a2 sin2 0 d02 + a2 (1 + cos 0)2 do3

Let c =

8ax.

the chord of the arc OQ; then, by the nature of the

Radius of Curvature of the Cycloid.

167. By Art. (132), we have

168. If a, ß, be the coordinates of any point of the evolute, then, by Art. (136),

and

adx + Bảy = x + ydy,

ad2x + ẞd2y = xd2x + yd3y + dx2 + dy3.

From these two equations there is

a (dxd3y – dyd2x) = x (dxd3y – dyd2x) – dy (dx2 + dy3).....(1),

and ẞ(dyd2x - dxd3y) = y (dyd2x – dxd3y) – dx (dy2 + dx3).....(2). From (1) we have, expressing x and y in terms of 0,

a (1 − cos 0) (1 + cos 0) − a (1 + cos 0). 2 (1+ cos 0),

hence a = a (1 − cos 0) + 2a (1 + cos 0) = a (3 + cos 0),

B (1 + cos 0) = a (0 + sin 0) (1 + cos 0) – sin 0 . 2a (1 + cos 0),

If we now change the origin to a point 2a, ± πa, by putting a = a' + 2a, ß = ß' ± πa, in (5), (6), we get

These results shew that the evolute of the cycloid HOK is a portion of the locus consisting of the infinite series of cycloids

denoted by the equations (7) and (8). Two of this series have their vertices at H and K, and their point of junction O', where they form a cusp, in the line OA produced, AO' being equal to AO. It is evident that each of the cycloidal evolutes is similar, and equal to the original cycloid, and similarly situated. HO'K is evidently the portion of the locus of (7) and (8) which constitutes the evolute of HOK.

THE END.

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