TRACT 38.

PROBLEMS.

S31

PROBLEM X.

Required to determine where a body, moving down the convexside of a cycloid, will fly off and quit the curve.

Let aveb represent the cy-cloid, tne properties of whichmay be seen at arts. 146 and 147vol. 2 of the Course, and vdc itsgenerating semicircle.

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y* i\

A C 15 a

Let e be the point where the motioncommences, whence it moves along the curve, its velocity-increasing both on the curve, and also in the horizontal di-rection df, till it come to such a point, F suppose, that thevelocity in the latter direction is become a constant quan-tity, then that will be the point where it will quit the cycloid,and afterwards describe a parabola fg, because the horizon-tal velocity in the latter curve is always the same constantquantity, by art. 76 vol. 2 of the Academy Course.

Put the diameter vc — d, vh — a , vi = .r ; then vd — Vdx,and id = ff(dx — x"). Now the velocity in the curve at f,in descending down ef, being the same as by falling throughHi or .r —a, will be = 8y(.r — a) ; but this velocity in thecurve at f, is to the horizontal velocity there, as vd to id,because vd is parallel to the curve or to the tangent at f.

that is ffdx : f(dx — x' 1 ) : : 8*/(x — a) : ^ - —>

which is the horizontal velocity at f, where the body is sup-posed to have that velocity a constant quantity; thereforealso \ / (.r — a) X Vld — x), as well as (.r — a) x [d — x) —axfdx — ad — x z is a constant quantity, and also ax-j-dx —x 1 ;but the fluxion of a constant quantity is equal to nothing,that is ax + dx — c 2xx = 0~a + d—2x, and hence x = la + ’d= vi, the arithmetical mean between vh and vc.

If the motion should commence at v, then x or vi wouldbe = \d, and I would be the centre of the semicircle.