B. To find the time of fall from rest down any arc of an inverted cycloid.

Let O be the point from which the particle commences its motion. Draw OA' parallel to CA, and on BA' describe

a semicircle. Let P, P', P" be corresponding points of the curve, the generating circle, and the circle just drawn, and let us compare the velocities of the particle at P, and the point P". Let P'T be the tangent at P".

time from A to B in circle = time from 0 to B in cycloid

COR. It is evident from the proof, that the particle descends half the vertical space to В in half the time it takes to reach B.

C. To find the nature of the brachistochrone, gravity being the only impressed force.

The following is founded on Bernoulli's original solution. (WOODHOUSE, Isoperimetrical Problems.)

From Art. 182 it is evident that the curve lies in the vertical plane which contains the given points. Also it is easy to see that if the time of descent through the entire curve is a minimum, that through any portion of the curve is less than if that portion were changed into any other curve.

And it is obvious that, between any two contiguous equal values of a continuously varying quantity, a maximum or

P

m

minimum must lie. [This principle, though excessively simple (witness its application to the barometer or thermometer), is of very great power, and often enables us to solve problems of maxima and minima, such as require in analysis not merely the processes of the Differential Calculus, but those of the Calculus of Variations. The present is a good example.]

Let, then, PQ, QR and PQ', Q'R be two pairs of indefinitely small sides of a polygon such that the time of descending through either pair, starting from P with a given velocity, may be equal. Let QQ' be horizontal and indefinitely small compared with PQ and QR. The brachistochrone must lie between these paths, and must possess any property which they possess in common. Hence if v be the velocity down PQ (supposed uniform) and v' that down QR, drawing Qm, Q'n perpendicular to RQ', PQ, we must have

Now if be the inclination of PQ to the horizon, ' that of QR, QnQQ' cos 0, Q'm = QQ' cos 0'. Hence the above equation becomes

COS 0

cos Ꮎ

This is true for any two consecutive elements of the required curve; therefore throughout the curve

v x cos 0.

But voc vertical space fallen through. (§ 171). Hence the curve required is such that the cosine of the angle it makes with the horizontal line through the point of departure varies as the square root of the distance from that line; which is easily seen to be a property of the cycloid, if we remember that the tangent to that curve is parallel to the corresponding chord of its generating circle. For in the fig. p. 163,

The brachistochrone then, in the case of gravity being the

only impressed force, is an inverted cycloid whose cusp is at the point from which the particle descends.

C. Were there any number of impressed forces we might suppose their resultant constant in magnitude and direction for two successive elements. Then reasoning similar to that in § 182 would shew that the osculating plane to the brachistochrone always contains the resultant force. Again we should have as in last Article,

where is now the complement of the angle between the curve and the resultant of the impressed forces.

Let that resultant = F, and let the element PQ=&s, and 0' =0+80. Then since F is supposed the same at P and Q, v" — v2 = 2Fds sin (by Chap. IV.),

But in the limit

=p, the radius of absolute curvature at Q, and F cos is the normal component of the impressed force. Hence in the general brachistochrone the pressure due to centrifugal force equals that due to the impressed forces, our result of § 189.

C. Now for the unconstrained path from P to R we have feds a minimum. Hence in the same way as before, being the angle corresponding to 0, v cos pvcos p' from element to element, and therefore throughout the curve, if the direction of the force be constant.

Now if the velocities in the two paths be equal at any equipotential surface, they will be equal at every other. Hence taking the angles for any equipotential surface

As an example, suppose a parabola with its vertex upwards to have for directrix the base of an inverted cycloid; these curves evidently satisfy the above condition, the one being the free path, the other the brachistochrone, for gravity, and the velocities being in each due to the same horizontal line. And it is seen at once that the product of the cosines of the angles which they make with any horizontal straight line which cuts both is a constant whose magnitude depends on that of the cycloid and parabola, its value being

4a

where is the latus rectum of the parabola, and a the diameter of the generating circle of the cycloid.

D. To shew that of two curves both concave in the sense of gravity, joining the same points in a vertical plane and not meeting in any other point, a particle will descend the enveloped in less time than it will the enveloping curve; the initial velocity being the same in both cases.

Take the axis of x as the line to a fall from which the initial velocity is due, and the axis of y in the sense of gravity, then