the circle is the same, lie in the circumferences of two circles. Mackenzie and Walton: Solutions of the Cambridge Problems of 1854, p. 124.

(15) Tangents are drawn to a vertical circle; find the locus of points in them, from which particles would descend in straight lines to the centre in the shortest time.

The required locus is an indefinite tangent at the highest point of the circle.

SECT. 3. Lines of slowest descent.

(1) To determine the straight line of slowest descent from a given point to a given circle, the point being without the circle and both being in the same vertical plane, and the highest point of the circle being lower than the given point.

Let P (fig. 118) be the given point, Q the highest point of the given circle: join PQ, and produce it to cut the circle in R. Then PR is the straight line of slowest descent from the point to the circle. For, let C be the centre of the circle. Join QC, RC, and produce RC to meet PO, drawn vertically from P, in the point 0. Then, PO being parallel to QC, and QC being equal to RC, PO must be equal to RO. Hence a circle described about O as a centre, with OP as radius, will pass through R: it will also touch the given circle at R.

Take any point R' in the given circle, not coinciding with R: join PR', and produce it to cut the subsidiary circle in Z.

Then, P being the highest point of the subsidiary circle, the time down PR is equal to that down PZ and therefore greater than that down PR'. Hence PR is the required straight line of slowest descent.

(2) Determine the line of slowest descent from a given circle to a given point without it, the point and circle being in the same vertical plane, and the point being lower than the lowest point of the circle.

From the given point draw an indefinite straight line, cutting the circle in its lowest point: then the distance between the

given point and the second intersection of the indefinite line and the circle is the required line of slowest descent.

(3) Find the line of slowest descent from one given circle to another given circle, both circles being in the same vertical plane and each being exterior to the other; the highest point of the latter circle being lower than the lowest of the former.

Produce the line, which joins the lowest point of the former circle and the highest point of the latter, to meet both circles again: then the distance between the second intersections is the required line of slowest descent.

SECT. 4. Oblique motion on inclined planes.

(1) A body is projected from a given point in a horizontal direction with a given velocity, and moves upon an inclined plane passing through the point: if the inclination of the plane vary, find the locus of the directrix of the parabola which the body describes.

If V be the velocity of projection, the locus of the directrix is a horizontal plane at an elevation above the point of projection

equal to

V2

2g

(2) A particle is projected horizontally, with a given velocity, along a plane inclined at a given angle to the horizon: find the velocity with which a body must be projected horizontally in free space, so that the parabolas described may be equal.

=

If u the given velocity and a = the given angle of inclination, the required velocity is equal to

(1) One end of a fine string is attached to an angular point B of a fixed regular polygon ABCD...on a fixed horizontal plane, its length being equal to the perimeter: a particle P, fixed to the other end of the string, which is stretched in the direction

AB, is projected in the plane of the polygon, perpendicularly to the string, with a given velocity, so that the string comes into contact with BC, CD,... successively to determine after what time the string will coincide with the perimeter of the polygon.

=

Let V the velocity of projection, which will be the velocity of the particle throughout the motion: let a = the length of each side of the polygon, n = the number of the sides. Let w1, w2, w1,........ be the angular velocities of the successively free portions BP, CP, DP,.... of the string. Then,

being the radii of the successive circular arcs described by P,

But, t1, ta, ta, ...t, being the times of the description of the suc

2π
n

cessive circular arcs, we have, being the angle subtended by

each of these arcs at their respective centres,

Hence the whole time occupied by the string in winding itself about the polygon is equal to

Let the length of the string. Then the whole time is equal to

COR. Suppose the number of the sides to be infinite, when the polygon becomes a circle: then the whole time is equal to

πι
V'

(2) A smooth tube, of uniform bore, is bent into the form of a circular arc, greater than a semicircle, and placed in a vertical plane with its open ends upwards and in the same horizontal line: find the velocity with which a ball, that fits the tube, must be projected along the interior from the lowest point, in order that it may pass out at one end and re-enter at the other.

=

If r the radius of the circle, h = the depth of the centre of the circle below the horizontal line through the two ends of the tube, and V the required velocity; then

=

(3) A particle slides from rest down a narrow smooth tube in the form of the thread of a screw, the axis of which is vertical: find the time in which it will make a complete revolution about the axis.

If a = the radius of the cylinder on which the helix is described, and a = the angle which the thread makes with a generating line of the cylinder, the required time is equal to

Επα
sin 2a/

(4) Shew that, if a particle descend down a cycloid, of which the axis is vertical and the vertex downwards, from an extremity of the base, the velocity at any point will be proportional to the radius of curvature at the point.

SECT. 6. Pendulums.

(1) If a clock pendulum lose 5′′ a day, to determine the alteration which must be made in its length.

Let be the length of a seconds pendulum, 7+ a of the pendulum under consideration. Then the number of seconds in the time of the oscillation of the latter pendulum is equal to

hence the number of oscillations which it performs in 24 hours is equal to

(1)

(1-2), nearly,

(1-2)

thus we see that the pendulum must be diminished by very nearly the (8640)th part of its length.

(2) A seconds pendulum was too long on a given day by a quantity a; it was then over-corrected so as to be too short by a during the next day: to prove that, 7 being the length of the seconds pendulum, the number of minutes gained in the two days was

The time of each oscillation in seconds during the first day was equal to