of the second point Q as the axis of x, and let a denote its position at the instant when the co-ordinates of the first, P, are x, y. The axis of y is chosen as that tangent to the curve of pursuit which is perpendicular to the axis of x, and the distance between the points in that position is a. e AP-OQ, and PQ is a tangent at P. These are our conditions, and lead to the following equa log (2)* = log [√/{1 + (1)} + dx]. Hence, (~)* = √/{1+ and therefore dx (da)"} + dy, dx dx dy (-)" - √ √ {1 + (dy)"} – dy, taking reciprocals. = This is true for all values of e except unity, a case to which we will presently recur. There are two cases of curves represented by equation (2). 1st, e> 1, 2nd, e<1. In the first case Q moves the faster, and P can never overtake it; the curve therefore never meets the axis of x, which indeed will be seen by (2) to be an asymptote. In the second case equation (2) becomes ае and for x= we have y=0, and also by (1) 2 1 dy Hence the curve touches the axis at this point. The remainder of the curve satisfies a modified form of the question, and is called the Curve of Flight. ае {It It is to be observed, however, that x= gives also y = ± a 1-e2 1+e\ 20 the only case in which we do not obtain an algebraic curve. Here again the axis of x is an asymptote. 31. A boat, propelled (relatively to the water) with uniform The constant velocity of the stream in this case communicated to P corresponds to the constant velocity of Qin the last example, but is in the opposite direction. In fact, if the earth were to be supposed moving in the direction xQ with uniform velocity v, the river would be at rest in space, and the actual motions of P and Q would be the same as in the last example. (See § 24.) M -X A To investigate the path, take as origin, Q, QA as the axes. Then the component velocities of P are v parallel to Qx and u along PQ, and the tangent to its path is in the direction of the resultant of these two. Putting 0 for PQx, This, being a homogeneous equation, is easily integrated, and we have, taking x = 0, y=a together, in polar co-ordinates. This evidently gives a parabola about Qas focus, if e = 1. To find the time of crossing the stream. This may easily be effected by considering the actual velocity parallel to the axis of y, Now taking quotients of y2 by both sides of (1), Taking the integral from a to 0, and putting T, for the time of crossing, But, if there had been no current, we should have had for the time of crossing, In the integration we have, of course, e< 1, else the boat could not reach Q. If e=1, the boat will reach the farther bank but not at Q. The solution of this case presents no special difficulty. 32. If the motion of a point in a plane be considered with reference to a fixed point in that plane, the rate of increase of the angle made by the line joining the two, with some fixed line in the plane, is called the Angular Velocity of the former point about the latter. Suppose this angle to be represented by 0 at time t; then at time t + St it has the value + 80, and it may be shewn as before (§ 7), that if a represent the angular velocity required, then de @= dt Ex. A point moves uniformly, with velocity v, in a straight line; to find at any instant its angular velocity about a fixed point whose distance from the straight line is a. Taking as initial line the perpendicular from the fixed point on the line of motion; the polar equation of the path is r = a sec 0. Also, if when t=0, 0=0, we have 33. A point describes a circle with uniform velocity; it is required to find the actual velocity, and the angular velocity (about the center) in any orthographic projection. Let ApA' be an ellipse and APA' the auxiliary circle. Then the former will be the orthographic projection if its axes be made in the ratio of the cosine of the angle (a) between the planes of projection. Also if PpM be perpendicular to AA', P and p will be corresponding points in the two. Draw the tangents pT, PT; then |