tion is again uniform circular motion. We have seen that ov is n. oq, where oq is the semi-conjugate diameter of op.

Of course the hodograph of every rectilinear motion is also a rectilinear motion; but in general a different one.

The velocity in the hodograph is called the acceleration of the moving point; thus the velocity of v is the acceleration of p. It is got from p in precisely the same way as p is got from p, and accordingly it is denoted by p. The acceleration is the flux of the velocity.

In the parabolic motion, since pẞ+2ty, we have p = 2y, or the acceleration is constant. In the case of a body falling freely in vacuo, this constant acceleration amounts at Paris to 981 centimeters a second per second; it is called the acceleration of gravity, and is usually denoted by the letter g. It varies from one place to another, for a reason which will be subsequently explained.

In elliptic harmonic motion p is to be got from p by the rule: Multiply by n, and increase the argument by π. Hence

p = n2 a cos (nt + e + π) + n2ß sin (nt + e +π)

=

==

n2p.

- n2 a cos (nt + e) — n2 ß sin (nt + e) : Thus the acceleration at p is n times po; that is, it is always directed towards the centre o, and proportional to the distance from it. It is clear from the figure that the tangent at v is parallel to po; and since the velocity of v is n times ou, which is itself n times po, this velocity is n2 times po.

Those motions in which the acceleration is constantly directed to a fixed point are of the greatest importance in physics: and we shall subsequently have to study them in considerable detail.

Acceleration is a quantity of the dimensions [L]: [T]3.

THE INVERSE METHOD.

So far we have considered the problem of finding the velocity when the position is given at every instant. We shall now shew how to find the position when the velocity is given. The problem is of two kinds: we may suppose

CURVE OF VELOCITIES.

69

the shape of the path given, and also the magnitude of the velocity at every instant; or we may suppose the hodograph given. For the present we shall restrict ourselves to the first case.

Let

Velocity being a continuous quantity, it can only be accurately given at every instant by means of a curve. a point t move along oX with unit velocity, and at every moment suppose a perpendicular tv to be set up which represents to a given scale the velocity of the moving point at that moment. Then the point v will trace out a curve which is

called the curve of velocities of the moving point.

In uniform motion the curve of velocities is a hori

zontal straight line, uv. In this case we can very easily

Y

น

m

n

find the distance traversed in a given interval mn; for we have merely to multiply the velocity by the time. Now the velocity being mu, and the time mn, the distance traversed must be represented on the same scale by the area of the rectangle umnv. The meaning of the words on the same scale is this. Time is represented on oX on the scale of one centimeter to a second; suppose that velocity is represented on oY on such a scale that a centimeter in length means a velocity of one centimeter per second; then length will be represented by the area umnu on the scale of one square centimeter to one linear centimeter. To find the length represented by a given area, we must convert it into a rectangle standing upon one centimeter; the height of this rectangle is the length represented. The breadth, one centimeter, which thus determines the scale of representation, is called the area-base.

It is true also when the velocity is variable that the distance traversed is represented by the area of the curve of

velocities (Newton). We shall prove this first for an interval in which the velocity is continually increasing; it will be seen that the proof holds equally well in the case of an interval in which it continually decreases, and as the whole time must be made up of intervals of increase and decrease, the theorem will be proved in general.

mn.

Y

Let uv be the curve of velocities during an interval Take a number of points a, b, c... between m and n, and draw vertical lines aA, bВ, CC,... through them to meet the curve of velocities in A, B, C... The length mn is thus divided into a certain number of parts, corresponding to divisions in the interval of time which it represents. Through A, B, C... draw hori- 0 zontal lines as in the figure, ug, fAk, hBl, etc. These will form as it were two staircases, one inside the curve of velocities, ug AkBlC..., and the other outside it, ufAhB... Let the horizontal line through u meet nu in r.

matc

We shall now make two false suppositions about the motion of the point, one of which makes the distance traversed too small, and the other too great. First, suppose the velocity in the intervals ma, ab, bc,... to be all through each interval what it actually is at the beginning of the interval; as the velocity is really always increasing, this supposition will make it too small, and therefore the distance traversed less than the real one. In the interval ma, according to this supposition, the velocity will be mu, and the distance traversed will be represented by the rectangle muga. So in the interval ab, the distance traversed will be a Akb. Hence the distance traversed in the whole interval mn will be represented by the area of the inside staircase mugAkBIC...qn. Secondly, suppose the velocity in each interval to be what it actually is at the end of the interval: then in the interval ma the velocity will be aA, and the distance traversed mfAa. So the distance traversed in the interval mn will be represented by the area of the outside staircase mufAhB...vn. But

CALCULATION OF AN AREA.

71

this supposition makes the velocity too great, excepting at the instants a, b, c...; therefore the actual distance traversed is less than the area of the outer staircase.

It appears therefore that the distance traversed in the actual motion is represented by an area which lies between the area of the outer and the area of the inner staircase. But the area of the curve of velocities lies between these two. Therefore the difference between the area of the curve of velocities and the area which represents the distance traversed is less than the difference between the areas of the outer and inner staircases. Now this last difference is less than a rectangle, whose height is r" and whose breadth is the greatest of the lengths ma, ab, ...; for it is made up of all the small rectangles on the curve like ufAg. But we may divide the interval mn into as many pieces as we like, and consequently we may make the largest of them as small as we like.

It follows that there is no difference between the area of the curve of velocities and that which represents the distance traversed. For if there is any, let it be called 8. Divide mn into so many parts that a rectangle of the height nv, standing on any of them, shall be less than this area 8. Then we know that the difference in question is less than a rectangle of height re standing on the greater of these parts, that is, less than 8; which is contrary to the supposition.

This demonstration indicates a method of finding the area of a curve, and, at the same time, of finding the distance traversed by a point moving with given velocity. The method is the same in the two problems (which, as we have just seen, are really the same) but has to be described in somewhat different language. For the area of the curve umnv, the rule is: Divide mn into a certain number of parts, and on each of these erect a rectangle whose height is the height of the curve at some point vertically over that part; then the sum of the areas of these rectangles will differ

from the area of the curve by a

quantity which can be made as small as we like by increasing the number of parts and diminishing the largest of them. For the distance traversed during a certain interval, the rule is: Divide the interval into a certain number of parts, and suppose a body to move uniformly during each of those parts with a velocity which the actual body has at some instant during that part of the interval; then the distance traversed by the supposed body will differ from that traversed by the actual body by a quantity which can be made as small as we like by increasing the number of parts and diminishing the largest of them.

For example (Wallis), suppose the velocity at time t to be th, and that we have to find the space described in the interval from t=a to t=b. Let this interval be divided into n parts in geometric progression, as follows. Let σ"=b: a, so that b=σ"a. Then the parts shall be the intervals between the instants a, σa, oa,...o"1a, b. The velocities of the moving body at the beginnings of these intervals are ak, okak, o2kak,...σ(n-1)kak, bk... Hence if a body move uniformly through each interval with the velocity which the actual body has at the beginning of that interval, it will describe the space

ak (oa− a)+okak (o2a - oa) +...+o(n−1)kak (ona - on-1a) =ak+1(σ-1) (1+σk+1 +σ2(k+1) + ... +σ (n−1)(k+1))

This

Now the larger n is taken, the more nearly σ approaches to unity, and consequently, the more nearly the denominator of this fraction approaches to the value k +1. Thus the space described from ta to t=b is (bk+1 — akt1) : k+1. By making a=0 and b=t in this formula, we find that the space traversed between 0 and t is t+1 : k+1. agrees with our previous investigation; for if (k+1) s=tk+1, we know that stk. As in the converse investigation, p. 55, it is easy to extend the method to the case in which is a commensurable fraction; for the quotient ok+1-1:σ-1 approaches also in that case the value k+1 when σ approaches unity.