0

m

given us, involving two or more letters, we may find its derived function in regard to any one of them. Thus of the quantity u=x+5y+3xy, if a represents the time, the derived function is 2x + 3y; but if y represents the time, the derived function is 10y+3x. If we suppose x and y to be horizontal and vertical components of a vector op=xi+yj, then for every point p in the plane there will be a value of x, a value of y, and consequently a value of u, = x2+5y2+3xy. If we make the point p move horizontally with velocity 1 centimeter per second, then x will represent the time, and y will not alter; so that i will be 2x+3y. This is called the flux of u with regard to x, or the x-flux of u; and it is denoted by u. Similarly if we make p move vertically with the unit velocity, x will be constant, and y will represent the time, so that i will be 3x+10y; this is called the flux of u with regard to y, or the y-flux of u, and is denoted by au. The characteristic may be supposed to stand for derived function.

We may now prove a very general rule for finding fluxes, namely one which enables us to find the flux of a function of functions. Let x and y be two variable quantities, and let it be required to find the flux of u which is a function of x and y; this is denoted thus: u=f(x, y). The method is the same as that used for a product. We find

and when we strike out common factors and omit the suffixes in this last expression, it becomes ƒ+ÿəvƒ ;

where ƒ has been shortly written instead of f(x, y). Or, substituting u for f, we have the formula

If a straight line ov be drawn through a fixed point o, to represent in magnitude and direction at every instant the velocity of a moving point p, the point v will describe some curve in a certain manner. This curve, so described, is called the hodograph of the motion of p. (ódòv ɣpáþeι, it describes the way.)

b

P

Thus in the parabolic motion p=a+tẞ+ty, we have ov=p=ß+2ty. Hence we see that the point v moves uniformly in a straight line. The hodograph of the parabolic motion, then, is a straight line described uniformly. Let ab be the initial velocity; draw through b a line parallel to the axis of the parabola. Then to find the velocity at any point p, we have only to draw av parallel to the tangent at p; the line av represents the

velocity in magnitude and direction. The straight line bv, described with uniform velocity 2y, is the hodograph. In the elliptic harmonic motion

we have

p = a cos (nt + e) +ẞ sin (nt + e)

ov = p = na cos (nt + e + 1) + nẞ sin (nt + e + 1⁄2π). Thus the point v moves harmonically in an ellipse similar and similarly situated to the original path, of n times its linear dimensions, being one quarter phase in advance. As a particular case, the hodograph of uniform circular mo

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 . Hence

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

-

=-n2 a cos (nt + e) — n2 ß sin (nt + e) = — n2p. 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]2.

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.

Velocity being a continuous quantity, it can only be accurately given at every instant by means of a curve.

Let

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

Y

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

X

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 umny 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.

Let uv be the curve of velocities during an interval mn. Take a number of points a, b, c... between m and n, and draw vertical lines aA, bВ,

Y

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 AkBIC..., and the other outside it, ufAhB... Let the horizontal line through u meet nu in r.

mab c

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 mug AkBIC...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