Зображення сторінки
PDF
ePub

or

(no – 1) (y + x) = C2 + 2Cx, which is the equation to a series of circles.

When n<1, the equation to the locus of the ultimate intersection of these is

(1 – no) y Enx = 0, giving two straight lines which obviously satisfy the condition.

[ocr errors]

(8) Find the curve in which the area is equal to the cube of the ordinate divided by the abscissa. The condition is

yi

d whence we have

(x+y + y) dx = 3xy*dy. This being homogeneous may be integrated in the usual way—the result is

x – 2y = Cx.

(9) Find the curve in which the normal is equal to the distance from the origin.

It is a circle or an equilateral hyperbola according as the two lines are on the same or opposite sides of the curve.

Trajectories.

A Trajectory may be defined to be a line which cuts, according to a given law, a series of curves expressed by one equation. Problems of this kind, at least with respect to Trajectories cutting curves at a constant angle were first proposed by John Bernoulli*, but they were also considered by several other writers, as James Bernoullit, and Leibnitz, who in 1715 proposed it as a challenge to English Analysts, “ad pulsum Anglorum Analystorum nonnihil tentandum;" a challenge which was answered by Newton | and by Taylor ||.

Com. Epis. Leib. et Bern. Vol. 1. p. 17, and Opera, Var. Loc.
+ Opera, Var. Loc.
| Phil. Trans. 1716.
| Id. 1717.

da

= C.

1+

The most elaborate discussion however of the problem is by Euler in three memoirs in the Novi Commen. Petrop. Vol. xiv. p. 46, Vol. xvii. p. 205; and Nova Acta Petrop. Vol. 1. p. 3.

Let F (it', y', a) - 0 be the equation to a series of curves which are to be cut by a trajectory at a constant angle: then if f(xy) = 0 be the equation to the proposed trajectory, and c be the tangent of the angle between the two curves, we have the condition

dy dy dx'

dy' dy

dx' do The differential equation to the trajectory is found by eliminating a between the equation to the curves, and this last equation, where, it is to be observed, x and y are to be

dy substituted for x' and y' in the value of If the trajectory

dx is to be rectangular, since in that case c= co, the condition becomes

dy dy 1+

= 0.

dx' d x The elimination of a may be more or less easily effected according to the way in which it is involved in the given equation to the series of curves: if that can be put under the form

(x, y) the elimination is readily effected.

(10) To find the curve which cuts at a constant angle a series of straight lines drawn from one point.

That point being taken as origin, the equation to the lines is

y' y' = ax or

= a,

= 0,

a being the variable parameter. The equation of condition becomes

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

therefore wdy - ydx = 0

= c (ydy + xdx). Dividing both sides by ạ? + y,

ady - ydx ydy + xd x

[ocr errors]
[ocr errors]

Integrating,

[merged small][ocr errors][merged small]

If we make

= tan 0, x2 + yo = qo?, this may be put under

the form

p = be,

which is the equation to the logarithmic spiral.

(11) Let the given curves be all the circles which pass through one point at which they all touch one straight line. Taking this line as the axis of y, the equation to the circles is

x' 2 + y'?

[ocr errors]

dx

y'a = 2ax' - x'? or

2 x'
dy (x? y')
Then

and the equation of condition is

2 x'y {c(x2 - y) + 2xy} dy + (x2 – yo – 2cxy) dx = 0. This being a homogeneous equation may be integrated by the appropriate method, and the result is

po? + y2 = b (cy x), b being the arbitrary constant introduced in the integration. This is evidently the equation to a circle.

(12) Find the orthogonal trajectory of the series of parabolas represented by the equation

y'? = 4a x'.

The equation of condition for orthogonal trajectories is

dy dy 1+

0. dx dvd

dy ยู่ Y In this case

-; therefore d x' 2 c'

9x dx + y du = 0, whence, by integration,

y?

= b,

[ocr errors]
[ocr errors]

2

b? being an arbitrary constant. This is evidently the equation to an ellipse. The equation to the lemniscate of Bernoulli is

(x2 + y)2 = a(x2 y'). That to the orthogonal trajectory is

(2x + yo) = bxy; which is the equation to a similar lemniscate, the axis of which is inclined at an angle of 45° to that of the former.

If one of the variables be given as a function of the other and the parameter, as if

y = f (it', a), we cannot eliminate a so readily. But let

dy' = Pdd' + Qda;

dy

curve

then for one

P, and the equation to the or

d x' thogonal trajectories is

dx + Pdy = 0; and as dy = Pdx + Qda, this becomes

(1 + P2) dx + P Qda = 0. As P and Q contain only x and a, this is an equation between two variables x and a, and to integrate it appropriate methods must be employed.

с

(14) Let the curves be a series of ellipses expressed by the equation

= (0 – 62). a being the variable parameter. Here

• ax

(c* – x')] P=

Q

c (c* – x)?? and the equation for the orthogonal trajectory is

ax (co – x”) da = {c* + (a' - co) x®} dx. To integrate this put (c* — x?)! = U, when it becomes

cu du au’da + apudu =

c-u Integrating, substituting for u its value, and eliminating a, the final equation is

y = 6 - 2 + c log æ*, where he is the arbitrary constant.

When the curves to be intersected are given by a differential equation, the trajectory can be investigated only under particular circumstances. For a detail of these the reader is referred to the memoirs of Euler quoted above.

Involutes of curves may be considered as Orthogonal Trajectories, since they cut at right angles all the tangents of the evolute. Let

v = f(u) be the equation to the evolute. The equation to a tangent at any point is

dv

(0 u), or y = x W +0 – Uw, du

du

[ocr errors]

if we put

du'

This, then, is the equation to a series of lines to which we wish to find the trajectory. The variable parameter may

« НазадПродовжити »