Since u is to be independent of c, the differential coefficient of u with respect to c must vanish. Now

This last integral then must vanish whatever c may be; hence we must have

This is the solution of a problem in Dynamics, which may be thus enunciated. Find a curve, such that the time of falling down an arc of the curve from any point to the lowest point may be the same. If s denote the arc of the curve

measured from the lowest point, then

ds

(x) and s = 2A √x ;

dx

so that the curve is a cycloid.

174

CHAPTER X.

ELLIPTIC INTEGRALS.

222. THE integrals √c2 sin3 0) ' √√(1 – c2 sin3 0) do,

ᏧᎾ

are called elliptic func

and (1 + a sin2 0) √/(1 - c sin' 0) c2 tions or elliptic integrals of the first, second, and third order respectively; the first is denoted by F(c, 0), the second by E(c, 0), and the third by II (c, a, 0). The integrals are all supposed to be taken between the limits 0 and 0, so that they vanish with 0; is called the amplitude of the function. The constant c is supposed less than unity; it is called the modulus of the function. The constant a, which occurs in the function of the third order is called the parameter. When the integrals are taken between the limits 0 and, they are called complete functions; that is, the amplitude of a complete function is.

2

223. The second elliptic integral expresses the length of a portion of the arc of an ellipse measured from the end of the minor axis, the excentricity of the ellipse being the modulus of the function. From this circumstance, and from the fact that the three integrals are connected by remarkable properties, the name elliptic integrals has been derived.

224. The subject of elliptic integrals is very extensive; we shall merely give a few of the simpler results, and refer the student for fuller investigations to Hymers's Integral Calculus, or to the writings of Legendre, Jacobi and Abel.

225. If 0 and are connected by the equation

F(c, 0)+ F(c, d) = F (c, μ),

where is a constant; then will

μ

cos e cos sin 0 sin √(1 − c2 sin2 μ)

= cos μ.

Consider and as functions of a new variable t, and differentiate the given equation; thus

Now as t is a new arbitrary variable, we are at liberty to

Now from the original given equation we see that if

F (c, 0) = F (c, μ);

0 =μ

therefore then =μ and x==μ;

thus from (3) (A – B) cos μ = C;

thus A cos (0-6)=B cos (0+ $) + (A – B) cos μ;

therefore

(3).

= 0

(A – B) cose cos & + (A + B) sin 0 sin = (A – B) cos μ... (4).

Substitute for A-B and A + B in (4);

thus

226.

cos e cos√(1-c2 sin2 μ) sin 0 sin = cos μ.

The relation just found may be put in a different form. Clear the equation of radicals; thus

(cos e cos - cos μ)2= (1 − c2 sin2 μ) sin2 0 sin2 ; therefore

cos2 + cos2 + cos2 μ 2 cose cos cos μ

= 1-c2 sin2 μ sin2 0 sin2 0.

Add cos2 cos2 μ to both sides and transpose; thus (cos - cos cos μ)2

=

1- cos2 - cos2 μ+ cos2 cos2 μ- c2 sinu sin2 0 sin2

therefore

cos = cos cos μ + sin o sin μ (1-c2 sin2 ).

The positive sign of the radical is taken, because when 0=0, we must have =μ.

227. We shall now shew how an elliptic function of the first order may be connected with another having a different modulus.

Let F(c, 0) denote the function; assume

tan 0

1

sin 24
c + cos 20

T. I. C.

12