COR. 1. Hence, if we know the equation between the radius vector, and the perpendicular upon the tangent in the spiral SP, we may find the corresponding equation in the spiral traced out by Y.

Ex. 1. Suppose the first spiral to be the logarithmic spiral, then

1

p = p. √1+{hyp. log. (a)}''

but p p p p

:: 1:√1+{hyp. log. (a)}2;

1

· · p' = P · √1+{hyp. log. (a)

(a)}2°

That is, the second spiral is also a logarithmic spiral similar to the first.

Ex. 2. Suppose the first spiral to be a parabola.

Here pap; hence, p'a; and, consequently, the line traced out by I, must be a straight line. Suppose the first spiral to be a circle, whose pole lies in the circumference. Let (a) = the diameter of this circle, then

Ex. 3.

Ex. 4. Let the first spiral be the equiangular hyperbola, having its centre for its pole. Here it

equation between the spiral angle, and the radius vector (p). In general,

Where, whether we are to take the positive or the negative sign, depends upon the position we assume for the initial line. Draw a straight line bisecting the right angle, contained by the asymptotes of the hyperbola, and assume this for the initial line. It is manifest, that when (p) coincides with this line, (p) will also coincide with it, and both (p) and (p) will

and, consequently, their difference is constant, now

-1

2

when pa, cos = 0; but at this point 0=0;

a2

and, consequently, this difference always =0; hence,

which is the equation to the lemniscate.

From this we may also find the equation to the lemniscate, considered as a curve, referred to rectangular co-ordinates.

Let the straight line, which we assumed for its initial line, as a spiral, be now assumed for its horizontal axis, and let the centre of the hyperbola be the first point, from which the co-ordinates are measured.

Let the abscissa = x, the ordinate =y, then we have the following equations,

To find the differential coefficient of the length of a spiral.

Let SPQ, (See Fig. Prop. 53. p. 164.) be the spiral, SP its radius vector = p, YPR a tangent to the point P, SQR another radius vector, making with SP an angle PSR= (1), join PQ, then the arc PQ is greater than the chord PQ; and, therefore, greater than PR- QR, and it is also less than PR+QR, but, (by Prop. 49. Cor. 2.).

.. PR+QR=pB1 (1) + (p B2 − R2) ;

and PR- QR=pB1 (ı)+(pB2— R2)

2

1.2

Now, if the arc SPs, we have the arc PQ, which is its increment, while the spiral angle increases by (4), equal to the series

But, since the arc PQ lies between PR+RQ, and PR-RQ, the series (C) lies between the series (A) and (B); hence, (by Prop. 13. Cor. 1.), we have

ds

do

= pB1 = p . sec2 0' =p. sec 0' √√1 + tan° 0′

COR. 1. Hence, ds=√/dp2+p3d02.

Ex. 1.

spiral.

To find the length of the logarithmic

Here pao; . 0. hyp. log. (a) = hyp. log. p ;

tions of (0), which have the same differential coefficient; and, therefore, their difference is constant. Now, if we suppose that when p=0, s=0, we have

Y