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(3) The equation to the evolute of the ellipse may be found in the same manner, but it is obtained more readily by considering it as the locus of the ultimate intersections of consecutive normals, as follows:
Let a, b, be the current co-ordinates of the normal, x, y of the curve; then if
be the equation to the ellipse, the equation to a normal passing through the point x, y is
O P = a? – 62.
Differentiating (2) and (1) with respect to x any y,
. xdx ydy
O, a + 2 = 0, x apa
times express y or æ in terms ß or a: then without finding the other variable it is sufficient to substitute the values
dß dx of y or æ in the equation the equation da=-dy' an
- , and we have a differential equation in a or B, which is the differential equation to the evolute.
y (2ay - y")
which is the equation to an equal and similar cycloid, but in an inverted position.
which is therefore the differential equation to the evolute.
and sa a+b + (B® – 8 a®)} = 0 is the equation to the evolute.
In curves referred to polar co-ordinates the most convenient mode of finding the equation to the evolute is by the relation between p and r. If p and r be the co-ordinates of the curve, P, and r, ......
......... evolute, p be the radius of curvature; then p = f (r) being the equation to the curve,
po? = gol + p - 2 pp,
Between these four equations we can eliminate p, r, p, and so find a relation between p, and r,, which is the equation to the evolute.
(10) Let p ? = go? - a'.
= q– p = a,
Hence p, and r, being both constants, the evolute is a circle.
and P,= mr,, the equation to a similar logarithmic spiral.
The logarithmic spiral may even be its own evolute; that is, one convolution of the curve may be the evolute of another convolution. To find the condition that this should be the case, let
r = er be the equation to the curve. Let P (fig. 52) be a point in the curve, PN the normal at that point touching a point Q in the convolution which is the evolute of the convolution AP. Then since the curve makes a constant angle with its radius vector, the angle SPT must be equal to the angle SQP; that is, PSQ must be a right angle. Hence the radius SQ is separated from the radius SP by some whole number of circumferences together with three right angles, or if