Thus, CP' being the involute of the OCQD, whose radius is unit,

let the ordinates be represented by p. Let also P'y be a tangent upon it. Then Dp' being the unwound line it curve, or 1 p'y. It is also AD. . p'y = AD.

at P' and Ay a

must be

Hence it appears that p always = and the construction is

.. obtained by unwinding the thread CD so as to describe the

involute, and taking AP always =

called the Complicated Tractrix.

23.

AC2
ΑΡ

1

The spiral is

ᎪᏢ

Let AB (See Fig. C. P. p. 209)
= x

BPY,

the radius of the = 1, and BC = y'.

The curve cuts the semicircle when y' y, or when

The at M is evidently an asymptote to the curve.

24.

Let NM (Fig. 16.) the line given in lengthm, pass through the point A given in position, and meet the straight line BC given in position. Required the locus of the extremity M. Let DAa be that position of NM which is BC, and since AD is given, let it = n.

the curve. diameter.

y

x
n + x

√ m2 − (n + x)2 which is the equation of

It is an oval similar and equal on each side the

25.

Given two radii vectores of a logarithmic spiral, and

the angle between them, to construct the spiral.

The property which distinguishes the logarithmic spiral is that it cuts all its radii vectores at the same angle. Let that angle PQR (AB, AB' are the given radii vectores, and AP, AQ are indefinitely near) = a, (Fig. 17). Also let AB = r, AB′ = r', ▲ BAB'ß, and BAP = 0, and AC, the radius of

Р

AP =
the Cpr
Then PR

1.

. de tan. a

the curve expressed in terms of 6, and given quantities.

The points of the curve corresponding to every possible value of, may therefore be found, by reference to the Tables; or the curve may be constructed.

26.

The locus of the intersections of the tangents at corresponding points of the common cycloid and its generating circle, is the involute of the generating circle.

Let PT, QT (Fig. 18,) the tangents at P and Q intersect in T. Join AQ, OQ, &c. &c.

Then, it is well known that

TP is parallel to AQ, and

TQ is OQ the radius.

Hence AQM = R. 2 − 4 O AQ = R. 2 - 4 AQO =AQT QTP,

27.

To find the locus of the intersections of the tangents to a circle with the perpendiculars to them, let fall from a given point in the circumference.

Let A (Fig. 19,) be the given point, M the intersection of the tangent at R and its from A, and let MR meet the diameter of

the produced in T.

Join CR (C being the centre). Then referring the locus BMA to A as a pole, and .. putting AM = and ≤ MAB = 0 (B being the point where

CR is parallel to AM we have

.. AM AN AC

the diameter of the circle), since = < RAN

Hence pr. (1 + cos.

the curve.

MAR = 2 ARC
CN = r . (1 + cos. 6.)

)....... (a) the polar equation of

Again, referring the curve to rectangular co-ordinates, by putting

Ap =x, PM = y, we get

√ x2 + y2 = AM'= AN (≤ RAN = 2 ARC = ≤ MAR) CN= + r. cos. 0.

=

Then x = 0 and 2r, .. the curve passes through the points

A and B.

To find the area of the curve, we have

= r √ 2 sd◊ √ 1 + cos. § = r√2fde sin.0 x

Then z = 2√2.rx √2=4r.

the whole length of the arc AMB 4r 4 times the radius of the circle.

The greatest ordinate is most easily found thus

У = sin. = r sin. 0. (1 + cos. 0)