Thus, CP being the involute of the O CQD, whose radius is unit, let the ordinates be represented by p. Let also P'y be a tangent at P'and Ay a I upon it. Then Dp' being the unwound line it must be I curve, or I p'y. It is also I AD. :: p'y = AD. Now from similar A, we have dó ✓p2 – 1 which is the same form as p 1 :: obtained by unwinding the thread CD so as to describe the AC2 involute, and taking AP always = The spiral is AP AP called the Complicated Tractriz. 23. Let AB = x (See Fig. C. P. p. 209) BP Y, the radius of the O = 1, and BC =y'. y'r Hence y = the equation of the curve APF. 2 2.x x The curve cuts the semicircle when y' = y, or when - APFMA – ACFM APFM = AGCFPA. 24. Let NM (Fig. 16.) the line given in length = m, 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 DAа be that position of NM which is I BC, and since AD is given, let it = n. Also let PM = y and AP = x. Then from similar A I : 1 :: AM : m AM .. : n + X :: AM - V x2 + y2 : m. Hence w m?-(n + x)? which is the equation of 1+ the curve. It is an oval similar and equal on each side the diameter. Pi 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', AP 2 BAB = B, and _BAP = 0, and AC, the radius of the o Cpr = 1. Then PR = QR, tan. Q= dp.tan, a also = p.do. de P Р 1. p B lor the equation of B to modulus Ir' – Ir the curve expressed in terms of 0, and given quantities. The points of the curve corresponding to every possible value of e, may therefore be found, by reference to the Tables ; or the curve may be constructed. or = log. (6) 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 I OQ the radius. Hence L AQM = R. L-LO AQ = R. L - L AQO E L AQT = L QTP, :: LQTP = L AQM = L TPQ .. TQ = QP = arc AQ 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 I from A, and let MR meet the diameter of the o produced in T. Join CR (C being the centre). Then referring the locus BMA to A as a pole, and : putting AM = p and Z MAB = 0 (B being the point where p = the diameter of the circle), since CR is parallel to AM we have < MAR = L ARC = Z RAN :. AM = AN = AC + CN = r.(1 + cos. 6.) Hence p=r.(1 + cos. 6) ....... (a) the polar equation of the curve. Again, referring the curve to rectangular co-ordinates, by putting Ap = x, PM = y, we get V x2 + y2 = AM'= AN (_ RAN = L ARC = { MAR) =rCN = r + r , cos. 0. sin. 0 = tan. 0 = Ni-cos. But y .. cos. Os √ x2 + y2 ✓ 2 + y2 Hence y* + (2.22 Zrr 92) y? = 28x3 x" the equation between the rectangular co-ordinates. Let y = 0. Then x = 0 and = 2r, :. the curve passes through the points A and B. To find the area of the curve, we have cos.' 0 T To find the length of the curve we have = I dp+ pod0% = $d6 + d42 ✓ 2 sdo 11 + cos. O = rm2 s do sin.8 x 1 + cos. O do sin. 0 NI - cos. 8 :: 2 = 2^2. rv 1 - cos. 8. Let 0 = " Then z = 2 2 x 12 = 4r. :: the whole length of the arc AMB = 4r = 4 times the radius of the circle. The greatest ordinate is most easily found thus y = p. sin. 6 = r sin. 8. (1 + cos. 6) dy =r cos. 8. (1 + cos. 8) – r sin.28. sin.2 0 = 0 1 or cos 20+ |