. Differential and integral calculus. ± V2)j*\ 4. Find by expansion the asymptotes to the curves in Exercises3, 4, 5, § 73- POLAR CURVES. 76. To find the slope of the tangent to a polar curve. From § 19, we have tan a = -z-dx for the slope of a curve when referred to rectangular we assume the pole coincident with the origin and the initialline coincident with the .#-axis we have, Analytical Geometry,Art. 34, (3), x = r cos 0, y = r sin 6. Geometric Application 85 Hence, dx d (r sin 9) r cos 9 dO -f sin 9 dr d (r cos 9)is the required expression. Cor. from § 18 (3), we have cos 9
. Differential and integral calculus. ± V2)j*\ 4. Find by expansion the asymptotes to the curves in Exercises3, 4, 5, § 73- POLAR CURVES. 76. To find the slope of the tangent to a polar curve. From § 19, we have tan a = -z-dx for the slope of a curve when referred to rectangular we assume the pole coincident with the origin and the initialline coincident with the .#-axis we have, Analytical Geometry,Art. 34, (3), x = r cos 0, y = r sin 6. Geometric Application 85 Hence, dx d (r sin 9) r cos 9 dO -f sin 9 dr d (r cos 9)is the required expression. Cor. from § 18 (3), we have cos 9 dr — r sin 9 dO ds = ^Jdx2 + dy1 ;.-. ds = V^(rcos0)j2+ {d(rsin9)\2 = V^2 + r8 dP. 77. Length of Subtangent. Length of Tangent. Let MS be any curve referredto O as pole and OX as initialline. Let P (r, 9) be any pointof the curve at which a tangentPB and a normal PA are drawn;then OB and CX4, segmentsof the perpendicular to theradius vector OP, are respec-tively the subtangent and sub-normal corresponding to thepoint P (r, 9).. 1. To find a value for OB, the subtangent. From the triangle OPB OB = r tan ; buttan = tan (a — 9) = tan Fig. 11. tan 9 1 + tan a tan 9r cos 9 dO -f- sin 0 */r sin 0 cos 9 dr sin 0 d# cos 9 r cos 9 d9 -\- sin 0 /Zr sin 0 Art. (76), 1 + *. tan = cos 9 dr — r sin 9 d9rd9 cos 9 dr dO Subtangent = r2 — 86 Differential Calculus 2. To find a value for PB, the the triangle OPB PB = Vr2 + OB2 = y/r2 + r4^ . .-. Tangent = ry i+r2^-2. 78. Length of Subnormal. Length of Normal. Perpendicularto Tangent. From triangle OAP, Fig. 11,OA °P tan OAP tan d> /# Subnormal = -^;du aiso, A4 = Vtf/» + 04a = y r2 + ^; dr2Normal = y r2 + -j™ • From triangle OCB (OC being perpendicular to BP), wehave OB dr dr L/C- .. Perpendicular sec COB Vi + tan2 <£ 4 / fW +^2_r2 J ^ + r*dff1 Geometric Application 87 EXAMPLES. 1. A circle whose diameter is a is referred to the lower ex-tremity of its vertical diameter as a pole, and to the tangent atthat po
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