254 DIFFERENTIAL CALCULUS . Asymptotes in polar coordinates. Let/(/3, 0)= be the equa- tion of the curve PQ having the asymptote CD. As the asymptote must pass within a finite distance (as OE} of the origin, and the point of contact is at an infinite distance, it is evident that the radius vector OF drawn to the point of contact is parallel to the asymptote, and the subtangent OE is perpendicular to it. Or, more precisely, the distance of the asymptote from the origin is the limiting value of the polar subtangent as the point of contact moves off an infinite distance. To determine the asymptotes to a polar curve, proceed as follows :. FIRST STEP. Find from the equation of the curve the values of 6 which make p = oo.* These values of 6 give the directions of the asymptotes. SECOND STEP. Find the limit of the polar subtangent de dp' by (7), p. 86 as 6 approaches each such value, remembering that p approaches oo at the same time. THIRD STEP. If the limiting value of the polar subtangent is finite, there is a corresponding asymptote at that distance from the origin and parallel to the radius vector drawn to the point of contact. When this limit is pos- itive the asymptote is to the right of the origin, and when negative, to the left, looking in the direction of the infinite radius vector. EXAMPLES . Examine the hyperbolic spiral p = - f or asymptotes.
Solution. When 0=0, p = oo. Also -!- = ; hence" dd u subtangent = p ^ = ^.- - = - a . It happens in this case that the subtangent is the same for all values of 0. The curve has therefore an asymptote BC parallel to the initial line OA and at a dis- tance a above it.
- If the equation can be written as a polynomial in p, these values of may be found by
equating to zero the coefficient of the highest power of />,
