有y X= ≈l 2RI 2RI 2R 6 R J =二xn≈l= x=0 Rl R R l、FA(x0,y0) 故在终端A的曲率为 0 C(x020) k R 3|x=x0 (1+y 4R2 l <<1,略去二次项 R 4R2得k,≈1 R 上页2 0 2 1 0 x Rl y 有  x= x = 2 2 1 l Rl  , 2R l = 0 1 0 x Rl y x= x = l Rl 1  , 1 R = 故在终端A的曲率为 0 2 3 2 (1 ) A x x y y k = +   = 2 3 2 2 ) 4 (1 1 R l R +   1, R l  . 1 R , 得 kA  4 2 2 R l 略去二次项 x y o R ( , ) 0 0 A x y ( ,0) 0 C x l