显然dx/dt=a(1-cost)dy/dt=a*sint那么dy/dx=sint /(1-cost)继续求二阶导就得到d(dy/dx)/dt *dt/dx=[(sint)' *(1-cost) -sint *(1-cost)']/(1-cost)^2 *1/ a(1-cost)=(cost-1)/(1-cost)^2 *1/ a(1-cost)= -1/ [a(1-cost)^2]
