dx/dt=3t^2+3dy/dt=3t^2-3y'=(dy/dt)/(dx/dt)=(t^2-1)/(t^2+1)=1-2/(t^2+1)dy'/dt=2/(t^2+1)*2t=4t/(t^2+1)^2y"=(dy'/dt)/(dx/dt)=4t/[3(t^2+1)^3]上凸，即y"<0, 得：t<0此时x=t^3+3t+1<1
