Question

设函数y=y(x)由参数方程x=1/3t3+t+1/3

设函数y=y(x)有参数方程{x=1-3t-(t^3) y=1-3t+(t^3)}确定,求该函数的极值

Answer 1

简单计算一下即可，答案如图所示

Answer 2

dx/dt=-3-3t²
dy/dt=-3+3t²
y'=(dy/dt)/(dx/dt)=(-3+3t²)/(-3-3t²)=(1-t²)/(1+t²)
由y'=0得：t=1,-1
dy'/dt=-4t/(1+t²)²
y"=-4t/(1+t²)²/(-3-3t²)=(4t/3)/(1+t²)³
t=-1时,y"