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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
人气:264 ℃ 时间:2020-05-21 16:13:30
解答
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"
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