设常数a>0,曲线上点(a,a,a)处的切线方程是__________.

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问题 设常数a>0,曲线上点(a,a,a)处的切线方程是__________.

选项

答案x+y=2a,z=a

解析 将x看成自变量,方程两边对x求导,得yz+xy'z+xyz'=0及x+yy'=az'.将(x,y,z)=(a,a,a)代入,得y'(a)+z'(a)=-1,y'(a)-z'(a)=-1,解得y'(a)=-1,z'(a)=0.
所以切线方程为,即x+y=2a,z=a.
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