函数u=ln根号x2+y2+z2在点(2,1,-2)处的梯度gradf(2,1,-2)

2025-05-17 13:32:27
推荐回答(3个)
回答1:

简单计算一下即可,答案如图所示

回答2:

回答3:

先求偏导 af/ax=2x/(x^2+y^2+z^2)将点代入=2/9 af/ay=2y/(x^2+y^2+z^2)将点代入=4/9 af/az=2z/(x^2+y^2+z^2)将点代入=-4/9 gradf|M=(2/9)i+(4/9)j+(-4/9)k