Question

求原点到曲面(x-y)^2-z^2=1的最短距离.

Answer 1

貌似是根号2/2

思路是对的呀

分别对x,y,z偏导

得

x/根号(x^2+y^2+z^2)+2к(x-y)=0

y/根号(x^2+y^2+z^2)-2к(x-y)=0

z/根号x^2+y^2+z^2+2кz=0

(x-y)^2-z^2=1

得x=-y z=0 即x=1/2 y=-1/2 z=0

此时距离为根号2/2