Calculus:tangent plane!
2010-06-18 4:31 am
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回答 (1)
2010-06-18 5:09 am
✔ 最佳答案
f(x,y,z)=x^2-y^2-z^2-1, grad(f)=<2x,-2y,-2z>//<x, -y, -z>Let a point P(a,b,c) on x^2-y^2-z^2=1 with the tangent plane parallels to z=x+y,
then <a, -b, -c>//<1,1,-1>
thus (a,b,c)=(t, -t, t) on x^2-y^2-z^2=1
t^2-t^2-t^2=1, t^2=-1, no solutions
so that, no tangent plane of x^2-y^2-z^2=1 parallels to z=x+y.
收錄日期: 2021-04-30 14:55:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100617000015KK07432