微積分lagrange

Question

find the points on the curve of intersection of the two surfacesx^2 y^2=1 and x^2-xy y^2-z^2=1which are nearest to the origin

我的作法是用f=x^2 y^2 z^2,g1=x^2 y^2-1,g2=x^2-xy y^2-z^2-1
然後▽f=a▽g1 b▽gx
找到(1/根號2 , -1/根號2 , 1/根號2)但是後來發現(1,0,0)帶進去會更小
請問這題要怎麼做謝謝

▽f=a▽g1 b▽g2這裡打錯了

Answer 1

我會這樣做
x2＋ y2=1 ...(1)
x2 - xy＋y2 - z2=1 => z2 = -xy ... (2)
設交點 (t, √(1 - t2), √[- t√(1 - t2)]), 0 <= t <= 1
與原點距離 => √[t2 ＋ (1 - t2) ＋ t√(1 - t2)]
找 f(t) = 1 ＋ t√(1 - t2)] 最短距離
f'(t) = √(1 - t2) - t2/√(1 - t2)
f'(t) = 0 => t = 1/√2
f(t) = 3/2
f(0) = 1, f(1) = 1 (0, 1 是端點)
=> (0, 1, 0) 是最接進近的點
又 x, y 有平方 且互為對稱
所以 (0, -1, 0), (1, 0, 0), (-1, 0, 0) 皆為所求
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.

Answer 2

ㄏㄏ 真會猜
ㄏㄏ 真夠閒
ㄏㄏ 真JJ

Answer 3

等一下我有時間再幫你