推 daoziwai : 感謝！我再慢慢看 04/09 19:41
※ 引述《daoziwai (daoziwai)》之銘言：
: 題目是這樣的
: Suppose you need to know an equation of the tangent plane to a surface S at th
: e
: r1(t)= (2+3t, 1-t^2, 3- 4t+t^3)
: r2(u)=(1+u^2 , 2u^3-1 , 2u+1)
: both lie on S. Find an equation of the tangent plane at P.
: 我只知道方程式會長成
: z=3+fx(x-2)+fy(y-1)
: 但是我不知道要怎麼從那兩個向量函數求出fx(x,y)跟fy(x,y)
: 老師連向量函數都沒介紹過...請求各位能夠深入淺出的讓我理解
: 感謝...
題目看不懂把他拿去google才知道你漏貼了
Suppose you need to know an equation of the tangent plane to a surface S at
the point P (2,1,3). ou don't have an equation for S but you know that the
curves:
r_1(t)= (2+3t, 1-t^2, 3- 4t+t^3)
r_2(u)=(1+u^2 , 2u^3-1 , 2u+1)
both lie on S. Find an equation of the tangent plane at P.
<sol>
idea：曲面S上的任一點P的切平面即是通過該點P的曲線的切線向量所形成的集合
即 T_P(S) := {α'(t)│α：[a,b]→S with α(t)=P}
經由計算 r_1'(t) = (3,-2t,4+3t^2), r_1'(0) = (3,0,4) and r_1(0) = P
r_2'(u) = (2u,6u^2,2), r_2'(1)=(2,6,2) and r_2(1) = P
因此T_P(S) = span{(3,0,4),(2,6,2)}
對了，以上是幾何定義切平面的方式(通過原點)，如果要通過P點，那答案就是P+T_P(S)
最後如果答案要求寫成ax+by+cz=d的形式，就...高中數學^^"
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