> 数学 >
求曲线x2+3y2+z2=9,z2=3x2+y2在点(1,1,2)处的切线与法平面方程.
人气:217 ℃ 时间:2020-04-12 01:28:45
解答
消去y得x^2+3(3x^2-z^2)+z^2=9对x求导得20x - 4z * z'(x) = 0所以z'(x) = 5x/z = 5/2消去x得z^2 = 3(9-z^2-3y^2)+y^2 =>4z^2 + 8y^2 = 27对y求导得z'(y) = -2y/z = -1所以切向量为(5,-2,2)切线为(x-1)/5=(1-y)/2=(z-...
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版