作者Honor1984 (希望愿望成真)
看板trans_math
标题Re: [微分] 问题好多好多...
时间Sat Nov 1 16:17:44 2008
※ 引述《victor7935 (victor)》之铭言：
: 求
: 在这个曲线上
: (x^2+y^2)^2 = x^2 - y^2
: 4个水平切线的点 >..<
2(x^2+y^2)(2x+2yy') = 2x -2yy'
y' = 0
2(x^2 + y^2) = 1
代入原式
(1/2)^2 = x^2 - y^2
解出 x=±√(3/8)
y=±√(1/8)
另解
此图对x,y轴对称
只需考虑第一象限
R^4 = R^2 - 2R^2 (sint)^2
= R^2 [ 1 - 2(sint)^2]
=> R^2 = cos2t t = -π/4 -> π/4 and 3π/4 -> 5π/4
x = Rcost
y = Rsint
dR
2R--- = -2sin2t
dt
dy -sin(2t)sint
-- = ------------- + √cos2t * cost
dt √cos2t
dx -sin(2t)cost
-- = --------------- - √cos2t * sint
dt √cos2t
cos(3t)
-------- = 0
-sin(3t)
3t = 3π/2, π/2, 5π/2, 7π/2,.....
=> t = π/6
x = √(3 /8)
y = √(1/8)
四个点为(±√(3/8) , ±√(1/8))
--
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◆ From: 122.124.102.144
1F：推 victor7935:(1/4)=x^2-y^2.....这我不会解!!= = '140.123.221.147 11/02 19:33
2F：→ Honor1984:和2(x^2 + y^2) = 1联立解x^2和y^2 122.124.102.16 11/08 01:41