作者LuisSantos ( )
看板trans_math
標題Re: [微分] tangent line
時間Wed Jul 2 13:33:28 2008
※ 引述《JULIKEBEN (JU)》之銘言：
: the length of the portion , cut off by x-axis and y-axis , of any tangent line to the curve
: x^(2/3)+y^(2/3)=4 is
: a.2 b.4 c.8 d.none of above
: 請問要怎麼下手呢??
: 謝謝
設點P(x_0 , y_0)為位於 x^(2/3) + y^(2/3) = 4 第一象限上的點
則 x_0 > 0 , y_0 > 0 , 且 (x_0)^(2/3) + (y_0)^(2/3) = 4
x^(2/3) + y^(2/3) = 4 , 將方程式等號兩邊對x微分得
2 -1 2 -1 dy
(---)(x^(---)) + (---)(y^(---))(----) = 0
3 3 3 3 dx
dy y^(1/3)
=> ---- = - --------
dx x^(1/3)
dy | (y_0)^(1/3)
=> ----| = - -----------
dx |(x_0 , y_0) (x_0)^(1/3)
所以過P之切線方程式為
(y_0)^(1/3)
y - y_0 = - (-----------)(x - x_0)
(x_0)^(1/3)
令 y = 0 => ((x_0)^(1/3))(y_0) = ((y_0)^(1/3))(x - x_0)
=> x - x_0 = ((x_0)^(1/3))((y_0)^(2/3))
=> x = ((x_0)^(1/3))((x_0)^(2/3) + (y_0)^(2/3)) = (4)((x_0)^(1/3))
令 A ((4)((x_0)^(1/3)) , 0)
令 x = 0 => y - y_0 = ((y_0)^(1/3))((x_0)^(2/3))
=> y = ((y_0)^(1/3))((x_0)^(2/3) + (y_0)^(2/3)) = (4)((y_0)^(1/3))
令 B (0 , (4)((y_0)^(1/3)))
線段AB的長度
= √(((4)((x_0)^(1/3)))^2 + ((4)((y_0)^(1/3)))^2)
= √((16)((x_0)^(2/3) + (y_0)^(2/3)))
= √(16*4) = √64 = 8
故選c
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