Re: [微分] tangent line

[LuisSantos ( )]

※ 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



--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 140.119.66.56
推 JULIKEBEN:謝謝  懂了