Re: [考古] 93台聯大填充第二題
※ 引述《losew (風城小菊花)》之銘言：
: Let L be the line tangent to the polar curve r(θ)＝sinθ-cosθ/sinθ+cosθ
: atθ＝0.The equation of L in x and y is _____.
: 我是想先解出dy/dx，但是超級複雜......
: 有高手會的嗎？？
sinθ-cosθ
r(θ) = -----------
sinθ+cosθ
sinθcosθ - (cosθ)^2
x(θ) = (r(θ))(cosθ) = ----------------------
sinθ+cosθ
dx (cos2θ+sin2θ)(sinθ+cosθ)-(cosθ-sinθ)(sinθcosθ - (cosθ)^2)
---- = ------------------------------------------------------------------
dθ (sinθ+cosθ)^2
(sinθ)^2 - (sinθ)(cosθ)
y(θ) = (r(θ))(sinθ) = --------------------------
sinθ+cosθ
dy (sin2θ-cos2θ)(sinθ+cosθ)-(cosθ-sinθ)((sinθ)^2 - sinθcosθ)
---- = ------------------------------------------------------------------
dθ (sinθ+cosθ)^2
dy (dy/dθ)
---- = ----------
dx (dx/dθ)
(sin2θ-cos2θ)(sinθ+cosθ)-(cosθ-sinθ)((sinθ)^2 - sinθcosθ)
= ------------------------------------------------------------------
(cos2θ+sin2θ)(sinθ+cosθ)-(cosθ-sinθ)(sinθcosθ - (cosθ)^2)
dy | (-1)(1) - (1)(0) -1
----| = ------------------- = ----
dx |θ=0 (1)(1) - (1)(0-1) 2
|
x = (r(θ))(cosθ) | = -1
|θ=0
|
y = (r(θ))(sinθ) | = 0
|θ=0
-1
因此所求直線方程式為 y = (---)(x+1)
2
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