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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク inverse curve ： ウィキペディア英語版 inverse curve In geometry, an inverse curve of a given curve ''C'' is the result of applying an inverse operation to ''C''. Specifically, with respect to a fixed circle with center ''O'' and radius ''k'' the inverse of a point ''Q'' is the point ''P'' for which ''P'' lies on the ray ''OQ'' and ''OP''·''OQ'' = ''k''2. The inverse of the curve ''C'' is then the locus of ''P'' as ''Q'' runs over ''C''. The point ''O'' in this construction is called the center of inversion, the circle the circle of inversion, and ''k'' the radius of inversion. An inversion applied twice is the identity transformation, so the inverse of an inverse curve with respect to the same circle is the original curve. Points on the circle of inversion are fixed by the inversion, so its inverse is itself. ==Equations== The inverse of the point (''x'', ''y'') with respect to the unit circle is (''X'', ''Y'') where :$X\; =\; \backslash frac,\backslash \; Y=\backslash frac,$ or equivalently :$x\; =\; \backslash frac,\backslash \; y=\backslash frac.$ So the inverse of the curve determined by ''f''(''x'', ''y'') = 0 with respect to the unit circle is :$f\backslash left(\backslash frac,\backslash \; \backslash frac\backslash right)=0.$ It is clear from this that inverting an algebraic curve of degree ''n'' with respect to a circle produces an algebraic curve of degree at most 2''n''. Similarly, the inverse of the curve defined parametrically by the equations :$x\; =\; x(t),\backslash \; y\; =\; y(t)$ with respect to the unit circle is given parametrically as :$X=X(t)=\backslash frac,\backslash \; Y=Y(t)=\backslash frac.$ This implies that the circular inverse of a rational curve is also rational. More generally, the inverse of the curve determined by ''f''(''x'', ''y'') = 0 with respect to the circle with center (''a'', ''b'') and radius ''k'' is :$f\backslash left(a+\backslash frac,\backslash \; b+\backslash frac\backslash right)=0.$ The inverse of the curve defined parametrically by :$x\; =\; x(t),\backslash \; y\; =\; y(t)$ with respect to the same circle is given parametrically as :$X=X(t)=a+\backslash frac,\backslash \; Y=Y(t)=b+\backslash frac.$ In polar coordinates, the equations are simpler when the circle of inversion is the unit circle. The inverse of the point (''r'', θ) with respect to the unit circle is (''R'', Θ) where :$R\; =\; \backslash frac,\backslash \; \backslash Theta=\backslash theta,$ or equivalently :$r\; =\; \backslash frac,\backslash \; \backslash theta=\backslash Theta.$ So the inverse of the curve ''f''(''r'', ''θ'') = 0 is determined by ''f''(1/''R'', Θ) = 0 and the inverse of the curve ''r'' = ''g''(θ) is ''r'' = 1/''g''(''θ''). 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「inverse curve」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.