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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 Orthogonal trajectory ： ウィキペディア英語版 Orthogonal trajectory In mathematics, orthogonal trajectories are a family of curves in the plane that intersect a given family of curves at right angles. The problem is classical, but is now understood by means of complex analysis; see for example harmonic conjugate. For a family of level curves described by $g(x,\; y)\; =\; C$, where $C$ is a constant, the orthogonal trajectories may be found as the level curves of a new function $f(x,\; y)$ by solving the partial differential equation :$\backslash nabla\; f\; \backslash cdot\; \backslash nabla\; g\; =\; 0$ for $f(x,\; y)$. This is literally a statement that the gradients of the functions (which are perpendicular to the curves) are orthogonal. The partial differential equation may be avoided by instead equating the tangent of a parametric curve $\backslash vec\; r(t)$ with the gradient of $g(x,\; y)$: :$\backslash frac\backslash left(\backslash vec\; r(t)\backslash right)\; =\; \backslash nabla\; g$ which will result in two possibly coupled ordinary differential equations, whose solutions are the orthogonal trajectories. Note that with this formula, if $g$ is a function of three variables its level sets are surfaces, and the family of curves $\backslash vec\; r(t)$ are orthogonal to the surfaces. ==Example 1: Circle== Suppose we are given the family of circles centered about the origin: :$x^2\; +\; y^2\; =\; c$ The orthogonal trajectories of this family are the family of curves such that intersect the circle at right angles. Given a line, the negative reciprocal of its slope is the slope of a perpendicular line. In calculus terms, if y and k are two perpendicular lines: :$$ \left\y} = f(x,y) \\ \frac = -\frac \end\right. Orthogonal trajectories of the given family of circles are no different. The tangent line of an orthogonal trajectory is perpendicular to the tangent line of the circle, and vice versa. By implicit differentiation, :$$ x^2 + y^2 = c \Rightarrow 2x + 2y\frac = 0 Solving the equation, one acquires the following: :$$ \frac = \frac Therefore, an orthogonal trajectory must satisfy the differential equation: :$$ \frac = \frac Solving the equation for y, we find the orthogonal trajectories: :$$ y = mx The orthogonal trajectories of a family of circles centered at the origin are linear equations containing the origin. In polar coordinates, the family of circles centered about the origin is the level curves of :$r\; -\; R\; =\; 0$ where $R$ is the radius of the circle. Then the orthogonal trajectories are the level curves of $f$ defined by: :$\backslash nabla\; f\; \backslash cdot\; \backslash nabla\; (r\; -\; R)\; =\; 0$ :$\backslash left(\backslash frac,\; \backslash frac\; \backslash frac\backslash right)\; \backslash cdot\; \backslash left(\backslash frac(r\; -\; R),\; \backslash frac\; \backslash frac(r\; -\; R)\backslash right)\; =\; 0$ :$\backslash frac\; =\; 0\; \backslash quad\; \backslash Rightarrow\; \backslash quad\; f\; =\; f(\backslash theta)$ The lack of complete boundary data prevents determining $f(\backslash theta)$. However, we want our orthogonal trajectories to span every point on every circle, which means that $f(\backslash theta)$ must have a range which at least include one period of rotation. Thus, the level curves of $f(\backslash theta)\; =\; 0$, with freedom to choose any $f$, are all of the $\backslash theta\; =\; constant$ curves that intersect circles, which are (all of the) straight lines passing through the origin. Note that the dot product takes nearly the familiar form since polar coordinates are orthogonal. The absence of boundary data is a good thing, as it makes solving the PDE simple as one doesn't need to contort the solution to any boundary. In general, though, it must be ensured that ''all'' of the trajectories are found. 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「Orthogonal trajectory」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.