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| In [[mathematics]], '''orthogonal trajectories''' are a family of curves in the plane that intersect a given family of curves at [[right angle]]s. The problem is classical, but is now understood by means of [[complex analysis]]; see for example [[harmonic conjugate]].
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| For a family of [[level curve]]s described by <math>g(x, y) = C</math>, where <math>C</math> is a constant, the orthogonal trajectories may be found as the level curves of a new function <math>f(x, y)</math> by solving the [[partial differential equation]]
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| :<math>\nabla f \cdot \nabla g = 0</math>
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| for <math>f(x, y)</math>. This is literally a statement that the [[gradient]]s of the functions (which are perpendicular to the curves) are orthogonal. Note that if <math>f</math> and <math>g</math> are functions of three variables instead of two, the equation above will be [[nonlinear]] and will specify orthogonal [[surface]]s.
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| The partial differential equation may be avoided by instead equating the tangent of a [[parametric curve]] <math>\vec r(t)</math> with the gradient of <math>g(x, y)</math>:
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| :<math>\frac{d}{d t}\left(\vec r(t)\right) = \nabla g</math>
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| which will result in two possibly coupled ordinary differential equations, whose solutions are the orthogonal trajectories. Note that with this formula, if <math>g</math> is a function of three variables its [[level set]]s are surfaces, and the family of curves <math>\vec r(t)</math> are orthogonal to the surfaces.
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| ==Example: circle==
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| In [[polar coordinates]], the family of circles centered about the origin is the level curves of
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| :<math>r - R = 0</math>
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| where <math>R</math> is the radius of the circle. Then the orthogonal trajectories are the level curves of <math>f</math> defined by: | |
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| :<math>\nabla f \cdot \nabla (r - R) = 0</math>
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| :<math>\left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}\right) \cdot \left(\frac{\partial}{\partial r}(r - R), \frac{1}{r} \frac{\partial}{\partial \theta}(r - R)\right) = 0</math>
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| :<math>\frac{\partial f}{\partial r} = 0 \quad \Rightarrow \quad f = f(\theta)</math>
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| The lack of complete boundary data prevents determining <math>f(\theta)</math>. However, we want our orthogonal trajectories to span every point on every circle, which means that <math>f(\theta)</math> must have a range which at least include one period of rotation. Thus, the level curves of <math>f(\theta) = 0</math>, with freedom to choose any <math>f</math>, are all of the <math>\theta = constant</math> 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 coordinates|orthogonal]].
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| 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.
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| ==External links==
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| * [http://www.bluffton.edu/~nesterd/java/OrthTraj.html Exploring orthogonal trajectories] - applet allowing user to draw families of curves and their orthogonal trajectories.
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| [[Category:Curves]]
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| [[Category:Complex analysis]]
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The writer is known by the name of Numbers Lint. Managing individuals is what I do and the salary has been really satisfying. To gather badges is what her family members and her appreciate. Minnesota is exactly where he's been residing for years.
My web-site; http://장원고시원.kr