Isogonal trajectory

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经过$(0,\pm a)$的圆组:\begin{equation}(x-r)^2+y^2=a^2+r^2\label1\end{equation}在点$(x,y)$处的切线斜率:\begin{equation}\frac{r-x}{y}\label2\end{equation}由\eqref{1}解得\begin{equation}r=\frac{-a^2+x^2+y^2}{2 x}\label3\end{equation}曲线的切线(斜率为$y'$)旋转一个固定的角度$\tan^{-1}k$, 斜率变为$\frac{y'-k}{1+k·y'}$. 将\eqref{3}代入\eqref{2}得\begin{equation}\label4\frac{y'-k}{1+k·y'}=\frac{\frac{-a^2+x^2+y^2}{2 x}-x}{y}\end{equation}当$k=1$时解得\begin{equation}\tanh ^{-1}\left(\frac{2 a y}{a^2+x^2+y^2}\right)-\tan ^{-1}\left(\frac{2 a x}{-a^2+x^2+y^2}\right)=C\end{equation}

右图$k=1$即45°交角 用ContourPlot画出来:

1. With[{a=0.5},Show[Graphics[Table[{Red,Circle[{c=(-a^2+b^2)/(2 b),0},b-c,{ArcTan[-c,a],-ArcTan[-c,a]}]},{b,0.1,0.5,0.1}]~Join~Table[{Red,Circle[{c=(-a^2+b^2)/(2 b),0},c-b,{Pi-ArcTan[c,a],Pi+ArcTan[c,a]}]},{b,-0.5,-0.1,0.1}]~Join~{Red,Line[{{0,a},{0,-a}}]}],ContourPlot[-ArcTan[(2 a x)/(-a^2+x^2+y^2)]+ArcTanh[(2 a y)/(a^2+x^2+y^2)],{x,y}\[Element]Disk[{0,0},a],Contours->20,ContourShading->None]]]

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Given a family of curves, assumed to be differentiable, an isocline for that family is formed by the set of points at which some member of the family attains a given slope.

Isoclines (blue), slope field (black), and some solution curves (red) of $y' = xy$.

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对于一般的$k$可以解出来\eqref{4}吗

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Isogonal trajectory A curve, which intersects any curve of a given pencil of (planar) curves by a fixed angle $ \alpha $ is called isogonal trajectory. Between the slope $ \eta ' $ of an isogonal trajectory and the slope $ y' $ of the curve of the pencil at a point $ (x,y) $ the following relation holds: $$ \eta '={\frac {y'+\tan(\alpha )}{1-y'\tan(\alpha )}}$$ This relation is due to the formula for $ \tan(\alpha +\beta ) $. For $ \alpha \rightarrow 90^{\circ } $ one gets the condition for the orthogonal trajectory. For the determination of the isogonal trajectory one has to adjust the 3. step of the instruction above: 3. step (isog. traj.) The differential equation of the isogonal trajectory is: $$y'={\frac {f(x,y)+\tan(\alpha )}{1-f(x,y)\;\tan(\alpha )}}$$ For the 1. example (concentric circles) and the angle $ \alpha =45^{\circ } $ one gets$$\ y'={\frac {-x/y+1}{1+x/y}}$$ This is a special kind of differential equation, which can be transformed by the substitution $ z=y/x $ into a differential equation, that can be solved by separation of variables. After reversing the substitution one gets the equation of the solution: $$ \arctan {\frac {y}{x}}+{\frac {1}{2}}\ln(x^{2}+y^{2})=C\ . $$ Introducing polar coordinates leads to the simple equation $$ C-\varphi =\ln(r)$$ which describes logarithmic spirals (see diagram).