# isogonal trajectory

Let a one-parametric family of plane curves^{} $\gamma $ have the differential equation^{}

$F(x,y,{\displaystyle \frac{dy}{dx}})=\mathrm{\hspace{0.33em}0}.$ | (1) |

We want to determine the *isogonal trajectories* of this family, i.e. the curves $\iota $ intersecting all members of the family under a given angle, which is denoted by $\omega $.
For this purpose, we denote the slope angle of any curve $\gamma $ at such an intersection point by $\alpha $ and the slope angle of $\iota $ at the same point by $\beta $. Then

$$\beta -\alpha =\omega \mathit{\hspace{1em}}(\text{or alternatively}-\omega ),$$ |

and accordingly

$$\frac{dy}{dx}=\mathrm{tan}\alpha =\frac{\mathrm{tan}\beta -\mathrm{tan}\omega}{1+\mathrm{tan}\beta \mathrm{tan}\omega}=\frac{{y}^{\prime}-\mathrm{tan}\omega}{1+{y}^{\prime}\mathrm{tan}\omega},$$ |

where ${y}^{\prime}$ means the slope of $\iota $. Thus the equation

$F(x,y,{\displaystyle \frac{{y}^{\prime}-\mathrm{tan}\omega}{1+{y}^{\prime}\mathrm{tan}\omega}})=\mathrm{\hspace{0.33em}0}$ | (2) |

is satisfied by the derivative ${y}^{\prime}$ of the ordinate of $\iota $. In other , (2) is the differential equation of all isogonal trajectories of the given family of curves.

Note. In the special case $\omega =\frac{\pi}{2}$, it’s a question of orthogonal trajectories.

Title | isogonal trajectory |

Canonical name | IsogonalTrajectory |

Date of creation | 2013-03-22 18:59:20 |

Last modified on | 2013-03-22 18:59:20 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Derivation |

Classification | msc 51N20 |

Classification | msc 34A26 |

Classification | msc 34A09 |

Related topic | AngleBetweenTwoCurves |

Related topic | OrthogonalCurves |

Related topic | ExampleOfIsogonalTrajectory |

Related topic | AngleBetweenTwoLines |

Defines | isogonal trajectory |