example of isogonal trajectory

Determine the curves which intersect the origin-centered circles at an angle of ${45}^{\circ }$.

The differential equation of the circles  $x^2+y^2=R^2$  is  $2xdx+2ydy=0$,  i.e.

$$\frac{x}{y}+\frac{dy}{dx}=\mathrm{\hspace{0.33em}0}.$$

Thus, by the model (2) of the parent entry (http://planetmath.org/IsogonalTrajectory), the differential equation of the isogonal trajectory reads

${\displaystyle \frac{x}{y}}+{\displaystyle \frac{y^{\prime }-\tan \frac{\pi }{4}}{1+y^{\prime }\tan \frac{\pi }{4}}}=\mathrm{\hspace{0.33em}0},$ (1)

which can be rewritten as

$$y^{\prime }=\frac{y-x}{y+x}=\frac{\frac{y}{x}-1}{\frac{y}{x}+1}.$$

Here, one may take  $\frac{y}{x}:=t$  as a new variable (see ODE types reductible to the variables separable case), when

$$y=xt,y^{\prime }=\frac{dy}{dx}=t+x\frac{dt}{dx},$$

and in the resulting equation

$$t+x\frac{dt}{dx}=\frac{t-1}{t+1}$$

one can separate the variables (http://planetmath.org/SeparationOfVariables):

$$\frac{1+t}{1+t^2}dt=-\frac{dx}{x}$$

Multiplying here by 2 and integrating then give

$$2\arctan t+\ln (1+t^2)=-2\ln x+\ln C^2\equiv -\ln \frac{x^2}{C^2},$$

or equivalently

$$\ln \frac{x^2+x^2{t}^2}{C^2}=-2\arctan t.$$

This is

$$\ln \frac{\sqrt{x^2+y^2}}{C}=-\arctan \frac{y}{x},$$

i.e.

$$\sqrt{x^2+y^2}=C{e}^{-\arctan \frac{y}{x}}.$$

Expressing this in the polar coordinates $r,\phi$ gives the family of the integral curves of the equation (1) in the form

$$r=C{e}^{-\phi }.$$

Consequently, the family of the isogonal trajectories consists of logarithmic spirals.

Title	example of isogonal trajectory
Canonical name	ExampleOfIsogonalTrajectory
Date of creation	2013-03-22 18:59:23
Last modified on	2013-03-22 18:59:23
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Example
Classification	msc 51N20
Classification	msc 34A26
Classification	msc 34A09
Synonym	isogonal trajectories of concentric circles
Related topic	IsogonalTrajectory