# 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  $2x\,dx+2y\,dy=0$,  i.e.

 $\frac{x}{y}+\frac{dy}{dx}\;=\;0.$

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

 $\displaystyle\frac{x}{y}+\frac{y^{\prime}-\tan\frac{\pi}{4}}{1+y^{\prime}\tan% \frac{\pi}{4}}\;=\;0,$ (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

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

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.

 $\sqrt{x^{2}\!+\!y^{2}}\;=\;Ce^{-\arctan\frac{y}{x}}.$

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

 $r\;=\;Ce^{-\varphi}.$

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

Title example of isogonal trajectory ExampleOfIsogonalTrajectory 2013-03-22 18:59:23 2013-03-22 18:59:23 pahio (2872) pahio (2872) 10 pahio (2872) Example msc 51N20 msc 34A26 msc 34A09 isogonal trajectories of concentric circles IsogonalTrajectory