Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by F(x,y,c)=0 where c denotes the parameter. The problem is to find another one-parameter family of curves, called the orthogonal trajectories of the family defined above and given analytically by G(x,y,k)=0 such that every curve in this new family intersects at right angles every curve in the original family.

We implicitly differentiate first equation with respect to x, then eliminate c between this derived equation and equation F(x,y,c)=0. This gives an equation connecting x, y, and y', which we solve for y' to obtain a differential equation of the form `(dy)/(dx)=f(x,y)` .

The orthogonal trajectories then are the solutions of `(dy)/(dx)=-1/(f(x,y))` .

For many families of curves, one cannot explicitly solve for `(dy)/(dx)` and obtain a differential equation of the form `(dy)/(dx)=f(x,y)` .

Example 1. Find the orthogonal trajectories of the family of curves `x^2+y^2=c^2` .

Rewrite as `x^2+y^2-c^2=` or `F(x,y,c)=x^2+y^2-c^2=0` . This family of curves is consists of circles with centers at the origin and radii c. Implicitly differentiating the given equation with respect to x, we obtain that `2x+2y(dy)/(dx)=0` or `(dy)/(dx)=-x/y`.

Here `f(x,y)=-x/y` so orthogonal trajectories satisfy equation `(dy)/(dx)=y/x` . This equation is separable: `(dy)/y=(dx)/x` or `ln(y)=ln(x)+a` which can be rewritten as `y=kx` where `k=e^a` .

Solution y=kx represents orthogonal trajectories (straight lines that pass through the origin).

Example 2. Find the orthogonal trajectories of the family of curves `y^2=4cx` .

Rewrite as `y^2-4cx=0` or `F(x,y,c)=y^2-4cx=0` . Implicitly differentiating the given equation with respect to x, we obtain that `2y(dy)/(dx)-4c=0` or `(dy)/(dx)=(2c)/y` .

Note that from initial equation `c=(y^2)/(4x)` , so `(dy)/(dx)=(2((y^2)/(4x)))/y` or `(dy)/(dx)=y/(2x)` .

Here `f(x,y)=y/(2x)` so orthogonal trajectories satisfy equation `(dy)/(dx)=-2x/y` . This equation is separable: it can be rewritten as `ydy=-2xdx` or `(y^2)/2=-x^2+k` .

So, orthogonal trajectories are `x^2+(y^2)/2=k` (k>0).