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Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by F(x,y,c)=0{F}{\left({x},{y},{c}\right)}={0}, where c{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{G}{\left({x},{y},{k}\right)}={0}, such that every curve in this new family intersects at right angles every curve in the original family.

We implicitly differentiate the first equation with respect to x{x}, then eliminate c{c} between this derived equation and the equation F(x,y,c)=0{F}{\left({x},{y},{c}\right)}={0}. This gives an equation connecting x{x}, y{y}, and y′{y}', which we solve for y′{y}' to obtain a differential equation of the form dydx=f(x,y)\frac{{{d}{y}}}{{{d}{x}}}={f{{\left({x},{y}\right)}}}.

The orthogonal trajectories then are the solutions of dydx=−1f(x,y)\frac{{{d}{y}}}{{{d}{x}}}=-\frac{{1}}{{{f{{\left({x},{y}\right)}}}}}.

For many families of curves, one cannot explicitly solve for dydx\frac{{{d}{y}}}{{{d}{x}}} and obtain a differential equation of the form dydx=f(x,y)\frac{{{d}{y}}}{{{d}{x}}}={f{{\left({x},{y}\right)}}}.

Example 1. Find the orthogonal trajectories of the family of curves x2+y2=c2{{x}}^{{2}}+{{y}}^{{2}}={{c}}^{{2}}.

Rewrite as x2+y2−c2=0{{x}}^{{2}}+{{y}}^{{2}}-{{c}}^{{2}}={0}, or F(x,y,c)=x2+y2−c2=0{F}{\left({x},{y},{c}\right)}={{x}}^{{2}}+{{y}}^{{2}}-{{c}}^{{2}}={0}. This family of curves consists of circles with centers at the origin and radii c{c}. Implicitly differentiating the given equation with respect to x{x}, we obtain that 2x+2ydydx=0{2}{x}+{2}{y}\frac{{{d}{y}}}{{{d}{x}}}={0}, or dydx=−xy\frac{{{d}{y}}}{{{d}{x}}}=-\frac{{x}}{{y}}.

Here, f(x,y)=−xy{f{{\left({x},{y}\right)}}}=-\frac{{x}}{{y}}; so, the orthogonal trajectories satisfy the equation dydx=yx\frac{{{d}{y}}}{{{d}{x}}}=\frac{{y}}{{x}}. This equation is separable: dyy=dxx\frac{{{d}{y}}}{{y}}=\frac{{{d}{x}}}{{x}}, or ln⁡(y)=ln⁡(x)+a{\ln{{\left({y}\right)}}}={\ln{{\left({x}\right)}}}+{a}, which can be rewritten as y=kx{y}={k}{x}, where k=ea{k}={{e}}^{{a}}.

The solution y=kx{y}={k}{x} represents the orthogonal trajectories (straight lines that pass through the origin).

Let's solve another example.

Example 2. Find the orthogonal trajectories of the family of curves y2=4cx{{y}}^{{2}}={4}{c}{x}.

Rewrite as y2−4cx=0{{y}}^{{2}}-{4}{c}{x}={0}, or F(x,y,c)=y2−4cx=0{F}{\left({x},{y},{c}\right)}={{y}}^{{2}}-{4}{c}{x}={0}. Implicitly differentiating the given equation with respect to x{x}, we obtain that 2ydydx−4c=0{2}{y}\frac{{{d}{y}}}{{{d}{x}}}-{4}{c}={0}, or dydx=2cy\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{2}{c}}}{{y}}.

Note that, from the initial equation, c=y24x{c}=\frac{{{{y}}^{{2}}}}{{{4}{x}}}; so, dydx=2(y24x)y\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{2}{\left(\frac{{{{y}}^{{2}}}}{{{4}{x}}}\right)}}}{{y}}, or dydx=y2x\frac{{{d}{y}}}{{{d}{x}}}=\frac{{y}}{{{2}{x}}}.

Here, f(x,y)=y2x{f{{\left({x},{y}\right)}}}=\frac{{y}}{{{2}{x}}}; so, the orthogonal trajectories satisfy the equation dydx=−2xy\frac{{{d}{y}}}{{{d}{x}}}=-{2}\frac{{x}}{{y}}. This equation is separable: it can be rewritten as ydy=−2xdx{y}{d}{y}=-{2}{x}{d}{x}, or y22=−x2+k\frac{{{{y}}^{{2}}}}{{2}}=-{{x}}^{{2}}+{k}.

So, the orthogonal trajectories are x2+y22=k{{x}}^{{2}}+\frac{{{{y}}^{{2}}}}{{2}}={k} (k>0)\left({k}>{0}\right).