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Orthogonal Trajectories
Definition and Examples
Let a family of curves be given by the equation $g\left( {x,y} \right) = C,$ where $$C$$ is a constant. For the given family of curves, we can draw the orthogonal trajectories, i.e. another family of curves $$f\left( {x,y} \right) = C$$ that cross the given curves at right angles.

For example, the orthogonal trajectory of the family of straight lines defined by the equation $$y = kx,$$ where $$k$$ is a parameter (the slope of the straight line), is any circle having centre at the origin (Figure $$1$$): ${x^2} + {y^2} = {R^2},$ where $$R$$ is the radius of the circle.
 Fig.1 Fig.2
Similarly, the orthogonal trajectories of the family of ellipses $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{c^2} - {a^2}}} = 1,\;\;\text{where}\;\;0 < a < c,$ are confocal hyperbolas satisfying the equation: $\frac{{{x^2}}}{{{b^2}}} - \frac{{{y^2}}}{{{b^2} - {c^2}}} = 1,\;\;\text{where}\;\;0 < c < b.$ Both families of curves are sketched in Figure $$2.$$ Here $$a$$ and $$b$$ play the role of parameters describing the family of ellipses and hyperbolas, respectively.
General Method of Finding Orthogonal Trajectories
The common approach for determining orthogonal trajectories is based on solving the partial differential equation: $\nabla f\left( {x,y} \right) \cdot \nabla g\left( {x,y} \right) = 0,$ where the symbol $$\nabla$$ means the gradient of the function $$f\left( {x,y} \right)$$ or $$g\left( {x,y} \right)$$ and the dot means the dot product of the two gradient vectors.

Using the definition of gradient, one can write: $\nabla f\left( {x,y} \right) = \mathbf{grad}\,f\left( {x,y} \right) = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right),$ $\nabla g\left( {x,y} \right) = \mathbf{grad}\,g\left( {x,y} \right) = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right).$ Hence, the partial differential equation is written in the form: ${\nabla f\left( {x,y} \right) \cdot \nabla g\left( {x,y} \right) = 0,}\;\; {\Rightarrow \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) \cdot \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right) = 0,}\;\; {\Rightarrow \frac{{\partial f}}{{\partial x}}\frac{{\partial g}}{{\partial x}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial g}}{{\partial y}} = 0.}$ Solving the last PDE, we can determine the equation of the orthogonal trajectories $$f\left( {x,y} \right) = C.$$
A Practical Algorithm for Constructing Orthogonal Trajectories
Below we describe an easier algorithm for finding orthogonal trajectories $$f\left( {x,y} \right) = C$$ of the given family of curves $$g\left( {x,y} \right) = C$$ using only ordinary differential equations. The algorithm includes the following steps:
1. Construct the differential equation $$G\left( {x,y,y'} \right) = 0$$ for the given family of curves $$g\left( {x,y} \right) = C.$$ See the web page Differential Equations of Plane Curves about how to do this.

2. Replace $$y'$$ with $$\left( { - \large\frac{1}{{y'}}\normalsize} \right)$$ in this differential equation. As a result, we obtain the differential equation of the orthogonal trajectories.

3. Solve the new differential equation to determine the algebraic equation of the family of orthogonal trajectories $$f\left( {x,y} \right) = C.$$

Example 1
Find the orthogonal trajectories of the family of straight lines $$y = Cx,$$ where $$C$$ is a parameter.

Solution.
We apply the algorithm described above.

$$1)$$ First, we construct the differential equation for the family of straight lines $$y = Cx.$$ By differentiating the last equation with respect to $$x,$$ we get: $y' = C = \text{const}.$ Eiminate the constant $$C$$ from the system of equations: $\left\{ \begin{array}{l} y = Cx\\ y' = C \end{array} \right.,\;\; \Rightarrow y' = \frac{y}{x}.$ We obtain the differential equation of the initial set of straight lines.

$$2)$$ Replace $$y'$$ with $$\left( { - \large\frac{1}{{y'}}\normalsize} \right).$$ This gives the differential equation of the orthogonal trajectories: $- \frac{1}{{y'}} = \frac{y}{x},\;\; \Rightarrow y' = - \frac{x}{y}.$ $$3)$$ Now we solve the last differential equation to find the algebraic equation of the family of orthogonal trajectories: ${y' = - \frac{x}{y},}\;\; {\Rightarrow \frac{{dy}}{{dx}} = - \frac{x}{y},}\;\; {\Rightarrow ydy = - xdx,}\;\; {\Rightarrow \int {ydy} = - \int {xdx} ,}\;\; {\Rightarrow \frac{{{y^2}}}{2} = - \frac{{{x^2}}}{2} + C,}\;\; {\Rightarrow \frac{{{x^2}}}{2} + \frac{{{y^2}}}{2} = C,}\;\; {\Rightarrow {x^2} + {y^2} = 2C.}$ By replacing $$2C$$ with $${R^2}$$ we see that the orthogonal trajectories for the family of straight lines are concentric circles (Figure $$1$$): ${x^2} + {y^2} = {R^2}.$
Example 2
A family of hyperbolic curves is given by the equation $$y = {\large\frac{C}{x}\normalsize}.$$ Find the orthogonal trajectories for these curves.

Solution.
$$1)$$ Determine the differential equation for the given family of hyperbolas. Differentiating the equation with respect to $$x$$ gives: $y' = - \frac{C}{{{x^2}}}.$ Now we eliminate the parameter $$C$$ from the system of two equations: $\left\{ \begin{array}{l} y = \frac{C}{x}\\ y' = - \frac{C}{{{x^2}}} \end{array} \right..$ It follows from the first equation that $$C = xy.$$ Substituting into the second equation yields: $y' = - \frac{{xy}}{{{x^2}}} = - \frac{y}{x}.$ $$2)$$ Replace $$y'$$ with $$\left( { - \large\frac{1}{{y'}}\normalsize} \right):$$ $- \frac{1}{{y'}} = - \frac{y}{x},\;\; \Rightarrow y' = \frac{x}{y}.$ $$3)$$ Now we integrate the differential equation of the orthogonal trajectories: ${y' = \frac{x}{y},}\;\; {\Rightarrow \frac{{dy}}{{dx}} = \frac{x}{y},}\;\; {\Rightarrow ydy = xdx,}\;\; {\Rightarrow \int {ydy} = \int {xdx} ,}\;\; {\Rightarrow \frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + C,}\;\; {\Rightarrow {x^2} - {y^2} = C.}$ In the last equation we replaced $$2C$$ with just a constant $$C.$$ Thus, we have obtained the equation of the family of orthogonal trajectories. As it can be seen, these orthogonal trajectories are also hyperbolas. Both the families of hyperbolas are shown schematically in Figure $$3.$$
 Fig.3 Fig.4
Example 3
Find the orthogonal trajectories of the family of curves given by the power function $$y = C{x^4}.$$

Solution.
$$1)$$ Determine the differential equation for the given family of power curves: $y = C{x^4},\;\; \Rightarrow y' = 4C{x^3}.$ By solving the system of two equations and eliminating $$C,$$ we get: ${C = \frac{y}{{{x^4}}},}\;\; {\Rightarrow y' = 4 \cdot \frac{y}{{{x^4}}} \cdot {x^3} = \frac{{4y}}{x}.}$ $$2)$$ Replacing $$y'$$ with $$\left( { - \large\frac{1}{{y'}}\normalsize} \right):$$ gives: $- \frac{1}{{y'}} = \frac{{4y}}{x},\;\; \Rightarrow y' = - \frac{x}{{4y}}.$ The last expression is the differential equation of the orthogonal trajectories.

$$3)$$ By integrating we can find the algebraic equation of the orthogonal trajectories: ${y' = - \frac{x}{{4y}},}\;\; {\Rightarrow \frac{{dy}}{{dx}} = - \frac{x}{{4y}},}\;\; {\Rightarrow 4ydy = - xdx,}\;\; {\Rightarrow 4\int {ydy} = - \int {xdx} ,}\;\; {\Rightarrow 4 \cdot \frac{{{y^2}}}{2} = - \frac{{{x^2}}}{2} + C,}\;\; {\Rightarrow 4{y^2} + {x^2} = 2C.}$ Divide both sides by $$2C:$$ ${\frac{{4{y^2}}}{{2C}} + \frac{{{x^2}}}{{2C}} = \frac{{2C}}{{2C}},}\;\; {\Rightarrow \frac{{{y^2}}}{{\frac{C}{2}}} + \frac{{{x^2}}}{{2C}} = 1,}\;\; {\Rightarrow \frac{{{y^2}}}{{{{\left( {\sqrt {\frac{C}{2}} } \right)}^2}}} + \frac{{{x^2}}}{{{{\left( {\sqrt {2C} } \right)}^2}}} = 1.}$ We obtain the equation of the family of ellipses, which are the orthogonal trajectories for the given family of power curves $$y = C{x^4}.$$ The ratio of the lengths of semiaxes for these ellipses is $\frac{{\sqrt {2C} }}{{\sqrt {\frac{C}{2}} }} = \frac{{\sqrt 2 }}{{\sqrt {\frac{1}{2}} }} = {\left( {\sqrt 2 } \right)^2} = 2.$ Schematically the graphs of the families of curves are shown in Figure $$4$$ above.

Example 4
Determine the orthogonal trajectories of the family of sinusoids $$y = C\sin x.$$

Solution.
$$1)$$ Differentiating the given equation with respect to $$x$$ gives: $y' = C\cos x.$ By substituting $$C = {\large\frac{y}{{\sin x}}\normalsize}$$ we find the differential equation of the given sinusoidal curves: $y' = \frac{y}{{\sin x}}\cos x = y\cot x.$ $$2)$$ Replace $$y'$$ with $$\left( { - \large\frac{1}{{y'}}\normalsize} \right)$$ to write the diferential equation of the orthogonal curves: ${- \frac{1}{{y'}} = y\cot x,}\;\; {\Rightarrow y' = - \frac{1}{{y\cot x}} = - \frac{{\tan x}}{y}.}$ $$3)$$ Now we can integrate this differential equation: ${y' = - \frac{{\tan x}}{y},}\;\; {\Rightarrow \frac{{dy}}{{dx}} = - \frac{{\tan x}}{y},}\;\; {\Rightarrow ydy = - \tan xdx,}\;\; {\Rightarrow \int {ydy} = - \int {\tan xdx} ,}\;\; {\Rightarrow \frac{{{y^2}}}{2} = \ln \left| {\cos x} \right| + \ln C,}\;\; {\Rightarrow \frac{{{y^2}}}{2} = \ln \left( {C\left| {\cos x} \right|} \right).}$ It follows from here that ${C\left| {\cos x} \right| = \exp \left( {\frac{{{y^2}}}{2}} \right),}\;\; {\Rightarrow \cos x = \pm \frac{1}{C}\exp \left( {\frac{{{y^2}}}{2}} \right).}$ By denoting $${C_1} = \pm {\large\frac{1}{C}\normalsize},$$ we obtain the final implicit equation of the orthogonal trajectories: $\cos x = {C_1}\exp \left( {\frac{{{y^2}}}{2}} \right).$