Orthogonal trajectory

In mathematics, orthogonal trajectories are a family of curves in the plane that intersect a given family of curves at right angles. The problem is classical, but is now understood by means of complex analysis; see for example harmonic conjugate.

For a family of level curves described by $g(x, y) = C$, where $C$ is a constant, the orthogonal trajectories may be found as the level curves of a new function $f(x, y)$ by solving the partial differential equation

$\nabla f \cdot \nabla g = 0$

for $f(x, y)$. This is literally a statement that the gradients of the functions (which are perpendicular to the curves) are orthogonal.

The partial differential equation may be avoided by instead equating the tangent of a parametric curve $\vec r(t)$ with the gradient of $g(x, y)$:

$\frac{d}{d t}\left(\vec r(t)\right) = \nabla g$

which will result in two possibly coupled ordinary differential equations, whose solutions are the orthogonal trajectories. Note that with this formula, if $g$ is a function of three variables its level sets are surfaces, and the family of curves $\vec r(t)$ are orthogonal to the surfaces.

Example 1: Circle

Suppose we are given the family of circles centered about the origin:

$x^2 + y^2 = c$

The orthogonal trajectories of this family are the family of curves such that intersect the circle at right angles. Given a line, the negative reciprocal of its slope is the slope of a perpendicular line. In calculus terms, if y and k are two perpendicular lines:

$\left\{\begin{matrix} \frac{\mathrm{d}y}{\mathrm{d}x} = f(x,y) \\ \frac{\mathrm{d}k}{\mathrm{d}x} = -\frac{1}{f(x,y)} \end{matrix}\right.$

Orthogonal trajectories of the given family of circles are no different. The tangent line of an orthogonal trajectory is perpendicular to the tangent line of the circle, and vice versa.

By implicit differentiation,

$x^2 + y^2 = c \Rightarrow 2x + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0$

Solving the equation, one acquires the following:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{y}$

Therefore, an orthogonal trajectory must satisfy the differential equation:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x}$

Solving the equation for y, we find the orthogonal trajectories:

$y = mx$

The orthogonal trajectories of a family of circles centered at the origin are linear equations containing the origin.

In polar coordinates, the family of circles centered about the origin is the level curves of

$r - R = 0$

where $R$ is the radius of the circle. Then the orthogonal trajectories are the level curves of $f$ defined by:

$\nabla f \cdot \nabla (r - R) = 0$

$\left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}\right) \cdot \left(\frac{\partial}{\partial r}(r - R), \frac{1}{r} \frac{\partial}{\partial \theta}(r - R)\right) = 0$

$\frac{\partial f}{\partial r} = 0 \quad \Rightarrow \quad f = f(\theta)$

The lack of complete boundary data prevents determining $f(\theta)$. However, we want our orthogonal trajectories to span every point on every circle, which means that $f(\theta)$ must have a range which at least include one period of rotation. Thus, the level curves of $f(\theta) = 0$, with freedom to choose any $f$, are all of the $\theta = constant$ curves that intersect circles, which are (all of the) straight lines passing through the origin. Note that the dot product takes nearly the familiar form since polar coordinates are orthogonal.

The absence of boundary data is a good thing, as it makes solving the PDE simple as one doesn't need to contort the solution to any boundary. In general, though, it must be ensured that all of the trajectories are found.

See also

• Isogonal trajectory, a generalization of orthogonal trajectories to angles other than right angles
• Apollonian circles, pairs of families of circles that are all orthogonal to each other

External links

• Exploring orthogonal trajectories - applet allowing user to draw families of curves and their orthogonal trajectories.