Orthogonal Trajectories

Orthogonal trajectories are a family of curves in the plane such that each curve in one family is orthogonal (perpendicular) to each curve in the other family at each point of intersection. This concept is often used in physics, engineering, and mathematics to describe phenomena where two related families of curves or surfaces intersect at right angles.

Understanding Orthogonal Trajectories

To find the orthogonal trajectories of a given family of curves, we typically follow these steps:

1. Find the differential equation of the given family of curves.
2. Modify the differential equation to represent the orthogonal trajectories.
3. Solve the modified differential equation to find the family of orthogonal trajectories.

Step 1: Differential Equation of the Given Family

Given a family of curves represented by an equation ( F(x, y, C) = 0 ), where ( C ) is a constant, we differentiate this equation with respect to ( x ) to find the differential equation that represents the family. This is typically done using implicit differentiation if the equation cannot be easily solved for ( y ).

Step 2: Modify the Differential Equation

Once we have the differential equation of the form ( \frac{dy}{dx} = f(x, y) ), we find the differential equation of the orthogonal trajectories by taking the negative reciprocal of ( f(x, y) ). This is because if two curves are orthogonal, the product of their slopes at the point of intersection must be ( -1 ).

So, the differential equation for the orthogonal trajectories is ( \frac{dy}{dx} = -\frac{1}{f(x, y)} ).

Step 3: Solve the Modified Differential Equation

Finally, we solve the modified differential equation to find the family of orthogonal trajectories.

Table of Differences and Important Points

Aspect	Given Family of Curves	Orthogonal Trajectories
Slope	( \frac{dy}{dx} = f(x, y) )	( \frac{dy}{dx} = -\frac{1}{f(x, y)} )
Relationship	Curves follow a specific pattern or rule	Curves intersect the given family at right angles
Example	( y = Cx ) (family of straight lines through the origin)	( y = -\frac{1}{C}x ) (family of straight lines orthogonal to the given family)
Application	Describes one set of related phenomena	Describes a related set of phenomena that interact with the first set at right angles

Examples

Example 1: Family of Circles

Consider the family of circles centered at the origin with radius ( r ): ( x^2 + y^2 = r^2 ).

1. Differentiate implicitly with respect to ( x ) to find the differential equation of the family: [ 2x + 2y\frac{dy}{dx} = 0 ] Simplifying, we get: [ \frac{dy}{dx} = -\frac{x}{y} ]

2. The differential equation for the orthogonal trajectories is the negative reciprocal of the slope: [ \frac{dy}{dx} = \frac{y}{x} ]

3. Solve this differential equation to find the orthogonal trajectories: [ \frac{dy}{y} = \frac{dx}{x} ] Integrating both sides, we get: [ \ln|y| = \ln|x| + \ln|C| ] Simplifying, we find the orthogonal trajectories are given by: [ y = Cx ] which represents a family of straight lines through the origin.

Example 2: Exponential Family

Consider the family of exponential curves given by ( y = Ce^{x} ).

1. Differentiate with respect to ( x ) to find the differential equation of the family: [ \frac{dy}{dx} = Ce^{x} ] Since ( y = Ce^{x} ), we can write: [ \frac{dy}{dx} = y ]

2. The differential equation for the orthogonal trajectories is the negative reciprocal of the slope: [ \frac{dy}{dx} = -\frac{1}{y} ]

3. Solve this differential equation to find the orthogonal trajectories: [ ydy = -dx ] Integrating both sides, we get: [ \frac{y^2}{2} = -x + C ] The orthogonal trajectories are given by: [ x + \frac{y^2}{2} = C ] which represents a family of parabolas opening to the left.

Orthogonal trajectories are a powerful tool in understanding the relationships between different families of curves and their interactions. They are particularly useful in fields such as electromagnetism, where electric and magnetic field lines are orthogonal, and in fluid dynamics, where streamlines and equipotential lines are orthogonal.