How to Find Orthogonal Trajectories: Step-by-Step Guide
Orthogonal trajectory is a set of curves that converge at right angles with another set of curves. Electrostatics and hydrodynamics are two examples of mutually orthogonal curve families. In electrostatics, the lines of force and lines of constant potential are orthogonal, and in hydrodynamics, the streamlines and lines of constant velocity are orthogonal.In this article we will learn about orthogonal trajectories and orthogonal family of curves. We will understand orthogonal trajectories differential equations.
Define Orthogonal Trajectories
An orthogonal trajectory of the family is a curve that cuts any member of a given family of curves at right angles. It is not appropriate for the curve to pass through each family member. When they intersect, however, the angle formed by their tangents at each point of intersection is 90o. Orthogonal trajectories differential equations are used to find curves that intersect a given family of curves at right angles.
Orthogonal Trajectories Calculator
Orthogonal trajectories calculator is shown below
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Find Orthogonal Trajectories
With the help of an example, we will be able to comprehend this.
Determine the orthogonal trajectories of the straight line family y = Cx, where C is a parameter.
Sol: First, we create the differential equation for the y = Cx family of straight lines. We get the following result by differentiating the last equation with respect to x.
\[y'=C=const\]
We have to eliminate the constant C from the equation system:
\[y=Cx\]
\[y'=C\]
\[y'=\frac{y}{x}\]
The differential equation for the initial set of straight lines is obtained.
Replace \[y'\]with \[\frac{-1}{y}\]. This yields the orthogonal trajectories' differential equation:
\[-\frac{1}{y'}=\frac{y}{x}\]
\[\Rightarrow y'=-\frac{x}{y}\]
To find the algebraic equation of the family of orthogonal trajectories, we solve the last differential equation:
\[y'=-\frac{x}{y}\]
\[\Rightarrow\frac{dy}{dx}=-\frac{x}{y}\]
\[\Rightarrow ydy=-xdx\]
\[\Rightarrow \int ydy=-\int xdx\]
\[\Rightarrow\frac{y^{2}}{2}=-\frac{x^{2}}{2}+C\]
\[\Rightarrow\frac{x^{2}}{2}+\frac{y^{2}}{2}=C\]
\[x^{2}+y^{2}=2C\]
We can see that the orthogonal trajectories for the family of straight lines are concentric circles by replacing 2C with R.
\[x^{2}+y^{2}=R^{2}\]
Orthogonal Trajectories of Family of Circles
The curves that are perpendicular to the family everywhere are called orthogonal trajectories. In other words, the orthogonal trajectories are another family of curves in which each curve is perpendicular to the curves in the original family. We'll see how to use calculus to find orthogonal trajectories in the example below, but we'll give away the answer for now so we can draw the family of orthogonal trajectories and see that they're perpendicular to the original family.
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The curves in blue are from the original family y = kx The curves in green are
\[x^{2}+y^{2}=1\]
\[x^{2}+y^{2}=2\]
\[x^{2}+y^{2}=3\]
\[x^{2}+y^{2}=4\]
which are four of their orthogonal trajectories, part of their whole family of orthogonal trajectories given by \[x^{2}+y^{2}=C\]. Where a green and blue curve converge, note how each green circle is perpendicular to every blue line.
Orthogonal Trajectories of Parabola
The orthogonal direction of the \[y^{2} = 4ax\] parabola family is as follows:
Given the equation of the family of parabolas is \[y^{2} = 4ax\]
The parameter in the ordinary differential equation is a, which is also an arbitrary constant.
Now, on both sides, differentiating the equation with respect to x yields,
\[\frac{dy}{dx} = \frac{2a}{y}\]
\[a = \frac{y}{2} (\frac{dy}{dx})\]
substituting in the equation of the curve family will give,
\[y^{2} = 2xy(\frac{dy}{dx})\] which is a differential equation of the family of parabolas.
Now, to find the equation of the orthogonal trajectories we need to replace \[\frac{dy}{dx}\] by \[-\frac{dy}{dx}\]
and we need to solve it back
\[y^{2} = 2xy(-\frac{dy}{dx})\]
Regrouping the terms and integrating gives,
\[\int ydy=\int(-2x)dx\]
\[\Rightarrow\frac{y^{2}}{2}=-x^{2}+c\]
Where c is integration constant
regrouping the terms gives,
\[2x^{2}+y^{2}=C^{2}\]
Where C is a constant
Applications of Orthogonal Trajectories
In mathematics, orthogonal trajectories are used as curved coordinate systems (i.e. elliptical coordinates), and they also appear in physics as electric fields and their equipotential curves.
An isogonal trajectory is one in which the trajectory intersects the given curves at an arbitrary (but fixed) angle.
FAQs on Orthogonal Trajectory Explained: Key Concepts & Uses
1. What is an orthogonal trajectory in mathematics?
An orthogonal trajectory is a curve that intersects every member of a given family of curves at a right angle (90 degrees). In essence, if you have one set of curves, their orthogonal trajectories form a second set of curves where each one is perpendicular to the original curves at every point of intersection.
2. What is the mathematical condition for two curves to be orthogonal?
Two curves are considered orthogonal if their tangents at the point of intersection are perpendicular to each other. If the slopes of the tangents to the two curves at their intersection point are m₁ and m₂, the condition for orthogonality is that their product must be -1, i.e., m₁ * m₂ = -1.
3. How are differential equations used to find orthogonal trajectories?
The process involves three main steps:
- First, determine the differential equation for the given family of curves. This equation will typically be in the form dy/dx = f(x, y).
- Next, to find the slope of the orthogonal curve, replace dy/dx in this equation with its negative reciprocal, which is -dx/dy.
- Finally, solve this new differential equation. The resulting solution represents the equation for the family of orthogonal trajectories.
4. What are some real-world applications or uses of orthogonal trajectories?
Orthogonal trajectories are important for modelling various physical phenomena. Key examples include:
- Electromagnetism: Electric field lines and equipotential lines are orthogonal to each other.
- Thermodynamics: Lines of constant temperature (isotherms) are orthogonal to the lines of heat flow.
- Fluid Dynamics: In two-dimensional ideal fluid flow, streamlines and equipotential lines form an orthogonal network.
- Cartography: Lines of latitude and longitude on a map are orthogonal.
5. What is the orthogonal trajectory for a family of concentric circles centred at the origin?
For a family of concentric circles described by the equation x² + y² = r², the orthogonal trajectories are a family of straight lines that pass through the origin. These lines are represented by the equation y = kx, where k is a constant. Each radial line intersects every circle at a perfect right angle.
6. Why is the concept of orthogonality so significant in physics and engineering?
Orthogonality is significant because it often represents a fundamental relationship between a force field and its corresponding potential. For example, moving along an equipotential line (a path of constant potential energy) requires no work to be done by the associated force field (like gravity or an electric field). This is because the direction of motion is always perpendicular, or orthogonal, to the direction of the force. This principle simplifies the analysis of complex systems.
7. How does an orthogonal trajectory differ from an isogonal trajectory?
The primary difference is the angle of intersection. An orthogonal trajectory is a specific type of trajectory that intersects a family of curves at a constant angle of exactly 90 degrees. An isogonal trajectory is a more general term for a curve that intersects a family of curves at any fixed angle, which may or may not be 90 degrees. Therefore, all orthogonal trajectories are isogonal, but not all isogonal trajectories are orthogonal.
8. How does the slope of a tangent relate to finding an orthogonal trajectory?
The entire method for finding orthogonal trajectories is built on the geometric relationship between the slopes of perpendicular lines. The derivative, dy/dx, gives the slope of the tangent to a curve at any point. To find a path that is always perpendicular, we need a slope that is the negative reciprocal of the original slope, i.e., -1 / (dy/dx). By creating and solving a new differential equation using this perpendicular slope, we are effectively tracing the path of the orthogonal trajectory.
