How to Find Orthogonal Trajectories with Formula and Solved Examples
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 of a Family of Curves
1. What is an orthogonal trajectory in mathematics?
An orthogonal trajectory is a curve that intersects every curve of a given family at a right angle (90°). In differential equations, if one family of curves has slope m, then its orthogonal trajectories have slope −1/m at the points of intersection. This concept is commonly used in coordinate geometry and calculus to study families of curves that cut each other perpendicularly.
2. How do you find the orthogonal trajectory of a given family of curves?
To find the orthogonal trajectory, replace the slope in the differential equation by its negative reciprocal and solve the new equation.
- Step 1: Differentiate the given family to obtain its differential equation.
- Step 2: Replace dy/dx with −dx/dy or use m → −1/m.
- Step 3: Solve the resulting differential equation.
3. What is the formula used in orthogonal trajectories?
The key formula in orthogonal trajectories is that perpendicular slopes satisfy m₁ × m₂ = −1. If the original family has slope dy/dx = f(x, y), then its orthogonal trajectories satisfy dy/dx = −1/f(x, y). This negative reciprocal relationship ensures the curves meet at 90°.
4. Can you give an example of orthogonal trajectories?
Yes, circles centered at the origin and straight lines through the origin are classic orthogonal trajectories.
- Family 1: x² + y² = c² (circles).
- Differentiating gives: 2x + 2y(dy/dx) = 0.
- Slope: dy/dx = −x/y.
- Orthogonal slope: y/x.
- Solving dy/dx = y/x gives y = kx.
5. Why do we use the negative reciprocal of the slope in orthogonal trajectories?
We use the negative reciprocal because two curves are perpendicular when the product of their slopes equals −1. If the original slope is m, then the perpendicular slope must be −1/m. This ensures the angle between the curves is exactly 90°, satisfying the orthogonality condition.
6. What is the difference between a family of curves and its orthogonal trajectories?
A family of curves is a set of curves defined by a parameter, while its orthogonal trajectories are another family that intersects the first at right angles.
- The original family satisfies one differential equation.
- The orthogonal family satisfies the modified equation with slope −1/m.
- Both families are mutually perpendicular at intersection points.
7. How do you find orthogonal trajectories in Cartesian coordinates?
In Cartesian form, orthogonal trajectories are found by modifying the differential equation of the given family.
- Differentiate the equation to eliminate the parameter.
- Express it in the form dy/dx = f(x, y).
- Replace dy/dx with −1/f(x, y).
- Solve the new differential equation.
8. How are orthogonal trajectories related to differential equations?
Orthogonal trajectories are obtained by solving a related differential equation formed using perpendicular slopes. If the original family satisfies F(x, y, c) = 0, differentiating removes the parameter and gives a first-order differential equation. Replacing its slope with the negative reciprocal produces another differential equation whose solution gives the orthogonal family.
9. What are the applications of orthogonal trajectories?
Orthogonal trajectories are used in physics, engineering, and coordinate geometry to model perpendicular fields and curves.
- Electric field lines and equipotential curves.
- Heat flow and level curves.
- Fluid flow and streamlines.
10. What are common mistakes when solving orthogonal trajectory problems?
The most common mistake is forgetting to use the negative reciprocal of the slope correctly.
- Not eliminating the parameter before forming the differential equation.
- Incorrectly replacing dy/dx without simplifying first.
- Forgetting that perpendicular slopes satisfy m₁m₂ = −1.
