Orthogonal Trajectory

In summary: Yes, that is correct for the orthogonal trajectories to the family of circles with center on the y-axis tangent to the x-axis.
  • #1
piano.lisa
34
0
I have to find the orthogonal trajectory to the family of circles tangents to the x-axis.

I eventually found that the derivative of the orthogonal trajectory would be:
[tex]\frac{dy}{dx}= \frac{-x^2 + y^2}{2xy} [/tex]

But how do i find the equation for "y" explicitly? (ps. This is not an assignment problem. I'm simply solving problems to get ready for my test)
ps#2: I recently discovered that this may be a solvable differential equation. Perhaps Bernoulli?
 
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  • #2
[tex] \frac{dy}{dx} + \frac{x^{2}+y^{2}}{2xy} = 0 [/tex]So [tex] \frac{dy}{dx} + \frac{x^{2}}{2xy} + \frac{y^{2}}{2xy} = 0 [/tex]

[tex] \frac{dy}{dx} + \frac{x}{2y} = -\frac{y}{2x} [/tex] which is the form: [tex] \frac{dy}{dx} + P(x)y = Q(x)y [/tex] (i.e. the Bernouli equation is linear)
 
Last edited:
  • #3
piano.lisa said:
I have to find the orthogonal trajectory to the family of circles tangents to the x-axis.
Do you mean with center on the y-axis also? You need another condition to have a one parameter family.

I eventually found that the derivative of the orthogonal trajectory would be:
[tex]\frac{dy}{dx}= \frac{-x^2 + y^2}{2xy} [/tex]
Yes, that is correct for the orthogonal trajectories to the family of circles with center on the y-axis tangent to the x-axis.

But how do i find the equation for "y" explicitly?


(ps. This is not an assignment problem. I'm simply solving problems to get ready for my test)
ps#2: I recently discovered that this may be a solvable differential equation. Perhaps Bernoulli?

This is a "homogeneous" equation (note that "homogeneous" for first order equations is not the same as "homogeneous" for linear equations!): If you divide both numerator and denominator by x2 you get
[tex]\frac{dy}{dx}= \frac{\frac{y^2}{x^2}- 1}{2\frac{y}{x}}[/itex]
Now let v= y/x so that y= xv and y'= xv'+ v. Your equation becomes
[tex]x\frac{dv}{dx}+ v= \frac{v^2-1}{v^2}= 1- \frac{1}{v^2}[/tex] That's a separable equation.

(Courtigrad's suggestion that this is linear is completely wrong: 1/y is not linear.)
 
  • #4
was I correct on what I did before that then? It is Bernoulli right? And its courtrigrad, not courtigrad.
 
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  • #5
made a mistake. should be:

[tex] \frac{dy}{dx} - \frac{x^{2}-y^{2}}{2xy} = 0 [/tex]
 

1. What is an orthogonal trajectory?

An orthogonal trajectory is a curve that intersects a given family of curves at right angles. In other words, it is a curve that is perpendicular to all of the curves in the given family.

2. How is the concept of orthogonal trajectories used in science?

Orthogonal trajectories are commonly used in physics and engineering to solve problems involving electric or magnetic fields. They are also used in mathematics to solve differential equations.

3. What is the relationship between orthogonal trajectories and curves with gradients?

Curves with gradients that are perpendicular to each other are orthogonal trajectories. This means that if two curves have gradients that are negative reciprocals of each other, they will be orthogonal trajectories.

4. How do you find the orthogonal trajectory of a given curve?

To find the orthogonal trajectory of a given curve, you can use the differential equation method. This involves finding the differential equation of the given curve and then solving for the orthogonal trajectory using the conditions that the gradients of the two curves are perpendicular.

5. Are orthogonal trajectories unique?

No, orthogonal trajectories are not unique. For a given family of curves, there can be multiple orthogonal trajectories that intersect the curves at right angles. The number of orthogonal trajectories depends on the complexity of the given family of curves.

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