Orthogonal Trajectories of Circles: Solving with Differentiation Method

In summary, the author found that x2+y2=Cx and x2+y2=y2-x2 are both circles that are orthogonal at (0,0). If you eliminate the 'a' from the equations, you get xyC=y2-x2 which is correct.
  • #1
zorro
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Homework Statement



Find the orthogonal trajectories of the circles x2 + y2 - ay = 0

The Attempt at a Solution



I differentiated the equation w.r.t. x.,
Replaced dy/dx with -dx/dy,
Solved the equation and got xyC = y2 - x2, where C is a constant.

I did not eliminate 'a' after differentiating for the first time. I did that after solving the differential equation with -dx/dy. Is this method correct?
The answer given is x2 + y2 = Cx
 
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  • #2
Right, you've found that:
[tex]
\frac{dy}{dx}=\frac{2x}{a-2y}
[/tex]
This represents the gradient at a given point, for the normal lines, you swap and make negative, so you have to solve:
[tex]
\frac{dy}{dx}=-\frac{a-2y}{2x}
[/tex]
Solve this equation to obtain you answer.
 
  • #3
But the answer obtained by that method doesnot match the given answer.
xyC = y2 - x2 is different from x2 + y2 = Cx (given)
 
  • #4
Are you sure the answer in the book is correct? Take a few examples and plot them and see if you get orthoganal lines at that point.
 
  • #5
I got the answer given in the book, but by a different method.
Eliminate 'a' from the equation of dy/dx in your post by substitution, and then integrate for orthogonal curve.

If you analyse carefully, xyC = y2 - x2 is a family of pair of straight lines
and x2 + y2 = Cx is a family of circles.

May be we have to find out an orthogonal curve not pair of lines. But I still don't get why the two answers are different.
 
  • #6
hunt_mat said:
Are you sure the answer in the book is correct? Take a few examples and plot them and see if you get orthoganal lines at that point.

The books answer is certainly correct. x^2+y^2=ay is a circle passing through (0,0) with it's origin on the the y-axis. x^2+y^2=cx is a circle passing through (0,0) with origin on the x-axis. They are certainly orthogonal at (0,0). Since they are circles, they are also orthogonal at the other intersection point. And yes, you do need to eliminate the 'a' before you integrate. a isn't fixed. It's parameter that describes a member of the family. If you do that, with some patience and substitution you can get the book answer.
 
  • #7
Dick said:
And yes, you do need to eliminate the 'a' before you integrate. a isn't fixed. It's parameter that describes a member of the family.

How does this affect the result - 'a isn't fixed' ? Its a constant.
 
  • #8
Abdul Quadeer said:
How does this affect the result - 'a isn't fixed' ? Its a constant.

Each value of 'a' describes a different circle. A curve in the orthogonal family may hit many curves with different values of a. Follow the books method of eliminating a and you should get an ODE to solve to get the books obviously correct answer. xyC=y^2-x^2 isn't correct.
 
  • #9
Thanks!
 

1. What are orthogonal trajectories?

Orthogonal trajectories are a set of curves that intersect each other at right angles. They are also known as perpendicular trajectories or cross-cutting curves.

2. How are orthogonal trajectories related to differential equations?

Orthogonal trajectories are closely related to differential equations. They are the solution curves to a family of differential equations, where the slope of the curve at any point is equal to the negative reciprocal of the slope of the corresponding curve in the family.

3. What is the significance of orthogonal trajectories in mathematics?

Orthogonal trajectories have various applications in mathematics, including in the study of electromagnetic fields, fluid dynamics, and optics. They are also used in curve fitting and optimization problems.

4. How do you find orthogonal trajectories?

To find orthogonal trajectories, you need to first determine the differential equation for the given family of curves. Then, you can use various methods such as substitution, implicit differentiation, or the method of undetermined coefficients to solve for the equation of the orthogonal trajectories.

5. Can orthogonal trajectories intersect more than once?

Yes, orthogonal trajectories can intersect multiple times. However, at each point of intersection, the two curves will be perpendicular to each other. This property is important in applications such as finding the shortest distance between two curves.

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