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Orthogonal Trajectories

by zorro
Tags: orthogonal, trajectories
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zorro
#1
Jan3-11, 10:11 AM
P: 1,394
1. The problem statement, all variables and given/known data

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


3. 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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hunt_mat
#2
Jan3-11, 01:26 PM
HW Helper
P: 1,583
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.
zorro
#3
Jan3-11, 02:40 PM
P: 1,394
But the answer obtained by that method doesnot match the given answer.
xyC = y2 - x2 is different from x2 + y2 = Cx (given)

hunt_mat
#4
Jan3-11, 03:03 PM
HW Helper
P: 1,583
Orthogonal Trajectories

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.
zorro
#5
Jan3-11, 04:21 PM
P: 1,394
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.
Dick
#6
Jan3-11, 09:12 PM
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Quote Quote by hunt_mat View Post
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.
zorro
#7
Jan4-11, 08:39 AM
P: 1,394
Quote Quote by Dick View Post
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.
Dick
#8
Jan4-11, 09:11 AM
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Quote Quote by Abdul Quadeer View Post
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.
zorro
#9
Jan4-11, 10:15 AM
P: 1,394
Thanks!


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