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Orthogonal Trajectories 
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#1
Jan311, 10:11 AM

P: 1,394

1. The problem statement, all variables and given/known data
Find the orthogonal trajectories of the circles x^{2} + y^{2}  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 = y^{2}  x^{2}, 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 x^{2} + y^{2} = Cx 


#2
Jan311, 01:26 PM

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P: 1,583

Right, you've found that:
[tex] \frac{dy}{dx}=\frac{2x}{a2y} [/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{a2y}{2x} [/tex] Solve this equation to obtain you answer. 


#3
Jan311, 02:40 PM

P: 1,394

But the answer obtained by that method doesnot match the given answer.
xyC = y^{2}  x^{2} is different from x^{2} + y^{2} = Cx (given) 


#4
Jan311, 03:03 PM

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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.



#5
Jan311, 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 = y^{2}  x^{2} is a family of pair of straight lines and x^{2} + y^{2} = 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
Jan311, 09:12 PM

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Thanks
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#7
Jan411, 08:39 AM

P: 1,394




#8
Jan411, 09:11 AM

Sci Advisor
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Thanks
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#9
Jan411, 10:15 AM

P: 1,394

Thanks!



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