# Thread: help with orthogonal trajectories

1. ## help with orthogonal trajectories

I tried solving for the orthogonal trajectories of the following three families:
1. $\displaystyle y = mx$
2. $\displaystyle xy = c^2$
3. $\displaystyle x^\frac{3}{2} + y^\frac{3}{2} = a^\frac{3}{2}$

here are my solutions: and my answers for the respective problems were:
1. $\displaystyle y^2 + x^2 = 2c$
2. $\displaystyle y^2 - x^2 = 2c$
3. $\displaystyle \sqrt{y} - \sqrt{x} = \frac{c}{2}$

But the answers from the book are, respectively:
1. $\displaystyle y^2 + x^2 = a^2$
2. $\displaystyle x^2 - y^2 = a^2$
3. $\displaystyle \sqrt{y} - \sqrt{x} = \sqrt{a}$

What am I doing wrong or not understanding or how do I make my answers more consistent with the proper format as per in the book? Especially in the 2nd problem on my left hand side i got $\displaystyle y^2 - x^2$ as opposed to the answer's left hand side which is $\displaystyle x^2 - y^2$ please help me
He is changing the constant 'c' into 'a' on purpose? and if so then how do you know what you're suppose to change 'c' into?

2. ## Re: help with orthogonal trajectories

You did all 3 fine. One constant is as good as another.

3. ## Re: help with orthogonal trajectories

I have a new set of roadblocks with this topic of orthogonal trajectories. Here are three problems that I tried to solve: The first one I was able to solve however my answer differs from the answer in the book and I cannot find a way to how that they are both equal so either I or the book have the wrong answer, the second one I am stuck in a part where the equations suddenly seem to become quadratic and I do not know how to deal with this in differential equations, the third one I simply do not know how to solve the resulting differential equation after dy/dx has been replace with its negative reciprocal. Any help would be appreciated. #### Search Tags

orthogonal, trajectories 