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Given the family of curves: $F(x,y,x_0)=0$ where $F(x,y,x_0) = (x-x_0)^2 + y^2 - R^2\ ,\ x_0 \in \mathbb{R}$ find the orthogonal family.

This is my attempt:

I first get the differential equation satisfied by the curves $F(x,y,x_0) = 0$ by supposing $x = x(y)$ then:

$\frac{\mathrm{d}}{\mathrm{d}x} F(x,y,x_0) = \frac{\partial}{\partial x}F(x,y,x_0) + \frac{\partial}{\partial y} F(x,y,x_0) \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{\frac{\partial}{\partial x} F(x,y,x_0)}{\frac{\partial}{\partial y} F(x,y,x_0)}$

Then

$y' = \frac{- 2 (x - x_0)}{2 y} = \frac{x_0 -x}{y} $

Using

$F(x,y,x_0) = (x-x_0)^2 + y^2 - R^2 = 0 \implies x_0^2 - 2 x x_0 + x^2 + y^2 - R^2 = 0 \implies x_0 = \frac{2x \pm \sqrt{4 x^2 - 4 x^2 - 4 y^2 + 4 R^2}}{2} = x \pm \sqrt{R^2 - y^2}$

We write $y'$ as: $y' = \frac{-y}{x_0 -x} = \frac{-y}{x \pm \sqrt{R^2 - y^2} - x} = \frac{-y}{\pm \sqrt{R^2 - y^2}} \implies \frac{\mp \sqrt{R^2 - y^2}}{y} \mathrm{d}y = \mathrm{d}x$

Solving the integral we get:

$R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{y} + \sqrt{R^2 - y^2} = \pm x + C_0$

But when I graph the solution, I can see I am missing some solutions, the complete solution set is

$R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{y} + \sqrt{R^2 - y^2} = \pm x + C_0$ and $R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{(-y)} + \sqrt{R^2 - y^2} = \pm x + C_0$

When did I loose those solutions?

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Remember that $\displaystyle \int\frac{dy}y = \log|y|$. It seems to be that easy.

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  • $\begingroup$ Such an easy bit... Thank you! That was indeed the mistake $\endgroup$ – José D. Dec 22 '13 at 18:11
  • $\begingroup$ @Trollkemada: You could note me the point of my mistake just by a kind comment not with an unfriendly action. At least, you could consider the time I took on your problem and respect it, However it was wrong and so was deleted. :-) $\endgroup$ – mrs Dec 22 '13 at 18:19
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    $\begingroup$ @B.S.It wasn't my intention to sound rude, so I am sorry if my comment was offensive to you (keep in mind english is not my mother tongue). I did downvotted your answer because i though i am supposed to downvote bad answers in order to keep the answers sorted. Sorry if you consider that an 'unfriendly action'. And I do appreciate your time and effort on my problem, thanks. $\endgroup$ – José D. Dec 22 '13 at 18:24
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    $\begingroup$ @Trollkemada: Thanks for you kind words. +1 and :-) $\endgroup$ – mrs Dec 22 '13 at 18:28

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