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Nikachu's question at Yahoo! Answers regarding orthogonal trajectories

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  • #1


Staff member
Feb 24, 2012
Here is the question:

Find the family of functions (circles) ?

I'm working on my Math homework and I'm quite stuck in a problem. I was only given a graph and the problem is..

Consider two points -1 and 1. Find the family of functions whose elements always intersect these two points. Then find the family of functions that are orthogonal trajectories with the graph shown.

The graph given had a lot of circles of different sizes (of course, the elements are supposed to be infinite but showed a few instead to make things clear). Each circle did have the two points on each of their graphs. One circle had the points very close to each other. A much smaller circle had the points on it at a distance of its diameter, etc. All their centers lie on the y-axis.
Here is a link to the question:

Find the family of functions (circles) ? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

edit: I see the question has been deleted at Yahoo!. (Dull)
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Feb 24, 2012
Hello Nikachu,

To find the required family of circles, please refer to the following diagram:

We have the center of the circle at $(0,k)$ where the radius is $r$. Thus the equation of the circle is:

\(\displaystyle x^2+(y-k)^2=r^2\)

By the Pythagorean theorem, we see that:

\(\displaystyle k^2=r^2-1\)


\(\displaystyle k=\pm\sqrt{r^2-1}\)

and so the family of circles is:

\(\displaystyle x^2+\left(y\mp\sqrt{r^2-1} \right)^2=r^2\)

Now to find the orthogonal trajectories, we will express the family of circles in the form:

\(\displaystyle F(x,y)=k\)

\(\displaystyle x^2+y^2\mp2\sqrt{r^2-1}y+r^2-1=r^2\)

\(\displaystyle x^2+y^2-1=\pm2\sqrt{r^2-1}y\)

\(\displaystyle \frac{x^2+y^2-1}{y}=\pm2\sqrt{r^2-1}\)

It can be shown that the orthogonal trajectories must satisfy:

\(\displaystyle \frac{\delta F}{\delta y}dx-\frac{\delta F}{\delta x}dy=0\)

So, we compute:

\(\displaystyle \frac{\delta F}{\delta y}=\frac{y(2y)-(x^2+y^2-1)(1)}{y^2}=\frac{y^2-x^2+1}{y^2}\)

\(\displaystyle \frac{\delta F}{\delta x}=\frac{2x}{y}\)

Hence, the ODE we need to solve is:

\(\displaystyle \frac{y^2-x^2+1}{y^2}dx-\frac{2x}{y}dy=0\)

Multiplying through by $y^2$, we obtain the equation:

\(\displaystyle (y^2-x^2+1)\,dx-(2xy)\,dy=0\)

This equation is not separable, exact nor linear, so let's look at finding an integrating factor that will make it exact. We have:

\(\displaystyle M(x,y)=y^2-x^2+1\,\therefore\,\frac{\delta M}{\delta y}=2y\)

\(\displaystyle N(x,y)=-2xy\,\therefore\,\frac{\delta N}{\delta x}=-2y\)

and so:

\(\displaystyle \frac{\frac{\delta M}{\delta y}-\frac{\delta N}{\delta x}}{N(x,y)}=\frac{2y-(-2y)}{-2xy}=\frac{4y}{-2xy}=-\frac{2}{x}\)

Since this is a function of $x$ alone, our integrating factor is:

\(\displaystyle \mu(x)=e^{-2\int\frac{dx}{x}}=e^{\ln(x^{-2})}=x^{-2}\)

Multiplying the ODE by this integrating factor, we obtain:

\(\displaystyle \frac{y^2-x^2+1}{x^2}dx-\frac{2y}{x}dy=0\)

Now we have an exact equation. If the solution is $F(x,y)=C$, then we must have:

\(\displaystyle F(x,y)=\int\frac{y^2-x^2+1}{x^2}\,dx+g(y)\)

\(\displaystyle F(x,y)=-\left(\frac{y^2+1}{x}+x \right)+g(y)\)

Next, to determine $g(y)$, we will take the partial derivative with respect to $y$ and substitute \(\displaystyle \frac{\delta F}{\delta y}=-\frac{2y}{x}\) and solve for $g'(y)$.

\(\displaystyle -\frac{2y}{x}=-\frac{2y}{x}+g'(y)\)

\(\displaystyle g'(y)=0\)

\(\displaystyle g(y)=C\)

And so we have:

\(\displaystyle F(x,y)=-\left(\frac{y^2+1}{x}+x \right)+C\)

Hence, the orthogonal trajectories are given by:

\(\displaystyle \frac{y^2+1}{x}+x=C\)

Multiplying through by $x$ and completing the square, and using \(\displaystyle c=\frac{C}{2}\), we find the orthogonal trajectories is given by the family of circles:

\(\displaystyle (x-c)^2+y^2=c^2-1\) where \(\displaystyle 1<|c|\)

To Nikachu and any other guests viewing this topic, I invite and encourage you to post other differential equations problems here in our Differential Equations forum.

Best Regards,

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