Translate $2xy=x^2-y^2$ to polar coordinates

Question

As per the title i need to translate $2xy=x^2-y^2$ into a polar coordinates function

Why it confuses me:

$$ x = r\cos\phi\\ y = r\sin\phi $$

Then:

$$ 2r^2\cos{\phi}\sin{\phi} = r^2\cos{\phi}-r^2\sin{\phi} = r^2(\cos^2{\phi} - \sin^2{\phi}) \\ r^2\sin{2\phi}=r^2\cos{2\phi} \\ sin{2\phi} = \cos{2\phi} $$

Which gives:

$$ \phi = \frac{\pi}{8} + \frac{\pi n}{2}, \;\;\;\; n \in \mathbb Z $$

But it's just a set of $\phi$ values. How do i plot such a function? In polar coordinates $r$ is a function of some $\phi$, isn't that?

Answer 1

It is $$ \phi= \frac{\pi}{8} + \frac{\pi n}{2}, \;\;\;\; n \in \mathbb Z $$

and not $x$.

So $$x= r\cos (\frac{\pi}{8} + \frac{\pi n}{2})=$$ $$ = r(\cos \frac{\pi}{8} \cos \frac{\pi n}{2}-\sin \frac{\pi}{8} \sin \frac{\pi n}{2})$$ Now try to put $n=0,1,2,3,4...$. We see that:

if we put $n=0,4,8...$ we get the same value. if we put $n=1,5,9...$ we get the same value. ...

So write $n= 4k+r$ where $r\in\{0,1,2,3\}$ and you will get 4 different families of solution.

Answer 2

After conversion to polar coordinates, $r$ disappears from the equations (but for $r=0$, which corresponds to the origin). This means that for a given $\phi$, $r$ is arbitrary and this describes an infinite straight line through the origin and with direction $\phi$.

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By the way, you can obtain the same result by plugging the equation of a straight line through the origin, $y=mx$, which gives

$$2mx^2=x^2-m^2x^2$$ or

$$2m=1-m^2$$ or

$$m=\pm\sqrt2-1.$$

Answer 3

The graph of $\phi =c$ where $c$ is a constant, is just a ray through the origin which makes an angle of $\phi$ with the positive direction of $x$-axis.

Answer 4

Accoring to the equation $$\sin 2\phi=\cos 2\phi,$$ we can obtain $$\phi=\frac{4n+1}{8}\pi,n \in \mathbb {Z}.$$ But in fact,

• $\phi=2k\pi+\dfrac{1}{8}\pi$,when $n=4k$;
• $\phi=2k\pi+\dfrac{5}{8}\pi$,when $n=4k+1$;
• $\phi=2k\pi+\dfrac{9}{8}\pi$,when $n=4k+2$;
• $\phi=2k\pi+\dfrac{13}{8}\pi$,when $n=4k+3$.

These are four rays，which are reverse pairwisely. In another word, they're two lines as follows

Another method

You can directly solve the equation $$y^2+2xy-x^2=0.$$ Therefore, $$y=\frac{-2x \pm \sqrt{(2x)^2-4 \cdot 1 \cdot (-x^2)}}{2}=(-1 \pm \sqrt 2)x.$$ This also represents the two lines.

Answer 5

What to expect:

$x^2-2xy-y^2= 0;$

$(x-y)^2-2y^2 =0;$

$(x-y-√2y)(x-y+√2y)= 0;$

1)$x = y(1+√2)$,or $y= (1+√2)^{-1}x$;

2) $y=(1-√2)^{-1}x;$

1)and 2) are 2 lines passing through the origin.

Slopes: $m_1= -1+√2$ ; $m_2=-1-√2$.

Perhaps not so surprising any more your result in polar coordinates.