I am trying to convert the equation $y=4/x$ into a polar equation.
I have done this work but I am not sure if it is right. I just subsituted $r\sin(\theta)$ for $y$ and $r\cos(\theta)$ for $x$ and solved for $r$ , which gave me $r = \sqrt{4\over (\sin(\theta)\cdot\cos(\theta))}$.
Is this correct? Or am I going about this all wrong? Thanks in advance.
$\endgroup$ 52 Answers
$\begingroup$Yes. You have $xy = 4$ in cartesian coordinates, so in polar coordinates that is indeed: $$r^2\sin(\theta)\cos(\theta) =4$$
You can leave it at that, or rearrange to suit. I recommend using $ 2\sin(\theta)\cos(\theta)=\sin(2\theta)$ .
$$r^2\,\sin(2\theta) = 8$$
$$r = +2\sqrt{2\csc(2\theta)}$$
$\endgroup$ $\begingroup$Yes, this is essentially correct; as usual, one must take care when taking square roots as one does here. Substituting as in the question gives $$r^2 \sin \theta \cos \theta = 4,$$ which is equivalent to $$r = \pm \sqrt{\frac{4}{\sin \theta \cos \theta}}.$$ In our case, however, we have (in polar coordinates) that \begin{align} \left(\sqrt{\frac{4}{\sin (\theta + \pi) \cos (\theta + \pi)}}, \theta + \pi \right)_p &= \left(\sqrt{\frac{4}{\sin \theta \cos \theta + \pi)}}, \theta + \pi \right)_p \\ &= -\left(\sqrt{\frac{4}{\sin \theta \cos \theta + \pi)}}, \theta\right)_p . \end{align} Taking the positive branch (the choice $+$ of $\pm$) defines a polar function $$r(\theta) = \sqrt{\frac{4}{\sin \theta \cos \theta}},$$ and the above computation shows that this function traces out all of the curve, i.e., that we don't miss any points by discarding the negative branch (the choice $-$ of $\pm$).
Note by the way that we can simplify our expression some using the double-angle identity $\sin 2 \theta = 2 \sin \theta \cos \theta$; rearranging gives $$r(\theta) = 2 \sqrt{2 \csc 2 \theta}.$$ This makes perhaps clearer, too, that $r(\theta)$ is only defined for $\theta$ corresponding to angles in the first and third quadrants.
$\endgroup$
