0
$\begingroup$

How would one sketch the graph of $$\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$$ in polar coordinates? I'm aware that polar coordinates involve a radius $r$ and angle $\theta$, such as $r=2-\cos(\theta)$, but in this case we have a cartesian equation. How would we convert this in terms of $r$ and $\theta$? Thanks.

$\endgroup$
  • 2
    $\begingroup$ $x^2+y^2 = r^2$ and $\arctan(y/x) = \theta$ Are you familiar with these formulae? $\endgroup$ – John Lou Sep 26 '17 at 13:59
  • $\begingroup$ @JohnLou Hi John, what process led to those derivations? (it seems straightforward but I still want to know, thanks!) $\endgroup$ – Programmer Sep 26 '17 at 14:02
  • $\begingroup$ The former is derived via the pythagorean theorem (or the distance formula). The same is true of $\theta$. Draw a triangle to help prove this. mathsisfun.com/polar-cartesian-coordinates.html $\endgroup$ – John Lou Sep 26 '17 at 14:03
  • 1
    $\begingroup$ It's also worth pointing out that $\arctan(y/x)$ is only $\theta$ when $-\pi/2 < \theta < \pi/2$. $\endgroup$ – Theo Bendit Sep 26 '17 at 14:03
  • $\begingroup$ So is it just convenient that my equation contains an $x^2+y^2$ and a trig function $\arctan(y/x)$? Otherwise you can't express a Cartesian equation in polar coordinates? $\endgroup$ – Programmer Sep 26 '17 at 14:05
1
$\begingroup$

This appears to be a logarithmic spiral. In the complex plane we have

$$z=e^{-\theta}e^{i\theta}\\ r=e^{-\theta}$$

Thus,

$$ x=\Re(z)=e^{-\theta}\cos\theta\\ y=\Im(z)=e^{-\theta}\sin\theta\\ x^2+y^2=e^{-2\theta}\\ \sqrt{x^2+y^2}=e^{-\theta}\\ \theta=\tan^{-1}\frac{y}{x}\\ $$

So that

$$\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.