Find asymptotes and sketch in polar

**DBA** · April 5th 2010, 08:39 PM

Find the asymptotes by expressing the rectangular coordinates x and y as function of theta, then sketch the curve in polar.

$ r= cot (\theta ) $

$ y= r*sin (\theta ) $
$ x= r*cos (\theta ) $

$ r= cot(\theta) $
$ r= \frac{cos(\theta)}{sin(\theta)}$ |multipy both sides by $sin(\theta)$
$ r sin(\theta) = \frac{cos(\theta)}{sin(\theta)} sin(\theta) $
$ y = cos(\theta) $

How do I get the horizontal asymptote from here?

Then
$ r= cot(\theta) $
$ r= \frac{cos(\theta)}{sin(\theta)}$ |multipy both sides by $cos(\theta)$
$ r cos(\theta) = \frac{cos(\theta)}{sin(\theta)} cos(\theta) $
$ x= \frac{cos^{2}(\theta)}{sin(\theta)} $

How do I get the vertical asymptote from here?
Thanks!

**TheEmptySet** · April 5th 2010, 09:28 PM

To change the equation into rectangular form remember that $r=\sqrt{x^2+y^2}$

Multiply both sides of the equation by $r$

$r=\frac{\cos(\theta)}{\sin(\theta)} \iff r\cdot r\sin(\theta)=r\cos(\theta) \iff \sqrt{x^2+y^2}\cdot y=x$

square one more time to get

$y^2(x^2+y^2)=x^2 \iff y^2=\frac{x^2}{x^2+y^2}$

To find the horizontal asymptotes take the limit as $x \to \pm \infty$

For the vertical take the implicit derivative; solve for $y'$ then set the denominator equal to 0.

**DBA** · April 5th 2010, 09:50 PM

Thanks for your answer, but I still do not knw how to get the asymptotes.

How do I do that with y^2?

And why do I have to take the implicit derivative for the vertical asymptote?

Thanks

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