# Thread: Find asymptotes and sketch in polar

1. ## Find asymptotes and sketch in polar

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!

2. 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.

3. 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