# Converting polar coordinates.

• March 17th 2009, 09:49 PM
Cheagles
Converting polar coordinates.
Absolutely lost on this:

r=8cotx (x should be in degrees).

I tried converting this (as well as others) but I seem to just get nowhere after I turned cot into cosx/sinx.

Thanks.

Edit: The equation above should be converted to rectangular form.
• March 18th 2009, 12:16 AM
mr fantastic
Quote:

Originally Posted by Cheagles
Absolutely lost on this:

r=8cotx (x should be in degrees).

I tried converting this (as well as others) but I seem to just get nowhere after I turned cot into cosx/sinx.

Thanks.

Edit: The equation above should be converted to rectangular form.

$\sqrt{x^2 + y^2} = 8 \, \frac{\frac{x}{r}}{\frac{y}{r}} = 8 \, \frac{x}{y}$

$\Rightarrow x^2 + y^2 = 64 \, \frac{x^2}{y^2}$

$\Rightarrow y^2 (x^2 + y^2) = 64 x^2$.
• March 18th 2009, 04:57 AM
HallsofIvy
I think I would have done this a slightly different way:

$r= 8 cot(\theta)= 8\frac{cos(\theta)}{sin(\theta)}$ so
$r sin(\theta)= 8 cos(\theta)$
Multiplying on both sides of the equation by r,
$r(r sin(\theta))= 8 r cos(\theta)$
and
$\sqrt{x^2+ y^2}y= 8x$
Now squaring both sides will give what Mr. fantastic got.