1. 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.

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

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

$\displaystyle \Rightarrow y^2 (x^2 + y^2) = 64 x^2$.

3. I think I would have done this a slightly different way:

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