# Polar to Cartesian Problem

• Oct 20th 2010, 03:43 AM
Paymemoney
Polar to Cartesian Problem
Hi
I am having trouble with the following question:
How would you change the following equation in r and $\theta$ to an equivalent equation in x and y?

$r=2sin2\theta$

Another question:

$r=tan\theta$

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

$x(x+y)=y$
$x^2+xy=y$

what have i done wrong

P.S
• Oct 20th 2010, 03:53 AM
Also sprach Zarathustra
Here:

Transformation from Polar to Cartesian Coordinates

In second one:

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

$x^2+y^2=\frac{y^2}{x^2}$

$\frac{x^4}{x^2-1}+y^2=0$
• Oct 20th 2010, 04:16 AM
Paymemoney
in the first question i am stuck on what does $sin2\theta$ equal to?

i know that sin\theta = $\frac{y}{r}$ so what is $sin2\theta$?
• Oct 20th 2010, 04:17 AM
Also sprach Zarathustra
sin(2t)=2sin(t)cos(t)
• Oct 20th 2010, 04:22 AM
Paymemoney
oh ok, knew it was an identity
• Oct 20th 2010, 04:31 AM
Paymemoney
i get for question 1: $(x^2+y^2)^2=4y^2x^2$ however the book's answer is $(x^2+y^2)^3=4y^2x^2$.
am i wrong or the book?
• Oct 20th 2010, 04:45 AM
Also sprach Zarathustra
$\sqrt{x^2+y^2}=4\frac{x}{\sqrt{x^2+y^2}}\frac{y}{\ sqrt{x^2+y^2}}$

$\sqrt{x^2+y^2} (x^2+y^2)=4xy$

$(x^2+y^2)^{\frac{3}{2}}=4xy$

$(x^2+y^2)^3=(4xy)^2$

$(x^2+y^2)^3=16x^2y^2$

Hence, you and your book mistaken...
• Oct 20th 2010, 05:01 AM
Paymemoney
srry my mistake i posted the wrong question, its suppose to be $r=sin2\theta$
• Oct 20th 2010, 05:06 AM
Paymemoney
just solved it
• Oct 20th 2010, 05:06 AM
Also sprach Zarathustra
Quote:

Originally Posted by Paymemoney
srry my mistake i posted the wrong question, its suppose to be $r=sin2\theta$

i get for question 1: $(x^2+y^2)^2=4y^2x^2$ however the book's answer is $(x^2+y^2)^3=4y^2x^2$.
$\displaystyle r = 2 \sin (\theta) \cos (\theta) = 2 \left( \frac{y}{r} \right) \left(\frac{x}{r}\right) = \frac{2xy}{r^2}$
$\Rightarrow r^3 = 2xy \Rightarrow (x^2 + y^2)^{3/2} = 2xy$