1. ## Polar to cartesian

Find the cartesian equation of $r^2 = 9\csc(20)$?
choices:
A. xy = 9
B. 3xy = 9
C. 2xy = 9
D. 4xy = 9

This is what ive done

$(r^2= \frac{9}{\sin(20)}) \frac{1}{r}$

$\sqrt{x^2 + y^2} = \frac{9}{y}$

$x^2 + y^2 = \frac{81}{y^2}$

$y^2(x^2 + y^2) = 81$

2. ## Re: Polar to cartesian

Originally Posted by TechnicianEngineer
Find the cartesian equation of $r^2 = 9\csc(20)$?
choices:
A. xy = 9
B. 3xy = 9
C. 2xy = 9
D. 4xy = 9

This is what ive done

$(r^2= \frac{9}{\sin(20)}) \frac{1}{r}$

$\sqrt{x^2 + y^2} = \frac{9}{y}$

$x^2 + y^2 = \frac{81}{y^2}$

$y^2(x^2 + y^2) = 81$
Why has 1/r appeared - out of nowhere - in the first line?

$r^2 = \frac{9}{\sin \theta} \Rightarrow r = \frac{9}{r \sin \theta} \Rightarrow \sqrt{x^2 + y^2} = \frac{9}{y} \Rightarrow ....$

I don't agree with any of the given options.

Edit: OK, this is what happens when someone uses poor notation to express an equation. No doubt the 20 is meant to be $2 \theta$, in which case you have $r^2 = \frac{9}{\sin (2 \theta)} \Rightarrow r^2 = \frac{9}{2 \sin (\theta) \cos (\theta)} \Rightarrow 1 = \frac{9}{2 r \sin (\theta) r \cos (\theta)} \Rightarrow ....$

3. ## Re: Polar to cartesian

Hello, TechnicianEngineer!

$\text{Find the cartesian equation of: }\:r^2 \:=\: 9\csc(2\theta)$

. . $(A)\;xy \,=\, 9 \qquad (B)\; 3xy \,=\, 9 \qquad (C)\; 2xy \,=\, 9 \qquad (D)\; 4xy \,=\, 9$

This is the way I would do it . . .

. . . . . . . . . . . . . $r^2 \:=\:9\csc(2\theta)$

. . . . . . . . . . . . . $r^2 \:=\:\frac{9}{\sin(2\theta)}$

m . . . . . . $r^2\sin(2\theta) \:=\:9$

. . . . $r^2(2\sin\theta\cos\theta) \:=\:9$

. . $2\overbrace{(r\cos\theta)}^{\text{This is }x}\underbrace{(r\sin\theta)}_{\text{This is }y} \;=\;9$

n . . . . . . . . . . $2xy \;=\;9 \quad\text{ answer (C)}$

4. ## Re: Polar to cartesian

i apologize that the questionnaire was not well photocopied
i myself thought it was csc(20)