1. ## polar coordinate equations

need a little help... i know how to convert given rectangular or polar cordinates.... bu ti have a feeling i should know how to do it with an equation too. help appreciated... might fail here...

instructions on this arent clear, so i need your help.

The letters x and y represent rectangular coordinates. Write in polar cordinates (r, theta)

$2x^2 + 2y^2 = 3$
$x^2 = 4y$
$2xy = 1$

The letter r and theta represent polar coordinates. Write using rectangular coordinates. (x,y)

$r = cos(theta)$
$r = 2$

a little explanation of the process of this would be much appreciated... im sure it can be very simple.
also does nay one know what the tag for theta is.

thanks, help much appreciated.

2. Originally Posted by cjmac87
need a little help... i know how to convert given rectangular or polar cordinates.... bu ti have a feeling i should know how to do it with an equation too. help appreciated... might fail here...

instructions on this arent clear, so i need your help.

The letters x and y represent rectangular coordinates. Write in polar cordinates (r, theta)

$2x^2 + 2y^2 = 3$
$x^2 = 4y$
$2xy = 1$

The letter r and theta represent polar coordinates. Write using rectangular coordinates. (x,y)

$r = cos(theta)$
$r = 2$

a little explanation of the process of this would be much appreciated... im sure it can be very simple.
also does nay one know what the tag for theta is.

thanks, help much appreciated.
$$\theta$$ gives $\theta$.

To go from Cartesian equations to polar equations, just substitute

$x = r \cos \theta$ .... (1)

$y = r \sin \theta$ .... (2)

Note: Square equations (1) and (2) and add: $x^2 + y^2 = r^2$.

Some simplification using double angle formulae etc. can sometimes be done.

$r = \cos \theta \Rightarrow r^2 = r \cos \theta \Rightarrow x^2 + y^2 = x$ ......

$r = 2 \Rightarrow r^2 = 2^2 \Rightarrow x^2 + y^2 = 2^2$, a circle as you might expect since r is a constant ....

3. Hello,

$x=r \cos(\theta)$

$y=r \sin(\theta)$

Thus ${\color{blue}2x^2 + 2y^2 = 3} \Longleftrightarrow 2r^2\cos^2(\theta)+2r^2\sin^2(\theta)=3$

$2r^2\underbrace{(\cos^2+\sin^2)}_{=1}=3$
$\boxed{r^2=\frac 32}$

${\color{blue}x^2 = 4y} \Longleftrightarrow r^2\cos^2(\theta)=4r\sin(\theta)$

$r \cos^2(\theta)=4\sin(\theta)$

${\color{blue}2xy = 1} \Longleftrightarrow 2r^2 \cos(\theta)\sin(\theta)=1$

And you know that
$2 \cos(\theta)\sin(\theta)=\sin(2 \theta)$

4. thank you, you both have been very helpful