# Polar Equations

• Nov 15th 2009, 07:09 PM
JSB1917
Polar Equations
Directions are simple either convert from polar to Cartesian or vice versa...

1. $\displaystyle r\sin(\theta - \frac{\pi}{4}) = 2$

Then....

$\displaystyle \sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{2}}{2}(sin\theta - cos\theta)$

$\displaystyle x - y = 2\sqrt{2}$

$\displaystyle y = x - 2\sqrt{2}$

Now, my book has $\displaystyle y = x + 2\sqrt{2}$ and I just can't see how they got it, maybe I'm skipping something or making a mistake, but I can't see it as of now.

And...

2. $\displaystyle r^2\sin2\theta = 2$

I got it to...

$\displaystyle r = \frac{1}{xy}$

But that doesn't seem right, so there's probably something obvious I'm missing.

Also, as intuitive as it seems

$\displaystyle \theta = \frac{\pi}{2}$ and $\displaystyle \theta = \frac{3\pi}{2}$ are both $\displaystyle x = 0$

$\displaystyle \theta = \pi$ and $\displaystyle \theta = 0$ are both $\displaystyle y = 0$

Thanks for the help.
• Nov 16th 2009, 03:01 AM
mr fantastic
Quote:

Originally Posted by JSB1917
Directions are simple either convert from polar to Cartesian or vice versa...

1. $\displaystyle r\sin(\theta - \frac{\pi}{4}) = 2$

Then....

$\displaystyle \sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{2}}{2}(sin\theta - cos\theta)$

$\displaystyle x - y = 2\sqrt{2}$

$\displaystyle y = x - 2\sqrt{2}$

Now, my book has $\displaystyle y = x + 2\sqrt{2}$ and I just can't see how they got it, maybe I'm skipping something or making a mistake, but I can't see it as of now.

[snip]

$\displaystyle r \frac{\sqrt{2}}{2}(\sin\theta - \cos\theta) = 2 \Rightarrow r (\sin\theta - \cos\theta) = \frac{4}{\sqrt{2}} \Rightarrow y - x = 2 \sqrt{2}$.
• Nov 16th 2009, 03:04 AM
mr fantastic
Quote:

Originally Posted by JSB1917
Directions are simple either convert from polar to Cartesian or vice versa...

[snip]
2. http://www.mathhelpforum.com/math-he...b9101739-1.gif

I got it to...

http://www.mathhelpforum.com/math-he...c8f6b112-1.gif

But that doesn't seem right, so there's probably something obvious I'm missing.

Also, as intuitive as it seems

$\displaystyle \theta = \frac{\pi}{2}$ and $\displaystyle \theta = \frac{3\pi}{2}$ are both $\displaystyle x = 0$

$\displaystyle \theta = \pi$ and $\displaystyle \theta = 0$ are both $\displaystyle y = 0$

Thanks for the help.

$\displaystyle r^2 2 \sin \theta \cos \theta = 2 \Rightarrow 2xy = 2$.
• Nov 16th 2009, 05:51 AM
JSB1917
Quote:

Originally Posted by mr fantastic
$\displaystyle r \frac{\sqrt{2}}{2}(\sin\theta - \cos\theta) = 2 \Rightarrow r (\sin\theta - \cos\theta) = \frac{4}{\sqrt{2}} \Rightarrow y - x = 2 \sqrt{2}$.

LOL. I see now. I guess it helps if I don't mix up sin and cos.

Quote:

Originally Posted by mr fantastic
$\displaystyle r^2 2 \sin \theta \cos \theta = 2 \Rightarrow 2xy = 2$.

Oh, ok. So it's more like thinking one r for each sin and cos in 2sincos. Oh yeah, because if it was only one r than I would still be left with a cos or sin.