Polar equations, limited information given

• May 23rd 2010, 04:45 AM
BG5965
Polar equations, limited information given
Hi, for polar equations, you usually need r and theta, in this case, I am only given one bit of information, how do I do it?

1) Convert these polar equations to Cartesian:
a] $\displaystyle r = 4$
b] $\displaystyle r = 2cos\theta$

2) Convert from Cartesian to polar:
a] $\displaystyle x^2 + (y - 1)^2 = 1$
b] $\displaystyle y = x + 1$

Thanks for any help :)
• May 23rd 2010, 05:14 AM
SpringFan25
There are 2 equations that define the relationship between cartesian and polar coordinates:

$\displaystyle x=r\cos{\theta}$
$\displaystyle y=r\sin{\theta}$

or, equivalently
$\displaystyle x^2 + y^2 = r^2$
$\displaystyle \theta = \tan{\frac{y}{x}}$

To do your conversions, just make the substitutions, eg:

1a
$\displaystyle r=4$
$\displaystyle r^2=16$
$\displaystyle x^2 + y^2=16$
• May 23rd 2010, 05:56 AM
BG5965
Okay, I've got that, but what about the others. For example, 2b) surely can't be:

$\displaystyle y = r sin\theta$ (formula)
$\displaystyle y = x + 1$ (given)

$\displaystyle x + 1 = r sin\theta$

$\displaystyle \therefore r = \frac{x + 1}{sin\theta}$

...could it?
• May 23rd 2010, 05:59 AM
Prove It
Quote:

Originally Posted by SpringFan25
There are 2 equations that define the relationship between cartesian and polar coordinates:

$\displaystyle x=r\cos{\theta}$
$\displaystyle y=r\sin{\theta}$

or, equivalently
$\displaystyle x^2 + y^2 = r^2$
$\displaystyle \theta = \tan{\frac{y}{x}}$

Actually, it's

$\displaystyle \tan{\theta} = \frac{y}{x}$

or $\displaystyle \theta = \arctan{\frac{y}{x}}$.

It's also important to take into account which quadrant you are working in.
• May 23rd 2010, 06:28 AM
SpringFan25
*ahem* yes i rushed it :D
• May 23rd 2010, 06:34 AM
SpringFan25
Quote:

Originally Posted by BG5965
Okay, I've got that, but what about the others. For example, 2b) surely can't be:

$\displaystyle y = r sin\theta$ (formula)
$\displaystyle y = x + 1$ (given)

$\displaystyle x + 1 = r sin\theta$

$\displaystyle \therefore r = \frac{x + 1}{sin\theta}$

...could it?

You need to substitute for the x as well.

so, to start you off:
$\displaystyle y = x + 1$
$\displaystyle r\sin{\theta} = r\cos{\theta} + 1$
• May 23rd 2010, 06:43 AM
BG5965
Is it:
$\displaystyle r = \frac{1}{sin\theta - cos\theta}$ ?
• May 23rd 2010, 01:42 PM
SpringFan25
i think so, but i fear the wrath of prove it if im wrong :D
• May 23rd 2010, 05:12 PM
Prove It
Quote:

Originally Posted by SpringFan25
i think so, but i fear the wrath of prove it if im wrong :D

Yes it's correct.

And yes, I'm always watching (Smirk)