Polar equations, limited information given

**BG5965** · May 23rd 2010, 04:45 AM

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] $r = 4$
b] $r = 2cos\theta$

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

Thanks for any help

**SpringFan25** · May 23rd 2010, 05:14 AM

There are 2 equations that define the relationship between cartesian and polar coordinates:

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

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

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

1a
$r=4$
$r^2=16$
$x^2 + y^2=16$

**BG5965** · May 23rd 2010, 05:56 AM

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

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

$x + 1 = r sin\theta$

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

...could it?

**Prove It** · May 23rd 2010, 05:59 AM

Originally Posted by SpringFan25

There are 2 equations that define the relationship between cartesian and polar coordinates:

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

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

Actually, it's

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

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

It's also important to take into account which quadrant you are working in.

**SpringFan25** · May 23rd 2010, 06:28 AM

*ahem* yes i rushed it

**SpringFan25** · May 23rd 2010, 06:34 AM

Originally Posted by BG5965

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

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

$x + 1 = r sin\theta$

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

...could it?

You need to substitute for the x as well.

so, to start you off:
$y = x + 1$
$r\sin{\theta} = r\cos{\theta} + 1$

**BG5965** · May 23rd 2010, 06:43 AM

Is it:
$r = \frac{1}{sin\theta - cos\theta}$ ?

**SpringFan25** · May 23rd 2010, 01:42 PM

i think so, but i fear the wrath of prove it if im wrong

**Prove It** · May 23rd 2010, 05:12 PM

Originally Posted by SpringFan25

i think so, but i fear the wrath of prove it if im wrong

Yes it's correct.

And yes, I'm always watching

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