# Polar equations, limited information given

#### BG5965

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

#### 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
$$\displaystyle r=4$$
$$\displaystyle r^2=16$$
$$\displaystyle x^2 + y^2=16$$

#### 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?

#### Prove It

MHF Helper
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.

SpringFan25

#### SpringFan25

*ahem* yes i rushed it

#### SpringFan25

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:
$$\displaystyle y = x + 1$$
$$\displaystyle r\sin{\theta} = r\cos{\theta} + 1$$

#### BG5965

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

#### SpringFan25

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

#### Prove It

MHF Helper
i think so, but i fear the wrath of prove it if im wrong
Yes it's correct.

And yes, I'm always watching (Smirk)