Polar equations, limited information given

May 2008
101
0
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 2010
1,034
272
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 2008
101
0
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
Aug 2008
12,894
5,000
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.
 
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May 2010
1,034
272
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 2008
101
0
Is it:
\(\displaystyle r = \frac{1}{sin\theta - cos\theta}\) ?
 
May 2010
1,034
272
i think so, but i fear the wrath of prove it if im wrong :D