Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

OK, so I have the following polar equation:

$r = Θ/20$

And I would like to translate this a little to the right, and down from the polar origin.

Now, I figure since I know cartesian coordinate equation translations are quite simple, the best way to do this would be to convert this polar equation to a rectangular equation, translate, and then convert back to polar again.

However, I was having some problems making the initial conversion.

I know we are supposed to use the following relationships between polar and rectangular equations:

$r^2 = x^2 + y^2$

$y = rsin(Θ)$

But, I cannot seem to convert my equation correctly...

Would someone be able to suggest a simpler way to make this translation?

Or, assist me in converting to polar and back again?

I'm quite new at this, so I apologize if this question is remedial.

Thanks!

share|improve this question
add comment

2 Answers 2

The other equation you need is $\theta = \arctan \frac {y}{x}$ and you have to worry about the branches of arctan. So now you have $\sqrt{x^2+y^2}=\frac{1}{20} \arctan \frac {y}{x}$. To translate right and down, you replace $x$ with $x+a$ and $y$ with $y-b$. And you have quite a mess. Taking $\sqrt{(x+a)^2+(y-b)^2}=\frac{1}{20} \arctan \frac {y-b}{x+a}$ back to polar is a lot of work and probably not illuminating.

share|improve this answer
    
Thanks for the help. So, is there a better way to translate the original polar equation without messing with conversions? –  Qcom Feb 27 '11 at 18:48
2  
I don't know of one. Polar coordinates are very good at things centered on the pole, and very bad at things that are not. –  Ross Millikan Feb 27 '11 at 18:50
    
OK, thanks again though. –  Qcom Feb 27 '11 at 18:51
add comment

If you want to do a translation in polar coordinates by $(r_0,\theta_0)$, the following formulas hold $$r' = \sqrt{r^2 + r_0^2 -2 r r_0 \cos(\theta - \theta_0)}$$ $$\theta' = \arctan([ r \sin(\theta) -r_0 \sin(\theta_0)]/[r \cos(\theta) -r_0 \cos(\theta_0)])$$ where $'$ denotes the new system. Note that there is a problem with the ambiguity of the $\arctan$ function. For this reason often arctan2 is implemented (or you can implement it yourself). Having arctan2 the second equation reads $$\theta' = \text{arctan2}( r \sin(\theta) -r_0 \sin(\theta_0),r \cos(\theta) -r_0 \cos(\theta_0)).$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.