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.

How to use polar coordinate to represent a $1 \times 1$ square rotated $45^\circ$ and translated to $(7,4)$? Does the $r(\theta)$ have discontinuous (such at jump from $+5$ to $-2$)?

Please help. Thanks!

share|improve this question
    
Do you really need to have it in polar coordinates? Translation's a pain in the arse to do in polar coordinates; maybe you'd be better satisfied with a (parametric) Cartesian form? –  J. M. Oct 30 '11 at 2:31
    
Actually, I want to know the theory. If I have a 1x1 square rotated 45deg, I guess there is unique +ve R in any given angle using polar coordinate in this case. If I translate the whole square to (7,4) , would some R jump from +ve to -ve?? –  Marco Oct 30 '11 at 4:08
    
$(7,4)$ is a pretty far-out point; you can certainly expect a rotated square that is translated that far to be contained entirely in the first quadrant. –  J. M. Oct 30 '11 at 4:12
    
So does it mean the R would jump? I am sorry that I am very bad in polar coordinate, but I am doing project using polar coordinate concept –  Marco Oct 30 '11 at 4:15

1 Answer 1

There is no the $r(\theta )$ for the translated square. In order for a shape to be described by a single polar equation $r(\theta)$, it must be star-shaped about the origin. A square centered at the origin has the property; your square does not.

Formally: instead of $\{(r,\theta):r\le r(\theta)\}$ you now have $\{(r,\theta):r_1(\theta)\le r\le r_2(\theta)\}$ where $r_1$ and $r_2$ are different functions of $\theta$: first for the part of the boundary that's visible from the origin, second for the rest. Since the boundary is piecewise linear, the building block is the polar equation of a line: $$r(\theta)=r_0\sec (\theta-\theta_0) \tag1$$ where $(r_0,\theta_0)$ are the polar coordinates of the orthogonal projection of the origin onto the line. (Formula (1) assumes that the line does not pass through the origin; otherwise the equation would be simpler.)

share|improve this answer

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.