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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!

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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
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1 Answer

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.)

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