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Question:

(i) Define in polar coordinates $r = f(\alpha)$ the origin-centred circle with radius $R$. Specify the domain range for the polar coordinate $\alpha$.

(ii) Define in polar coordinates $r = f(\alpha)$ a circle with radius $R$ and the centre at the Cartesian coordinates $(R, 0)$. Specify the domain range for the polar coordinate $\alpha$.

My Answer for (i):

$x = r \cdot \cos \alpha$

$y = r \cdot \sin \alpha$

$x^2 + y^2 = R^2$ (equation for circle, radius R centered at origin)

$r^2(\cos^2 \alpha + \sin^2 \alpha) = R^2 $

$r^2 = R^2$

$r = R$

The domain range of $\alpha$ is $(-\infty, \infty) $?

Is my answer correct? I don't think it's correct because my RHS doesn't have $\alpha$ at all. I've learnt that domain is the set valid input values and range is the set valid output values. However, I don't understand what the question means by "domain range" of $\alpha$?

For part (ii) I don't understand how do you shift a circle centered at origin to become the radius.

I am sorry if this seems like a trivial question, please don't close this question. I haven't done any serious math for the last 2 years, especially geometry. The textbook that he had written (and made us buy) didn't really ease us into the topic but expected us to already know all these formulas, I only managed to do these much after 6 hours of googling and reading up on functions, polar equations, and so on. Please please please help, thanks!!

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  • $\begingroup$ For the first one you're right that $\alpha$ has no restriction, but the question is really asking what is the smallest set of angles you could get away with, but still retrieve the whole circle (so $(-\infty,\infty)$ is definitely overkill because you repeat the circle many, many times) $\endgroup$ – Ninad Munshi Aug 13 '20 at 5:37
  • $\begingroup$ Thanks Ninad, does that mean that the domain range is (0, 2$\pi$), which means all the angles from 0 to 360 degrees? $\endgroup$ – IDontUnderstandMath Aug 13 '20 at 6:03
  • $\begingroup$ Yes it does, and zkutch has an answer for the other one below. $\endgroup$ – Ninad Munshi Aug 13 '20 at 6:04
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For first think about period for trigonometry functions.

For second you have $(x-R)^2+y^2=R^2$ circle. Inserting polar coordinates gives $$(x-R)^2+y^2=R^2 \Leftrightarrow r-2\cos \alpha=0$$ From here you obtain also restrictions for $\alpha$: $\cos \alpha \geqslant 0$ and its enough to find one continuous segment for it.

And, at last little gift for second: consider $x-R = r \cdot \cos \alpha$ and $y = r \cdot \sin \alpha$. This is little "extended" polar coordinates, but sometimes such shifting center creates big simplicity.

If I can advise something, then, it is to try to "look at world with polar eyes": disk bounded by circle in polar coordinates becomes rectangle. Disk with shifted center, as in your 2nd example, becomes area under $\cos$ function, but if you use "gift", then it again will be rectangle. Try to draw every figure you work with in polar coordinates so, if they are simple cartesian $(r, \alpha)$. Same world with another eyes.

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