Linear function in polar

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Linear function in polar

Postby moonlord » Sat Mar 04, 2006 4:01 pm

This question has bothered me for a while... What function type draws a straight line when represented in polar coordinates? My derivation suggested that tangent should be fine... However, the plotter showed something else. I'm quite out of ideas now. Anyone can help?
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Postby bo198214 » Sat Mar 04, 2006 11:13 pm

For drawing a parallel line to the x-axis through (0,b), simply use
r(phi) = b/sin(phi)

If you want to have a general linear function ax+b in poloar coordinates, simply note
r^2 = x^2 + (ax+b)^2
x = r cos(phi)

This results in a polynom of second degree in r, which you can already solve for r (as I have heard ;) )

For verification let a=0, then
r^2 = (r cos(phi))^2 + b^2
r^2(1-cos^2(phi)) = b^2
r^2 sin^2(phi) = b^2
r = b/sin(phi)
and we have the first given formula.

Or did I completely misinterpret your question :?
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Postby moonlord » Sun Mar 05, 2006 2:34 pm

Thank you. It's exactly what I was asking about... Is it something you learn as-is or did you derive it someway?

Yes, I can solve 2nd, but also 3rd and 4th degree polynomials ;).

EDIT: Right. After actually reading your post, it seems logical that a circle in polar should be a straight line in cartesian...
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Postby bo198214 » Sun Mar 05, 2006 2:53 pm

Its purely geometrical derivation:

r^2 = x^2 + y^2
x = r cos(phi)

these are the relations in the right angled triangle (0,0),(x,0),(x,y)

Code: Select all
^
|
|          /|
|        /  |
|    r /    |
|    /      | y
|  / \      |
|/ ph |     |
-----------|----->
       x
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Postby moonlord » Sun Mar 05, 2006 3:21 pm

Yes, yes, I see now. :P
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Postby thigle » Thu Mar 09, 2006 8:02 am

looking into glass-sphere, or crystal-ball, one sees clearly that all physical-world parallels are drawn as circle(arcs) except for the ones whose image pass the centre.

spherical-lens is an exciting model of elliptic geometry to explore, especially in movement.
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