Shifting spiral of archimedes over to the right and down

• Feb 26th 2011, 09:16 PM
qcom
Shifting spiral of archimedes over to the right and down
So, I'm graphing a simple spiral of archimedes, $r= Θ/20$.

Now, I'm trying to figure out how to shift this graph over to the right, and down.

But, it doesn't appear to abide by the standard rules of translation with cartesian coordinate systems.

Considering this is a polar graph, I would assume it's different.

If anyone has any suggestions, I'm listening.
• Feb 27th 2011, 12:28 AM
ragnar
My only advice is to convert to rectangular, shift, try to convert back to polar. Unfortunately, since you have just $\theta$ you'll have to use arcsine/arccosine and that's not good because it's only defined when the argument is in the appropriate interval. But if you do it you get $r = a + b\theta = \sqrt{x^{2} + y^{2}} = a + b \sin^{-1}\Big( \frac{x}{\sqrt{x^{2} + y^{2}}}\Big)$. Shifting by $c,d >0$, $\sqrt{(x-c)^{2} + (y+d)^{2}} = a + b \sin^{-1}\Big( \frac{x-c}{\sqrt{(x-c)^{2} + (y+d)^{2}}}\Big)$ and you can have fun trying to convert this back to just $r$ and $\theta$.