How do I get an Archimedean spiral that decreases from an initial radius?

Question

So, where the equation of an archimedean spiral is: $$r = a + b\theta$$

I want to be able to use the equation in this form to then have a function where r decreases in exactly the same way by an angle I will call $\phi$ from an initial radius. Does this make sense? So on the initial radius, $\phi= 0$ and when $r = 0, \theta = 0$

It might be something really simple I am overlooking here, but any help would be appreciated!

Also, could someone edit this and add the appropriate tags? I'm kind of novice, and I don't know of a tag that fits this.

Answer 1

I think I understand what you mean.

You want to start with an initial non-zero radius $a$.

So $r(\theta=0)=a$

The general equation is $r=a+b\theta$ and you want at the end a zero angle with the $x$ axis.

I will then state that $r(\theta=2k\pi)=0$, with $k$ being the number of "turns" of your graph before $r$ reaches $0$.

Then you choose $b=-\dfrac{a}{2k\pi}$ and you final equation is

$r=a-\dfrac{a}{2k\pi}\theta$ ou encore $r=a(1-\dfrac{\theta}{2k\pi})$