Equation for arc with decaying radius Hoping for some insight into the equation and mechanics of an arc with a decaying radius.
Say at 0 degrees / 0 rad, the radius is 1.  But by 90 degrees / π/2 rad the radius is 1/2.  During that sweep from 0 --> 90 degrees, the radius is decaying at a constant rate.
As I was thinking about this, I was surprised -- though I shouldn't be -- that the arc would "balloon out" a bit on its way to 1/2r from 1r.  And I dig this, and would like to understand the equation that would graph this.
Attached is a sketch of what I'm attempting to graph:
 
Thanks.
 A: What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = \frac1\pi\theta$ (with $\theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
A: You should look into polar coordinates. If you have a point at angle $\theta$ from the $Ox^\to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, \theta)$.
In your case, $r$ is a function of $\theta$: $r(\theta) = 1 - \frac{1}{\pi}\theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(\frac{\pi}{2}) = 1 - \frac{1}{2} = \frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = r\cos(\theta) = (1 - \frac{1}{\pi}\theta)\cos(\theta)$$
$$y = r\sin(\theta) = (1 - \frac{1}{\pi}\theta)\sin(\theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 \leq \theta \leq \frac{\pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $\sin(\theta) = \sqrt{1 - \cos^2(\theta)}$. Then
$$y = (1 - \frac{1}{\pi}\theta)\sqrt{1 - \cos^2(\theta)} =
(1 - \frac{1}{\pi}\theta)
\sqrt{1 - \Bigg(\frac{x}{1 - \frac{1}{\pi}\theta}\Bigg)^2}$$
