Domain of a Bounded Archimedian Spiral???

Question

So I have a question about a bounded Archimedian Spiral: In one context I get that an Archimedian Spiral's domain and range are all Reals. Thus if I'm looking at what appears to be a bounded spiral: its domain and range must too be bounded i.e. have restrictions. If I drew this spiral curve: assuming we're on the polar coordinate plane--As the spirals angle increases so does its distance from the origin. Would any of that matter when I'm asked to find domain and range? What is the domain of this polar function? I apologize my picture is not very precise: but let's say it was bounded between -3 and 6 once again assuming that this was asked of me on an exam--should I assume this is the polar plane and thus think of the domain in terms of angles and radii?

Answer 1

In an Archimedean spiral the distance from the origin grows at a rate proportional to the change in the polar angle, i.e. in polar coordinates one writes $\rho = c\theta$. But your picture looks as if it starts out going upward rather than rightward from the origin, so with the usual conventions it would be $\rho=c(\theta-\pi/2)$. If it's $\rho=c\theta$ the domain is $\theta\in[0,\infty)$ and if it's $\rho=c(\theta-\pi/2)$ then the domain is $\theta\in[\pi/2,\infty)$. If we take "range" to mean the set of values of $\rho$ in polar coordinates, then that is $[0,\infty)$, i.e. the distance from the origin starts at $0$ and grows without bound.

It looks as if your spiral goes through one-and-a-quarter full turns to get from $0$ to a distance $3$ units from the origin. Therefore it would take another one-and-a-quarter full turns for the distance to grow from $3$ to $6$. In that detail it's not a correct picture of an Archimedean spiral.

For your bounded spiral, it looks as if $\theta$ goes from $\pi/2$ to $3\pi$. I.e. first it's going straight upward, rather than rightward (as, I think, the spiral is usually pictured) so that's $\pi/2$. Then it goes around one-and-three-quarters full circles to reach $3\pi$. That's the domain. But I wonder if the picture you drew is really drawn as it should be for what you're doing. The radial distance $\rho$ goes from $0$ to $6$ in your picture, so that's the range.