By ``basic topology of $\mathbb{R}^n$'' I assume that you are familiar with the notions of openness, closedness, connectedness, and compactness. If you are unclear on these notions (I found compactness hard to get used to), you should remedy that before attempting to learn differential geometry.
If you understand these, then you're probably already prepared to read an introductory book on differential geometry, such as do Carmo's Differential Geometry of Curves and Surfaces or O'Neill's Elementary Differential Geometry. Apart from the concepts I mentioned above, all the necessary topology is developed alongside the geometry in these books (e.g. homeomorphism, homotopy, Euler characteristic, and so on).
If you want to learn quickly about the topology of smooth manifolds without having to learn about general topological spaces, there is probably no better place to look than Milnor's Topology from the Differentiable Viewpoint. A more in-depth treatment along the same lines is Guillemin and Pollack's Differential Topology.
I have two answers. The first one is very specific.
Answer 1: The curve is a variant of the circle involute.
The curve can be rewritten as
$$
\alpha(t) = c_2 \gamma(t) - \frac{c_1}{\omega} t \gamma'(t),
$$
where $\gamma(t)=(-\sin \omega t, \cos \omega t)$ parametrizes the circle. If we "ignore" the values of $c_1$, $c_2$ and $\omega_1$ (assume they are all 1), then $\alpha$ is the standard parametrization of "the" circle involute. Involutes have a very nice visualization. Imagine there is a piece of string tightly glued on to the curve. When you unravel this string, and keep it taut while pulling, the endpoint of the string describes the path of the involute. See the image of the circle involute in the link.
The second answer is more advanced, i.e. you need to know a bit of theory to understand why it works. But the actual calculation is easy, no matter what the parametrization is.
Answer 2: Use the Tait-Kneser theorem. In my old differential geometry class, spirals (or arcs of spirals) were defined as (parts of) curves who have non-zero and monotonous curvature $\kappa$.
The curvature is given by
$$
\kappa(t) = \frac{\det(\alpha'(t)\; \alpha''(t))}{\|\alpha'(t)\|^3}.
$$
So basically, you only have to calculate $\kappa$, and show that $\kappa$ doesn't vanish and that $\kappa'(t)$ is positive or negative everywhere.
This definition of a spiral is very general, but the definition is motivated by the
Tait-Kneser Theorem. It states that the osculating circles of a curve with monotonous curvature are disjoint and nested into each other. Two immediate consequences:
- A curve with monotonous curvature does not intersect itself, and
- Let $c:[a,b]\to\mathbb{R}^2$ be a curve with non-zero, increasing curvature. If $B_{\alpha(s)}$ is the osculating circle of $c$ at $s \in [a,b]$, then $\alpha(t)$ lies outside the circle $B_{\alpha(s)}$ for all $t\in[a,s)$ and $\alpha(t)$ lies inside the circle $B_{\alpha(s)}$ for all $t\in(s,b]$.
To summarize things in a very simple way: The Tait-Kneser theorem just says that a curve with monotonous curvature indeed looks like (a part of) a spiral.
Best Answer
Let $r(\theta)=a+b\theta$ the equation of the Archimedean spiral.
The cartesian coordinates of a point with polar coordinates $(r,\theta)$ are $$\left\{\begin{align} x(r,\theta)&=r\cos\theta\\ y(r,\theta)&=r\sin\theta \end{align}\right. $$ so a point on the spiral has coordinates $$\left\{\begin{align} x(\theta)&=r(\theta)\cos\theta=(a+b\theta)\cos\theta\\ y(\theta)&=r(\theta)\sin\theta=(a+b\theta)\sin\theta \end{align}\right. $$ Differentiating we have $$ \left\{\begin{align} x'(\theta)&=b\cos\theta-(a+b\theta)\sin\theta\\ y'(\theta)&=b\sin\theta+(a+b\theta)\cos\theta \end{align}\right. $$ The parametric equation of a line tangent to the spiral at the point $(r(\theta_0),\theta_0)$ is $$ \left\{\begin{align} x(\theta)&=x(\theta_0)+x'(\theta_0)(\theta-\theta_0)=x(\theta_0)+[b\cos\theta_0-(a+b\theta_0)\sin\theta_0](\theta-\theta_0)\\ y(\theta)&=y(\theta_0)+y'(\theta_0)(\theta-\theta_0)=x(\theta_0)+[b\sin\theta_0+(a+b\theta_0)\cos\theta_0](\theta-\theta_0) \end{align}\right. $$ or in cartesian form $$ y(\theta)-y(\theta_0)=\frac{y'(\theta_0)}{x'(\theta_0)}[x(\theta)-x(\theta_0)]=\frac{b\sin\theta_0+(a+b\theta_0)\cos\theta_0}{b\cos\theta_0-(a+b\theta_0)\sin\theta_0}[x(\theta)-x(\theta_0)] $$ where the slope of the line is $$ \frac{y'(\theta)}{x'(\theta)}=\frac{\operatorname{d}y}{\operatorname{d}x}=\frac{b\sin\theta+(a+b\theta)\cos\theta}{b\cos\theta-(a+b\theta)\sin\theta}=\frac{b\tan\theta+(a+b\theta)}{b-(a+b\theta)\tan\theta} $$