How can I argue that the trace of this curve is a spiral? I'm solving a problem where I need to describe what kind of behavior certain curves have. I have the curve
$$
\alpha(t) = c_1 t (\cos(wt), \sin(wt)) + c_2(-\sin(wt), \cos(wt)) \qquad t \in \mathbb{R}
$$
where $c_1, c_2,w$ are some real arbitrary constants and $c_1, w \neq 0$. By using graphing software I have determined that the trace of the curve is a spiral, however, I have not found a way to argue why this is the behavior the curve has using only analytic methods.

I noticed that the vectors $ (\cos(wt), \sin(wt))$ and $(-\sin(wt), \cos(wt))$ which compose $\alpha$ both describe a unit circle, so in a sense, I could see $\alpha$ as $2$ "circles" being added, where one has constant radius $\lvert c_2 \rvert$ and the other one has an increasing radius. However, It's not obvious to me why adding those $2$ curves should give a spiral.
I also tried to write the curve in polar form, since if the curve $\alpha$ can be written as
$$
r = r(\theta)
$$
and $r(\theta)$ is a continuous monotonic function, then I can argue it's a spiral. The furthest I got with this was that
$$
r =\sqrt{x^2 + y^2} = \sqrt{c_1^2 [t(\theta)]^2 + c_2^2}
$$
but I wasn't able to solve for $t$ in terms of $\theta$, so this didn't seem to work either.
Lastly, I also thought about calculating the curvature of $\alpha$, but this seemed to be a tedious process and I wasn't sure how I could conclude spiral-like behavior just from the curvature of $\alpha$.

Does anyone have any ideas to determine that the trace of $\alpha$ is a spiral without needing to graph the curve explicitly? Any ideas are welcome. Thank you!
 A: 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.
A: A spiral is defined here as a polar curve described by any equation (complex version form)
$$z=r(t)e^{i\omega t}$$ where the modulus $r$ is a real function with two properties: it has to be monotonous and continuous.
"Our" spiral can be described under complex form like this:
$$z=(c_1t+c_2i)e^{it}=i(c_2-i c_1t)e^{it}$$
(I have assumed $\omega = 1$ WLOG).
Let us drop the front "$i$" factor accounting for a $\pi/2$ rotation, which doesn't change the nature of the curve. We are now with :
$$z=(c_2-i c_1t)e^{it}=\underbrace{\sqrt{(c_1t)^2+c_2^2}}_{r(t)}e^{-i \operatorname{atan}\frac{c_1 t}{c_2}}e^{it}$$
where  function $r$ (that you had considered) is clearly continuous and increasing. The previous relationship can be transformed into:
$$z=r(t)e^{if(t)} \ \text{where} \ f(t):=t-\operatorname{atan} (at), \ \text{with} \ a:=c_1/c_2\tag{1}$$
The derivative of $f$ is:
$$f'(t)=1-\frac{a}{1+a^2t^2}=\frac{(1-a)+a^2t^2}{1+a^2t^2}\tag{2}$$
First case: $a<1$; in this case, as a consequence of (2), $f$ is strictly increasing, which means that $f^{-1}$ exists and is as well continuous and strictly increasing.
This allows to switch to a new parameterization of the curve by transforming polar equation (1) in the following way (using notations $s=f(t) \iff f^{-1}(s)=t$)
$$(1) \ \iff \ z=r(f^{-1}(s))e^{is}$$
The composition $r \circ f^{-1}$ of two increasing and continuous functions remains increasing and continuous. We have established that it is a spiral in this case.
Second case: $a \ge 1$.
In this case, one has to decompose the issue in two parts, according to the value of $t$.
Indeed, considering (2), seting $t_0:=\frac{1}{a}\sqrt{a-1}$

*

*if $t>t_0$, we find back the condition in which we were in case 1.


*if $t<t_0$, we have to reverse the orientation followed by the curve, which amounts to operate on it a symmetry which doesn't change the nature of the curve : it is still a spiral.
As a conclusion, in this second case: the curve is the union of two "patented" spirals...
A: Why not just plot it and be done with it?

