In what sense does the unit vector $\hat{\theta}$ have length 1? Consider the 2-dimensional polar coordinate system describing Euclidean space where points in the space are given by the coordinates $(r,\theta)$. In what sense does the unit vector $\hat{\theta}$ have length 1? If one considers the unit circle centred on the origin of the space, is a step in the $\theta$-direction of unit length just the distance around the circle corresponding to an angle of 1 radian?
Edit for further clarification
My background is in physics, so an example in terms of a physics problem might help highlight my confusion.
From studying physics, I do have some degree of understanding and intuition of vectors defined in a polar coordinate system (or any kind of curvilinear coordinate system, I suppose). I can make sense of the angular velocity vector for circular motion, to pick just one example.
My understanding here is that if the force keeping a particle constrained to a circular trajectory was suddenly removed then the particle would fly off tangent to the circle (i.e. in the instantaneous direction of the angular velocity vector) with a speed given by the magnitude of the angular velocity vector. Essentially, my understanding in this case is that the unit vector in the $\theta$-direction defines one of the two coordinate axes in the tangent space to the circular path, and it is in this direction that the particle would move if suddenly freed of its confining force.
Once freed I'd understand $\hat{\theta}$ to correspond to this tangent space, and some step of unit length in the $\theta$-direction to correspond to a step of unit length in the tangent space to the circle. However, before the particle is freed, it's confined to stay on the circular path and I'm not sure that I can make sense of what it means to make a step of unit length in the $\theta$-direction. My understanding is that if one travels some distance in the $\theta$ direction then they've literally travelled along the circumference of a circle of fixed $r$. This understanding might be wrong. If so, what is the resolution? If not, what does it mean to make a step of unit length in the $\theta$-direction when confined to a circle?
 A: Maybe one way that will help the understanding is to consider the integral
$$\int_0^{\theta_0} {\hat \theta} d{\theta}$$
I am assuming $\hat \theta$ defined in this way:
$${\hat \theta} = \begin{bmatrix}-\sin(\theta)\\\cos(\theta)\end{bmatrix}$$
It will be unit vector because of the trigonometric identity: $\sin(\theta)^2+\cos(\theta)^2 = 1$. We can also convince ourselves that it will be orthogonal to $\begin{bmatrix}r\cos(\theta)\\r\sin(\theta)\end{bmatrix}$ which is often defined to be the radial component.
If we approximate the integral using a computer, we see a relation between the angle integrated, the unit circle and the line integral result:

The end points of the blue line are angles $0$ and $\theta_0$ on the unit circle. And the red vector is the vector from point $(1,0)$ pointing onto the unit circle at angle $\theta_0$. (I could not find any arrowhead to indicate the direction).
A: Given polar coordinates $(r, \theta)$
for a vector
$$
(x,y) = r \, e_r
$$
we have the radial unit vector
$$
e_r = (\cos \theta, \sin \theta) \\
$$
and the perpendicular
$$
e_\theta = (- \sin \theta, \cos \theta)
$$
The length of $e_\theta$ is
$$
\lVert e_\theta \rVert 
= \sqrt{(-\sin \theta)^2 + (\cos \theta)^2} 
= \sqrt{1} = 1
$$
