Let $\alpha(s)$ be a unit speed curve in $R^2$. Show $\kappa=|\frac{d\theta}{ds}|$

Question

I'm lost on solving the following problem.

Let $\alpha(s)$ be a unit speed curve in $R^2$. Show $\kappa=|\frac{d\theta}{ds}|$, where $\theta$ is the angle between the positive $x$-axis and the tangent line to the curve $\alpha$ measured in the counterclockwise sense.

I'm used to curves in $R^3$ and I'm not sure how to tackle this problem, which is restricted to $R^2$. I don't know what property of curvature I should be using. I would greatly appreciate any solutions, hints or suggestions.

Answer 1

If $\theta(s)$ is the angle ´twixt the $x$-axis and the tangent line at the point $\alpha(s)$, the unit tangent vector $\mathbf T(s) = \alpha'(s)$ is given by

$\mathbf T(s) = (\cos \theta(s), \sin \theta (s)); \tag{1}$

then we have

$\mathbf T'(s) = (-\sin \theta(s), \cos \theta (s)) \theta'(s); \tag{2}$

setting

$\mathbf n(s) = (-\sin \theta(s), \cos \theta(s)), \tag{3}$

we see that

$\mathbf n(s) \cdot \mathbf n(s) = \sin^2 \theta(s) + \cos^2 \theta(s) = 1, \tag{4}$

whilst

$\mathbf T(s) \cdot \mathbf n(s) = (\cos \theta(s))(-\sin \theta(s)) + (\sin \theta(s))(\cos \theta(s)) = 0; \tag{5}$

by (4) and (5), $\mathbf n(s)$ is a unit normal vector field along the curve $\alpha(s)$. We also have the unit normal field $\mathbf N(s)$ which occurs in the Frenet-Serret equation for a planar curve, i.e., in the equation

$\mathbf T'(s) = \kappa(s) \mathbf N(s)\tag{6}$

(which in fact is also the first equation in the three-dimensional Frenet-Serret equations for space curves); we recall that in (6) $\kappa(s)$ is always taken to be positive; this convention ensures $\mathbf N(s)$ is uniquely defined via (6) and we have that $\kappa(s)$ is in fact the magnitude $\Vert \mathbf T'(s) \Vert$ of $\mathbf T'(s)$:

$\Vert \mathbf T'(s) \Vert = \Vert \kappa(s) \mathbf N(s) \Vert = \kappa(s) \Vert N(s) \Vert = \kappa(s). \tag{7}$

The problem at hand is now resolved by equating the two expressions for $\mathbf T'(s)$ (2) and (6): noting that by (3), (2) may be written

$\mathbf T'(s) = \theta'(s) \mathbf n(s), \tag{8}$

we see that

$\theta'(s) \mathbf n(s) = \kappa(s) \mathbf N(s); \tag{9}$

taking norms yields

$\vert \theta'(s) \vert = \Vert \theta'(s) \mathbf n(s) \Vert = \Vert \kappa(s) \mathbf N(s) \Vert = \kappa(s), \tag{10}$

again since $\kappa(s) > 0$ as in (6).

Answer 2

Let $\alpha: \mathbb{R} \rightarrow \mathbb{R}^2$ be a non-linear, non-stop, curve parametrized by arclength $s$. By assumption, $||\alpha'(s)||=1$, but, we'd rather write this as $T \cdot T=1$ where $T = \alpha'$. Ok, so differentiate, $T' \cdot T+T \cdot T'=0$ and we find $T'$ is perpendicular to $T$. Look familiar? I hope, the three dimensional argument is the same. Now, we have a vector $T'$ which is perpendicular to $T$. To get an orthonormal framing of the curve in two dimensions it suffices to normalize $T'$ hence the normal and curvature: $||T'||= \kappa$ and $T' = \kappa N$ follows. Thus, to calculate the curvature in two dimensions we still calculate: $$ \kappa = \left|\left| \frac{dT}{ds} \right|\right| $$ Continuing, I think we can define $\theta$ by trigonometry... but, I'll stop here since you probably can do the rest.

Answer 3

$$ t:= \alpha'(s) \in \mathbb{R}^2,\ t'=\kappa n\in \mathbb{R}^2 $$ And $|t|=1$ implies that $$ t\cdot n=0$$

Then $$ \cos\ \theta = t\cdot e_1,\ -\sin\ \theta =n\cdot e_1 $$ so that $$ -\sin\ \theta \theta ' = t'\cdot e_1=\kappa n\cdot e_1 $$