# Understanding curvature as rate of change of angle between neighbouring tangents

When introducing the concept of curvature in "Differential Geometry of Curves and Surfaces", do Carmo makes the statement

the norm $|\alpha''(s)|$ of the second derivative measure the rate of change of the angle which neighbouring tangents make with the tangent at $s$.

Here $\alpha(s)$ is a curve parametrized by arc length.

I have read similar things elsewhere, e.g. Wikipedia states

the curvature of a plane curve at any point is the limiting ratio of $d\theta$, an infinitesimal angle (in radians) between tangents to that curve at the ends of an infinitesimal segment of the curve, to the length of that segment $ds$, i.e., $d\theta/ds$. If the tangents at the ends of the segment are represented by unit vectors, it is easy to show that in this limit, the magnitude of the difference vector is equal to $d\theta$, which leads to the given expression in the second definition of curvature.

Accompanying this is the following potentially useful diagram:

I cannot understand how to make this interpretation rigorous. In particular, interpreting do Carmo at face value, if we define $\theta_{s_0}(s)$ to be the angle between $\alpha'(s)$ and $\alpha'(s_0)$ for some fixed $s_0$, then

$$\theta_{s_0}(s) = \arccos\left[\alpha'(s) \cdot \alpha'(s_0)\right]$$

(since $||\alpha'(s)|| = 1$). However, this has derivative

$$\theta_{s_0}'(s) = -\frac{\alpha'(s_0) \cdot \alpha''(s)}{\sqrt{1 - (\alpha'(s) \cdot \alpha'(s_0))^2}},$$

which I do not believe is equal to $||\alpha''(s_0)||$.

Can anyone explain do Carmo's comment or the Wikipedia passage in precise terms?

• This page might help: brickisland.net/cs177fa12/?p=187 – WalterJ May 31 '17 at 15:29
• It’s common when working with “infinitesimal” angles to replace $\sin d\theta$ by $d\theta$ and $\cos d\theta$ by $1$. Perhaps that will get to you an expression that looks more like the formula for curvature. – amd May 31 '17 at 19:25
• I've just asked the same question here: math.stackexchange.com/questions/2336751/… – user42912 Jun 26 '17 at 13:27
• But $\theta'_{s_0}(s)$ is exactly $\Vert \alpha''(s)\Vert$. You almost proved that! – Rafa Budría Jun 28 '17 at 16:04

You can see the tangent $\alpha'(s)$ as the function $\alpha'(s)=(\cos\theta(s),\sin\theta(s))$ where $\theta(s)$ is the angle the tangent at $s$ makes with the horizontal line.

Let $n(s)$ be the normal vector as defined by Do Carmo, i.e, $n(s)$ is the unit vector perpendicular to $\alpha'(s)$ at $s$ such that $n(s)$ and $\alpha'(s)$ has a positive orientation. Note we have $|\alpha''(s)|=\langle\alpha''(s),n(s)\rangle$, since $\alpha''(s)$ is a multiple of the vector $n(s)$ at $s$.

Thus we have

$\alpha'(s)=(\cos\theta(s),\sin\theta(s))$

$\implies \alpha''(s)=\theta'(s)(-\sin\theta(s),\cos\theta(s))=\theta'(s)n(s)$

$\implies |a''(s)|=\langle \alpha''(s),n(s)\rangle=\theta'(s)$

$\theta_{s_0}'(s) = -\dfrac{\alpha'(s_0) \cdot \alpha''(s)}{\sqrt{1 - (\alpha'(s) \cdot \alpha'(s_0))^2}}$

By definition of $\theta$, $\alpha'(s)\cdot\alpha'(s_0)=\Vert\alpha'(s)\Vert\Vert\alpha'(s_0)\Vert\cos\theta=\cos\theta$

Now, $\alpha'(s_0) \cdot \alpha''(s)=\Vert\alpha'(s_0)\Vert \Vert\alpha''(s)\Vert\cos\gamma$

But $\alpha'(s)$ and $\alpha''(s)$ are perpendicular and $\alpha''(s)$ positive for $\theta'_{s_0}$ positive, then, $\gamma=\pi/2+\theta$ and $\cos(\pi/2+\theta)=-\sin\theta$

$\alpha'(s_0) \cdot \alpha''(s)=-\Vert\alpha'(s_0)\Vert \Vert\alpha''(s)\Vert\sin\theta=-\Vert\alpha''(s)\Vert\sin\theta$.

Substituting in the expression for $\theta'_{s_0}(s)$

$\theta'_{s_0}(s)=-\dfrac{-\Vert\alpha''(s)\Vert\sin\theta}{\sqrt{1-\cos^2\theta}}=\Vert\alpha''(s)\Vert$