TITLE: About the definition of curvature
QUESTION [10 upvotes]: In Do Carmo's differential geometry book, he says for a curve $\alpha: I=(a,b)\rightarrow\mathbb{R}^3$ parametrized by arc length, "since the tangent vector $\alpha'$(s) has unit length, the norm $|\alpha''(s)|$ of the second derivative measures the rate of change of the angle which neighboring tangents make with the tangent at $s$. 
Why does the unit length of the tangent vector imply this geometric meaning of $|\alpha''(s)|$?

REPLY [6 votes]: It is easier to think of it in two dimensions.  Suppose $\alpha: I \rightarrow \mathbb{R}^2$.  We can encode the derivative with polar coordinates.  There are two functions $r:I\rightarrow\mathbb{R}$ and $\theta:I\rightarrow\mathbb{R}$ such that
$$
\alpha'(s) = (r(s)\cdot \cos \theta(s), r(s)\cdot \sin \theta(s)).
$$
Notice that 
$$
\begin{align}
\alpha''(s) &= (r'(s)\cdot \cos \theta(s) - r(s)\theta'(s) \sin \theta(s), r'(s)\cdot \sin \theta(s) + r(s)\theta'(s) \cos \theta(s)) \\
&= r'(s) (\cos \theta(s), \sin \theta(s)) + r(s)\theta'(s) (-\sin \theta(s), \cos \theta(s)).
\end{align}
$$
The first term is the forward acceleration and the second term is the centripetal acceleration.  If we only want the rate that the angle is changing, $\theta'(s)$, then we could force $r(s)$ to be 1 by reparametrizing the curve.  If we set $r(s)=1$, then 
$$
\begin{align}
\alpha''(s) &= r'(s) (\cos \theta(s), \sin \theta(s)) + r(s)\theta'(s) (-\sin \theta(s), \cos \theta(s)) \\
&= \theta'(s) (-\sin \theta(s), \cos \theta(s)). 
\end{align}
$$
Taking the norm of both sides gives,
$$
\begin{align}
||\alpha''(s)||&= ||\theta'(s) (-\sin \theta(s), \cos \theta(s)) || \\
&= |\theta'(s)|\cdot ||(-\sin \theta(s), \cos \theta(s)) || \\
&= |\theta'(s)|.
\end{align}
$$