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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 $|\alpha''(s)|$?

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Imagine that you're driving on a road and you're sitting in the car. By Newton's laws of motion whenever there's an acceleration, you will feel something pulls you toward the seat, or more generally in the opposite direction of acceleration. Since velocity is a vector and acceleration is caused by a change in velocity, two things can cause acceleration: a change in the direction of the velocity vector or a change in its magnitude.

While we're driving on a straight road, the direction is always the same. The only kind of acceleration that we feel is caused by a change in the magnitude of velocity (speed). On the other hand, we all have felt that when we're making a U-turn or driving on a non-straight (curved) road, even with constant speed, some mysterious force pulls us toward the center of the circle that fits our path the best at that point. This kind of acceleration is caused by a change in the direction of velocity and it is caused by the curvature of the road.

In geometry, we're interested in this second type of change. We don't want the change in the magnitude of velocity counts because we want a straight line to have zero curvature. Therefore, we must first do something to ensure that the velocity of our curve is always constant, preferably equal to $1$. This can be achieved by reparametrizing our curve using the arc length as you said. See here for more information about reparametrizing by the arc length.

Also, the idea of measuring curvature using acceleration is important and it is the basis of defining many important concepts in future such as geodesics, covariant differentiation, parallel transport, etc.

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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, $$ \begin{align} ||\alpha''(s)||&= ||\theta'(s) (-\sin \theta(s), \cos \theta(s)) || \\ &= |\theta'(s)|\cdot ||(-\sin \theta(s), \cos \theta(s)) || \\ &= |\theta'(s)|. \end{align} $$

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