# Don't understand this small step in deriving curvature

When defining curvature my lecture notes say that for $\gamma (s)$ a curve parameterised by arc length.

Let $T(s) = \gamma' (s)$ be the unit tangent.

Let $N(s)$ be the unit normal.

Now $T . T = 1$ and so $T . T' = 0$.

Thus we can write $T'(s) = \kappa (s) N(s)$ for $\kappa(s) \in \mathbb{R}$.

I'm probably missing something obvious but I don't understand the small stop from $T . T = 1$ to $T . T' = 0$ and also how this then means we can derive $T'(s) = \kappa (s) N(s)$.

Any explanation would be appreciated!

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You have $(T\cdot T)'=0$, but $(T\cdot T)'=T\cdot T'+T'\cdot T=2T\cdot T'$. –  David Mitra Dec 14 '12 at 17:03
Has $N$ been defined previously? Normally, one defines the unit normal vector as $N={dT/ds\over |dT/ds|}$. That $T\cdot T'=0$ insures $N$ is indeed perpendicular to $T$. $\kappa$ is defined to be $|dT/ds|$; and from this you have $T'(s)=\kappa N$. –  David Mitra Dec 14 '12 at 17:26
It seems they are just saying that since $T'$ is perpendicular to $T$, $T'$ must be a scalar multiple of the unit normal $N$, and defining $\kappa$ to be that multiple. –  David Mitra Dec 14 '12 at 17:34

The idea is given as follows. If the derivative of the dot product of $T\cdot T = 1$ which when taking the derivative we use the chain rule to get $T\cdot T'=0$. This just tells you that the angle between the $T$ unit tangent vector, and the $T'$ which is the change of unit tangent vector are perpendicular.
So this gives information about $T'$. What is $T'$?
It is the rate of change of $\theta$ with respect to $s$. Which is another way of saying "the rate of change with how the Unit tangent vector is deviating from a straight line with respect to $s$".
However this vector is not unit length and needs to be normalized, so to create a vector we have to normalize it with "how much" the unit tangent is changing. This "how much" is a scalar magnitude of $T'$ which is precisely $K(s)$.