The tangent vector $\TT$ has constant magnitude, and changes only in direction. Thus, its derivative measures how much the curves bends — this is the curvature $\kappa$: \begin{equation} \kappa = \left| {d\TT\over ds} \right| \end{equation} If $\kappa\ne0$, then the curve bends in a particular direction, the principal unit normal vector $\NN$, given by \begin{equation} \NN = {1\over\kappa} {d\TT\over ds} \end{equation}

It is clear that $\NN$ is a unit vector; we now show it is normal (perpendicular) to the curve in the sense that $\NN\perp\TT$: \begin{equation} 1 = \TT\cdot\TT \quad\Longrightarrow\quad 0 = {d\over ds}(\TT\cdot\TT) = 2 \TT \cdot {d\TT\over ds} = 2\kappa \TT \cdot \NN \end{equation} One also defines the unit binormal vector $\BB=\TT\times\NN$.

Since it is often difficult in practice to determine $s$, it is useful to replace derivatives with respect to $s$ by derivatives with respect to $u$ by suitably multiplying by $v=ds/du$. We obtain \begin{equation} {d\TT\over du} = \kappa v \, \NN \end{equation}

We can now rewrite velocity and acceleration in terms of their tangential and normal components, i.e. we can expand $\vv$ and $\aa$ in terms of the orthonormal vectors $\TT$ and $\NN$. We have first of all \begin{equation} \vv = v \, \TT \end{equation} from which it follows that \begin{eqnarray} \aa &=& {dv\over du} \, \TT + v \, {d\TT\over du} \\ &=& {dv\over du} \, \TT + \kappa v^2 \, \NN \\ &=&: a_T \, \TT + a_N \, \NN \\ \end{eqnarray} where the last line is a definition. It is now easy to see that \begin{equation} \kappa = {|\vv\times\aa| \over v^3} \end{equation} and this formula provides a way to calculate $\kappa$ directly.

If the parametric curve lies in the plane, so that $z=0$, the above formula for the curvature simplifies considerably. We have \begin{eqnarray} \vv &=& \dot{x}\xhat + \dot{y}\yhat = {dx\over du}\xhat + {dy\over du}\yhat \\ \aa &=& \ddot{x}\xhat + \ddot{y}\yhat = {d^2x\over du^2}\xhat + {d^2y\over du^2}\yhat \\ \end{eqnarray} so that \begin{equation} \kappa = {|\dot{x}\ddot{y}-\dot{y}\ddot{x}| \over    (\dot{x}^2+\dot{y}^2)^{3/2}} \end{equation} If furthermore $y=f(x)$, then we have $x=u$, $y=f(u)$, so that \begin{equation} \kappa = {|\ddot{f}| \over    (1+\dot{f}^2)^{3/2}} \end{equation}

The above formulas for the curvature all require you to compute a cross product, and the original definition requires you to differentiate $\TT$, and hence $v$, which typically involves a messy square root. There is yet another computational method which avoids both of these techniques. This method is particularly useful if you are asked to find all of $\{\TT,\NN,\kappa\}$, and/or if you are asked to find these quantities at a particular point, rather than as functions of $u$.

1. Compute $\displaystyle\vv={d\rr\over du}$, $\displaystyle\aa={d^2\rr\over du^2}$, and $v=|\vv|$. (Evaluate at the given point $t=t_0$ if appropriate.)
2. Compute $\TT = {\displaystyle {\vv \over v}}$.
3. Compute $a_T=\aa\cdot\TT$.
4. Compute $a_N=\sqrt{|\aa|^2-a_T^2}$ and then $\kappa = {\displaystyle {a_N \over v^2}}$.
5. Finally, compute $\displaystyle\NN={\aa-a_T\TT \over a_N}$.