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Section 11.8 Curvature

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 \(\kappa\text{:}\)

\begin{equation} \kappa = \left| {d\TT\over ds} \right| .\tag{11.8.1} \end{equation}

If \(\kappa\ne0\text{,}\) then the curve bends in a particular direction, the principal unit normal vector \(\NN\text{,}\) given by

\begin{equation} \NN = {1\over\kappa} {d\TT\over ds} .\tag{11.8.2} \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 \(\NN\perp\TT\text{:}\)

\begin{equation} 1 = \TT\cdot\TT \Longrightarrow 0 = {d\over ds}(\TT\cdot\TT) = 2 \TT \cdot {d\TT\over ds} = 2\kappa \TT \cdot \NN .\tag{11.8.3} \end{equation}

One also defines the unit binormal vector \(\BB=\TT\times\NN\text{.}\)

Since it is often difficult in practice to determine \(s\text{,}\) it is useful to replace derivatives with respect to \(s\) by derivatives with respect to \(u\) by suitably multiplying by \(v=ds/du\text{.}\) We obtain

\begin{equation} {d\TT\over du} = \kappa v \, \NN .\tag{11.8.4} \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 \(\NN\text{.}\) We have first of all

\begin{equation} \vv = v \, \TT\tag{11.8.5} \end{equation}

from which it follows that

\begin{align*} \aa \amp= {dv\over du} \, \TT + v \, {d\TT\over du}\\ \amp= {dv\over du} \, \TT + \kappa v^2 \, \NN\\ \amp = : a_T \, \TT + a_N \, \NN \end{align*}

where the last line is a definition. It is now easy to see that

\begin{equation} \kappa = {|\vv\times\aa| \over v^3}\tag{11.8.6} \end{equation}

and this formula provides a way to calculate \(\kappa\) directly.

If the parametric curve lies in the plane, so that \(z=0\text{,}\) the above formula for the curvature simplifies considerably. We have

\begin{align*} \vv \amp= \dot{x}\xhat + \dot{y}\yhat = {dx\over du}\xhat + {dy\over du}\yhat\\ \aa \amp= \ddot{x}\xhat + \ddot{y}\yhat = {d^2x\over du^2}\xhat + {d^2y\over du^2}\yhat \end{align*}

so that

\begin{equation} \kappa = {|\dot{x}\ddot{y}-\dot{y}\ddot{x}| \over ~~~(\dot{x}^2+\dot{y}^2)^{3/2}} .\tag{11.8.7} \end{equation}

If furthermore \(y=f(x)\text{,}\) then we have \(x=u\text{,}\) \(y=f(u)\text{,}\) so that

\begin{equation} \kappa = {|\ddot{f}| \over ~~~(1+\dot{f}^2)^{3/2}} .\tag{11.8.8} \end{equation}

The above formulas for the curvature all require you to compute a cross product, and the original definition requires you to differentiate \(\TT\text{,}\) and hence \(v\text{,}\) 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 \(\{\TT,\NN,\kappa\}\text{,}\) and/or if you are asked to find these quantities at a particular point, rather than as functions of \(u\text{.}\)

  1. Compute \(\displaystyle\vv={d\rr\over du}\text{,}\) \(\displaystyle\aa={d^2\rr\over du^2}\text{,}\) and \(v=|\vv|\text{.}\)

    (Evaluate at the given point \(t=t_0\) if appropriate.)

  2. Compute \(\TT = {\displaystyle {\vv \over v}}\text{.}\)

  3. Compute \(a_T=\aa\cdot\TT\text{.}\)

  4. Compute \(a_N=\sqrt{|\aa|^2-a_T^2}\) and then \(\kappa = {\displaystyle {a_N \over v^2}}\text{.}\)

  5. Finally, compute \(\displaystyle\NN={\aa-a_T\TT \over a_N}\text{.}\)

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