# Deriving curvature formula

How do you derive the formula for unsigned curvature of a curve $\gamma (t) = (x(t), y(t)$ which is not necessarily parameterised by arc-length.

The formula given is $$|\kappa (t)| = \left|\frac{x'' y' - y'' x'}{((x')^2 + (y')^2)^{3/2}}\right|$$

All the definitions of curvature in my notes give it in terms for an arc-length parameterisation $T'(s) = \kappa(s) N(s)$. This is apparently meant to be 'easily derivable', am I missing something?

• What are you taking as your definition of curvature? Typically it is defined as the magnitude of the derivative of the unit tangent vector with respect to arc length, right? – JohnD Jan 10 '13 at 17:00
• So you have to arc length parameterise $\gamma (t)$ first in order to derive the above? – user53076 Jan 10 '13 at 18:19
• I don't know about have to, but looking here and here, it's the way curvature is defined. If you are not going to do it this way, the question is how are you defining curvature? – JohnD Jan 10 '13 at 18:23
• I fixed a mistake: you had $x''y''$ when you meant $x''y'$. Also, it's common to use the prime for differentiation with respect to arc-length. We use dot, e.g. $\dot{x}$ for differentiation with respect to $t$. You don't need to parametrise with arc-length first. The derivation is simply a double application of the chain rule. (Assuming arc-length $s$ is a function of $t$.) See my answer below. – Fly by Night Jan 10 '13 at 18:26
• @JohnD The OP has defined curvature in the normal way: ${\bf T}' = \kappa {\bf N}$ (and so $||{\bf T}'|| = |\kappa|$.) His/her question is about how to derive the general formula that works for any parametrisation, not just arc-length. – Fly by Night Jan 10 '13 at 18:30

Assume that your curve $\gamma(t) = (x(t),y(t))$ is parametrised by a parameter of your choice. We need to relate $t$ to the arc-length parameter $s$ and, more importantly, relate their derivatives.

The arc-length parameter is defined as follows:

$$s(t) = \int ||\dot{\gamma}|| \ dt = \int \sqrt{\dot{x}^2+\dot{y}^2} \ dt \, ,$$

where $\dot{\gamma}$ (etc) represents $d\gamma/dt$. From this, we can see that:

$$\frac{ds}{dt} = \sqrt{\dot{x}^2+\dot{y}^2} \, .$$

Using the chain rule, we see that:

$$\begin{array}{cccCC} \frac{d\gamma}{dt} &=& \frac{ds}{dt} \, \frac{d\gamma}{ds} &=& \frac{ds}{dt}{\bf T} \\ \frac{d^2\gamma}{dt^2} &=& \frac{d^2s}{dt^2} \, \frac{d\gamma}{ds} + \left(\frac{ds}{dt}\right)^{\! \! 2} \, \frac{d^2\gamma}{ds^2} &=& \frac{d^2s}{dt^2} {\bf T} + \left(\frac{ds}{dt}\right)^{\! \! 2} \kappa{\bf N} \end{array}$$

We can take the scaler (dot) product of both sides of this last expression with ${\bf N}$. Since ${\bf T}$ and ${\bf N}$ are both unit length and are perpendicular we have $\langle {\bf T},{\bf T}\rangle = \langle {\bf N},{\bf N}\rangle = 1$ and $\langle {\bf T},{\bf N}\rangle = 0$. Thus:

$$\left\langle \frac{d^2\gamma}{dt^2},{\bf N} \right\rangle = \kappa \left(\frac{ds}{dt}\right)^{\! \! 2} \, .$$

We know that $ds/dt = (\dot{x}^2+\dot{y}^2)^{1/2}$ and so $(ds/dt)^2 = \dot{x}^2+\dot{y}^2$. We now have:

$$\begin{array}{ccc} \kappa(\dot{x}^2+\dot{y}^2) &=& \left\langle \frac{d^2\gamma}{dt^2},{\bf N} \right\rangle \\ \kappa &=& \frac{1}{\dot{x}^2+\dot{y}^2} \left\langle \left(\ddot{x},\ddot{y}\right),\frac{(-\dot{y},\dot{x})}{\sqrt{\dot{x}^2+\dot{y}^2}} \right\rangle \\ &=& \frac{1}{\dot{x}^2+\dot{y}^2} \left( \frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}} \right) \\ &=& \frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\left(\dot{x}^2+\dot{y}^2\right)^{3/2}} \end{array}$$

If you want the unsigned curvature then just take the absolute value of both sides.

Let $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)$$ be a regular $C^2$-parametrization of a curve $\gamma$. Then $s'(t)=\sqrt{x'^2(t)+y'^2(t)}>0$ for all $t$. The function $$\theta(t):={\rm arg}\bigl(x'(t),y'(t)\bigr)$$ gives the polar angle of the tangent vector along $\gamma$. From $$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},\>{x\over x^2+y^2}\right)$$ we obtain $$\theta'(t)={-y'\over x'^2+y'^2}x''+{x'\over x'^2+y'^2}y''={x'y''-x''y'\over x'^2+y'^2}$$ and therefore $$\kappa:={d\theta\over ds}={\theta'\over s'}={x'y''-x''y'\over (x'^2+y'^2)^{3/2}}\ .$$