# 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? Commented Jan 10, 2013 at 17:00
• So you have to arc length parameterise $\gamma (t)$ first in order to derive the above? Commented Jan 10, 2013 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? Commented Jan 10, 2013 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. Commented Jan 10, 2013 at 18:26
• Sure, but ${\bf T}' = \kappa {\bf N}$ there means ${\bf T}'(s) = \kappa(s) {\bf N}(s)$ and so curvature is defined in terms of arc length. It sounds to me like he is asking for a derivation divorced from arc length, but maybe I am hearing him wrong. Commented Jan 10, 2013 at 19:47

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.

• please, let me know which is the T's argument or N's argument? $T(s)$ or $T(t)$? thank you!
– Iuli
Commented May 19, 2020 at 13:26
• @luli I'm not sure so, take it with a grain of salt: I think because T and N are unit vectors that we always re-normalize whatever parametrization we choose to compute them has no effect and so T(s) ~= T(t) as well as N(t) ~= N(s). which is why people usually drop their arguments as it has no importance. At least that's the conclusion I've reached so far. Commented Feb 19, 2023 at 12:58

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 $$\kappa:={d\theta\over ds}={\theta'\over s'}={x'y''-x''y'\over (x'^2+y'^2)^{3/2}}\ .$$