# User:Asyndeton

(Redirected from User:Algebra man)

Let $\psi$ be the tangential angle, s be arc length and $\phi$ be a parameter.

From the definition of curvature,

$|k|=|\frac{d\psi}{ds}|$

$=|\frac{\frac{d\psi}{d\phi}} {\frac{ds}{d\phi}}|$
$=\left|\frac{\frac{d\psi}{d\phi}} {\sqrt{(\frac{dx}{d\phi})^2+(\frac{dy}{d\phi})^2}}\right|$
$=\left|\frac{\frac{d\psi}{d\phi}} {\sqrt{{x'}^2+{y'}^2}}\right|$.

One way of evaluating the derivative of $\tan\psi$ with respect to $\phi$ is

\begin{align} \frac{d }{d\phi}\tan\psi &=\frac{d }{d\phi} \left(\frac{dy}{dx}\right)\\ &=\frac{d }{d\phi} \left(\frac{\frac{dy}{d\phi}}{\frac{dx}{d\phi}}\right)\\ &=\frac{d }{d\phi} \left(\frac{y'}{x'}\right)\\ &=\frac{x'y''-y'x''}{{x'}^2}.\\ \end{align}

Another is

\begin{align} \frac{d }{d\phi}\tan\psi &=\sec^2\psi \frac{d\psi}{d\phi}\\ &=(1+\tan^2\psi) \frac{d\psi}{d\phi}\\ &=\left(1+\frac{{y'}^2}{{x'}^2}\right) \frac{d\psi}{d\phi}\\ &=\left(\frac{{x'}^2+{y'}^2}{{x'}^2}\right) \frac{d\psi}{d\phi}.\\ \end{align}

Equating the two gives

$\frac{d\psi}{d\phi} = \frac{x'y''-y'x''}{{x'}^2+{y'}^2}$.

Substituting this into the original equation for $|k|$ gives

$|k|=\left|\frac{x'y''-y'x''}{\left({x'}^2+{y'}^2\right)^\frac{3}{2}}\right|$.

$|k|=\frac{| \mathbf{x'} \times \mathbf{x''} |} {|\mathbf{x'}|^3}$