User:Asyndeton

From Wikipedia, the free encyclopedia
  (Redirected from User:Algebra man)
Jump to: navigation, search

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}