Curvature of plane parametric curves What is the neatest way to derive the following formula for the curvature of a parametric curve? $$\kappa =\frac{\|y'x''-y''x'\|}{(x'^2+y'^2)^{\frac{3}{2}}}  $$
 A: In polar coordinates:
$$
\vec{OM} = r\hat e_r\\
\vec v = r' \hat e_r + r \omega \hat e_\theta \simeq  r \omega \hat e_\theta \\
\vec a = [r'' - r\omega^2] \hat e_r +
 [2r' \theta' + r\omega'] \hat e_\theta
\simeq - r\omega^2 \hat e_r +  r\omega' \hat e_\theta
\\
\implies r = \frac{\|\vec v\|^2}
{\| \vec a \wedge \frac{\vec v}{\|\vec v\|} \|}
$$
Write everything in cartesian coordinates gives the result:
$$
\|\vec v\|^2 = {(x')^2 + (y')^2} \\
\vec u := \frac{\vec v}{\|\vec v\|}
= \frac{x' \hat e_x + y' \hat e_y}{\sqrt {(x')^2 + (y')^2}}\\
\| \vec a \wedge \vec u \|
= \frac{|x' y''-  x'' y'|}{\sqrt {(x')^2 + (y')^2}} \\
\implies 
r = \frac{\|\vec v\|^2}
{\| \vec a \wedge \vec u \|} = 
\frac  {\left[{(x')^2 + (y')^2}\right]^{3/2}}  {|x' y''-  x'' y'|}
$$
A: Here is another approach, which stays entirely within Cartesian coordinates.
Given:
$$x=f(t), y=g(t)$$
It follows that:
$${dy\over dx}={g'(t)\over f'(t)}$$
Let $T$ be any specific value of $t$.
Since the normal of a curve is perpendicular to the tangent of that curve, and the slopes of perpendicular lines are negative reciprocals of each other, we have:
$$Y=-{f'(T)\over g'(T)}(X-f(T))+g(T)$$
(This is just point-slope form, with the $y_0$ moved to the other side.)
Now we introduce $\Delta T$, in order to use the difference quotient. The new line equation is:
$$Y=-{f'(T+\Delta T)\over g'(T+\Delta T)}(X-f(T+\Delta T))+g(T+\Delta T)$$
Next, find the intersection of the two lines:
$$-{f'(T+\Delta T)\over g'(T+\Delta T)}(X-f(T+\Delta T))+g(T+\Delta T)+{f'(T)\over g'(T)}(X-f(T))-g(T)=0$$
Solving for $X$, we get
$$X={g'(T+\Delta T)(f(T)f'(T)+g(T)g'(T))-g'(T)(f(T+\Delta T)f'(T+\Delta T)+g(T+\Delta T)g'(T+\Delta T))\over f'(T)g'(T+\Delta T)-f'(T+\Delta T)g'(T)}$$
But we want $\Delta T=0$, so we take the limit.
$$X=\lim_{\Delta T\to 0}{g'(T+\Delta T)(f(T)f'(T)+g(T)g'(T))-g'(T)(f(T+\Delta T)f'(T+\Delta T)+g(T+\Delta T)g'(T+\Delta T))\over f'(T)g'(T+\Delta T)-f'(T+\Delta T)g'(T)}$$
As the result is $\frac00$ by direct substitution, we use L'Hôpital's rule:
$$X=\lim_{\Delta T\to 0}{g''(T+\Delta T)(f(T)f'(T)+g(T)g'(T))-g'(T)(f(T+\Delta T)f''(T+\Delta T)+f'(T+\Delta T)^2+g(T+\Delta T)g''(T+\Delta T)+g'(T+\Delta T)^2)\over f'(T)g''(T+\Delta T)-f''(T+\Delta T)g'(T)}$$
Now we can substitute $\Delta T=0$ and simplify to get:
$$X=f(T)-g'(T){f'(T)^2+g'(T)^2\over f'(T)g''(T)-f''(T)g'(T)}$$
Substituting that in for the original equation of the normal, we get:
$$Y=g(T)+f'(T){f'(T)^2+g'(T)^2\over f'(T)g''(T)-f''(T)g'(T)}$$
You might recognize the two above equations as those defining the evolute of a curve, the locus of all centers of curvature. But we want the curvature itself, which is given by the reciprocal of the distance to the center of curvature:
$$\frac1{\sqrt{(X-f(T))^2+(Y-g(T))^2}}$$
Substituting in known values for $X$ and $Y$, the expression simplifies to:
$${f'(T)g''(T)-f''(T)g'(T)\over{\sqrt{f'(T)^2+g'(T)^2}^3}}$$
The specific $t$ requirement can be dropped, leading to:
$${f'(t)g''(t)-f''(t)g'(t)\over{\sqrt{f'(t)^2+g'(t)^2}^3}}$$
as desired (within a plus or minus sign).
A: The definition of curvature $\kappa$ is
$$
\kappa = \frac{d\alpha}{ds}
$$
where $\alpha = \arctan(\frac{dy}{dx})$ and $s$ is distance along the curve. $ds = dx\sqrt{(1 + (\frac{dy}{dx})^2)}\ $. Curvature has the units angle per length.
Start with the simplest parametric curve in two dimensions: $y = y(x)$. We then have:
$$
\begin{align*}
\kappa &= \frac{d\alpha}{ds} \\
&= \frac{d\alpha/dx}{ds/dx} \\
&= \frac{\frac{1}{1+(\frac{dy}{dx})^2} \frac{d^2y}{dx^2} }{\sqrt{(1 + (\frac{dy}{dx})^2)}} \\
&= \frac{\frac{d^2y}{dx^2}}{\left(1 + (\frac{dy}{dx})^2\right)^{3/2}}
\end{align*}
$$
Now consider a two-dimensional parameterized curve, given by $x = x(t)$ and $y = y(t)$. Denote $\frac{dx}{dt} = x'$ and $\frac{dy}{dt} = y'$ Then
\begin{align*}
\frac{dy}{dx} &= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'}{x'}\\
\end{align*}
\begin{align*}
\frac{d^2y}{dx^2} &= \frac{\frac{d}{dt}\left( \frac{dy}{dt} / \frac{dx}{dt} \right) }{\frac{dx}{dt}} \\
&= \frac{1}{\left( \frac{dx}{dt} \right)^3} \left[ \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right] \\
&= \frac{x'y'' - x''y'}{x'^3}
\end{align*}
Substituting $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the definition of $\kappa$, we get
$$
\kappa = \frac{x'y'' - x''y'}{\left( x'^2 + y'^2 \right)^{3/2}}
$$
