Curvature of plane parametric curves

Question

What is the neatest way to derive the following formula for the curvature of a parametric curve? $$\kappa =\frac{\|y'x''-y''x'\|}{(x'^2+y'^2)^{\frac{3}{2}}} $$

Answer 1

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:

$$ \|\vec v\|^2 = {(x')^2 + (y')^2} \\ \vec u := \frac{\vec v}{\|\vec v\|} = \frac{x' \hat e_x + y' \hat e_y}{\sqrt {(x')^2 + (y')^2}}\\ \| \vec a \wedge \vec u \| = \frac{|x' y''- x'' y'|}{\sqrt {(x')^2 + (y')^2}} \\ \implies r = \frac{\|\vec v\|^2} {\| \vec a \wedge \vec u \|} = \frac {\left[{(x')^2 + (y')^2}\right]^{3/2}} {|x' y''- x'' y'|} $$

Answer 2

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).

Answer 3

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 $$ \kappa = \frac{x'y'' - x''y'}{\left( x'^2 + y'^2 \right)^{3/2}} $$