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

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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:

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

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  • $\begingroup$ An interesting derivation. What. exactly do you mean by "$\simeq$"? $\endgroup$ – Robert Lewis Apr 24 '15 at 23:28
  • $\begingroup$ in the limit where the trajectory is circular, you can consider that the terms containing $r'$ or $r''$ factors contribute much less than terms containing $r$ factors. $\endgroup$ – mookid Apr 24 '15 at 23:32
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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).

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