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all.

So with a parametric curve $\vec{r}=\langle x(t),y(t)\rangle$, curvature is given by $$\kappa=\frac{|x'y''-x''y'|}{(x'^2+y'^2)^{3/2}}.$$ When we have constant arc-length, an alternate expression is $$\kappa=|\vec{r}''(t)|=\sqrt{x''^2+y''^2}.$$ So I see why these are both valid expressions (I can derive them both), but when we have constant arc-length, I don't see why, when we have constant arc-length $$|x'y''-x''y'|=x''^2+y''^2$$ is true. Care to enlighten me?

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  • $\begingroup$ The first equation for curvature you gave holds generally. The second holds in the case of constant arc-length. So, $$\dfrac{|x'y''-x''y'|}{(x'^2+y'^2)^{3/2}}=x''^2+y''^2\\ \mbox{If the statement is true, then one has }(x'^2+y'^2)^{3/2}=1\\ \implies x'^2+y'^2=\dfrac{dx}{dt}^2+\dfrac{dy}{dt}^2=1$$ Substituting into the equation for arc length gives $$\int^b_a\sqrt{\dfrac{dx}{dt}^2+\dfrac{dy}{dt}^2}dt=\int^b_adt=b-a,\mbox{ which is constant.}$$ $\endgroup$ – user122283 Mar 12 '14 at 0:03
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    $\begingroup$ Did you mean $\kappa = |\vec{r}''(t)| = \sqrt{(x'')^2 + (y'')^2}$...? $\endgroup$ – Andrew D. Hwang Mar 12 '14 at 2:47
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For a unit speed curve, $1 = (x')^2 + (y')^2$. Differentiating and dividing by $2$ gives $0 = x'' x' + y'' y'$. Geometrically, the velocity $(x', y')$ and acceleration $(x'', y'')$ are orthogonal. Since the velocity is non-vanishing by hypothesis, there exists a real $\kappa$ such that $(x'', y'') = \kappa(-y', x')$; taking magnitudes shows $|\kappa|$ is in fact the curvature. Substituting $x'' = -\kappa y'$ and $y'' = \kappa x'$ gives $$ |x'y'' - y'x''| = |\kappa(x')^2 + \kappa(y')^2| = |\kappa| = \sqrt{(x'')^2 + (y'')^2}. $$

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