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Let $\alpha(t)=(x(t), y(t))$ be a regular curve (not necessarily an unit speed curve) in $\mathbb R^2$. How to show the curvature of $\alpha$ is given by $$\displaystyle \kappa=\frac{\langle \alpha^{''}, J(\alpha^{'})\rangle}{\langle \alpha^{'}, \alpha^{'}\rangle^{3/2}},$$ where $J$ is the operator given by $J(a, b)=(-b, a)$? Any help will be valuable.

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  • $\begingroup$ Note that the numerator is the 2-dimensional 'cross product' of $\alpha''$ and $\alpha'$, a measure of the orthogonality between the two. What's your definition of the curvature that you're trying to show equal, the radius of the osculating circle or something else? (The expression you've written often is taken as the definition...) $\endgroup$ – Steven Stadnicki May 2 '14 at 21:28
  • $\begingroup$ That is a problem because my textbook defines $\kappa$ only for unit speed curves in $\mathbb R^2$ as $\kappa=\langle T^{'}, N\rangle$. $\endgroup$ – PtF May 2 '14 at 21:38
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By definition, curvature $\kappa$ is given by the equation $T' = \kappa N$ for an arclength parametrization (where $T$, $N$ are, respectively, the unit tangent and unit normal). Note that $N=J(T)$.

In general, $\alpha' = \upsilon T$, where $\upsilon = \|\alpha'\|$ is the speed of the curve. Differentiating, $\alpha'' = \upsilon' T + \kappa\upsilon^2 N$, and so $$\kappa = \frac{\alpha''\cdot N}{\upsilon^2} = \frac{\alpha''\cdot J(T)}{\|\alpha'\|^2}.$$ Your formula is actually wrong. Since $\alpha'$ has length $\upsilon$, the correct formula should be $$\kappa = \frac{\alpha''\cdot J(\alpha')}{\|\alpha'\|^3}.$$

EDIT: OK, the formula is no longer wrong ... lest we confuse readers in the future.

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  • $\begingroup$ Good catch on the slip - note that one easy way of seeing this is to try a scaling reparameterization $t=\mu t$ and then use the chain rule to see what effect it has on the formula; $\kappa$ should remain unchanged under such a reparameterization. $\endgroup$ – Steven Stadnicki May 2 '14 at 21:40
  • $\begingroup$ Because $J$ is linear and $T=\alpha'/\upsilon$ :) $\endgroup$ – Ted Shifrin May 2 '14 at 21:51
  • $\begingroup$ All right, I deleted the comment because I had seen it =) $\endgroup$ – PtF May 2 '14 at 21:56
  • $\begingroup$ You troublemaker! $\endgroup$ – Ted Shifrin May 2 '14 at 21:57
  • $\begingroup$ @TedShifrin can you check this another question math.stackexchange.com/questions/778725/… it is related to this one.. $\endgroup$ – PtF May 2 '14 at 21:58

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