To find curvature in polar coordinates, is it the same formula as in rectangular but like with $dr/d\theta$ instead of $dy/dx$?
2026-04-02 04:34:42.1775104482
Polar curvature
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No. For $\gamma(\theta)=(x(\theta),y(\theta))$ you know that $$\kappa(\theta) =\frac{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}{(x'(\theta)^2+y'(\theta)^2)^{3/2}}.$$Write $x(\theta)=r(\theta)\cos\theta$, $y(\theta)=r(\theta)\sin\theta$, differentiate and plug above to get the correct formula for the curvature in polar coordinates.