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It is easy to determine concave, convex curve in $xy$ coordinate. But I am placing a question that I only have a polar polynomial equation like $r(θ)=a_4⋅θ^4 + a_3⋅θ^3 + \dots+ a_0$;

How I can tell whether it is convex shape by given a equation like this? Convert back to $xy$ coordinate is NOT an allowable option.

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  • $\begingroup$ Polar equations that don't have trigonometric functions in them often look spiral-like... $\endgroup$ Oct 28, 2011 at 16:51
  • $\begingroup$ You are right, but assume I only care 1 complete turn (360 deg) and the discontinuous at 0deg/360deg can be ignored. Now how I can find whether the shape is convex? $\endgroup$
    – Marco
    Oct 28, 2011 at 17:21

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In general for a simple smooth closed curve ${\bf R} = {\bf R}(t)$ traversed counterclockwise, let ${\bf v} = \frac{d{\bf R}}{dt}$ be the velocity vector and ${\bf a} = \frac{d{\bf v}}{dt}$ the acceleration. The curve is convex if $({\bf v} \times {\bf a})_3 \ge 0$ everywhere. For the polar curve $r = r(\theta)$ parametrized by $\theta$, if ${\bf u}_r$ and ${\bf u}_\theta$ are the unit vectors in the radial and counterclockwise directions, we have ${\bf v} = r' {\bf u}_r + r {\bf u}_\theta$ and ${\bf a} = (r'' - r) {\bf u}_r + 2 r' {\bf u}_\theta$, so ${\bf v} \times {\bf a} = (r^2 + 2 (r')^2 - r r'') {\bf u}_r \times {\bf u}_\theta$, and the condition is $r^2 + 2 (r')^2 - r r'' \ge 0$ for all $\theta$. Note that we can't just "ignore" a discontinuity in $r'$, which could correspond to a sharp corner that makes the curve non-convex.

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you set $u(t) = 1 / r(t)$. then the curve is convex iff $u''+u \geq 0$.

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