Everybody knows the second derivative test for the convexity of Cartesian curves . What is the convexity test for polar curves
? Google search brought up Robert Israel’s answer on Math.SE: the relevant inequality is
But when using it, one should be aware of the singularity at the origin. For example, satisfies
but the curve is not convex: it’s the cardioid.

The formula (1) was derived for ; the points with
must be investigated directly. Actually, it is easy to see that when
has a strict local minimum with value
, the polar curve has an inward cusp and therefore is not convex.
As usual, theoretical material is followed by an exercise.
Exercise: find all real numbers such that the polar curve
is convex.
All values are ruled out by the cusp formed at
. For
we get a circle, obviously convex. When
, some calculations are in order:
For this to be nonnegative for all , we need
. Which is equivalent to two inequalities:
and
. Since
, the first inequality says that
, which is the opposite of the second.
Answer: and
.
This exercise is relevant to the problem from the previous post Peanut allergy and Polar interpolation about the search for the “right” interpolation method in polar coordinates. Given a set of values, I interpolated
with a trigonometric polynomial, and then raised that polynomial to power
.
If the given points had Cartesian coordinates then this interpolation yields
where
depend on
and satisfy
. Using the exercise above, one can deduce that
is the only nonzero power for which the interpolated curve is convex for any given points of the form
.
In general, curves of the form , with
a trigonometric polynomial, need not be convex. But even then they look more natural than their counterparts with powers
. Perhaps this is not surprising: the equation
has a decent chance of being algebraic when the degree of
is low.
Random example at the end:
