Points of maximal curvature

In Calculus I students are taught how to find the points at which the graph of a function has zero curvature (that is, the points of inflection). The points of maximal curvature are usually not discussed. This post attempts to rectify this omission.

The (signed) curvature of {y=f(x)} is

{\displaystyle \kappa = \frac{y''}{(1+(y')^2)^{3/2}}}

We want to maximize the absolute value of {\kappa}, whether the function is positive or negative there (so both maxima and minima of {\kappa} can be of interest). The critical points of {\kappa} are the zeros of

{\displaystyle \kappa' = \frac{y''' (1+(y')^2) - 3 y' (y'')^2 }{(1+(y')^2)^{5/2}}}

So we are lead to consider a polynomial of the first three derivatives of {y}, namely {p := y''' (1+(y')^2) - 3 y' (y'')^2 }.

Begin with some simple examples:

{y = x^2} has {p = -24x} so the curvature of a parabola is maximal at its vertex. No surprise there.

{y = x^3} has {p = 6(1 - 45x^4)}, indicating two symmetric points of maximal curvature, {x\approx \pm 0.386}, pretty close to the point of inflection.

x3
y = x^3

 

{y=x^4} has {p = 24 x (1 - 56 x^6)}. This has three real roots, but {x=0} actually minimizes curvature (it vanishes there).

x4
y = x^4

More generally, {y=x^n} with positive integer {n} yields {\displaystyle p = n(n-1)x^{n-3} (n - 2 - (2n^3-n^2) x^{2n-2})} indicating two points {\displaystyle x = \pm \left(\frac{n-2}{2n^3-n^2} \right)^{1/(2n-2)}} which tend to {\pm 1} as {n} grows.

The graph of a polynomial of degree {n} can have at most {n-2} points of zero curvature, because the second derivative vanishes at those. How many points of maximal curvature can it have? The degree of expression {p} above is {3n-5} but it is not obvious whether all of its roots can be real and distinct, and also be the maxima of {|\kappa|} (unlike {x=0} for {y=x^4}). For {n=2} we do get {3n-5 = 1} point of maximal curvature. But for {n=3}, there can be at most {2} such points, not {3n-5 = 4}. Edwards and Gordon (Extreme curvature of polynomials, 2004) conjectured that the graph of a polynomial of degree {n} has at most {n-1} points of maximal curvature. This remains open despite several partial results: see the recent paper Extreme curvature of polynomials and level sets (2017).

A few more elementary functions:

{y = e^x} has {p = e^x(1-2e^{2x})}, so the curvature is maximal at {x=-\log(2)/2}. Did I expect the maximum of curvature to occur for a negative {x}? Not really.

exp
y = exp(x)

{y=\sin x} has {p = (\cos 2x - 3)\cos x}. The first factor is irrelevant: the points of maximum curvature of a sine wave are at its extrema, as one would guess.

{y = \tan x} has {p = -6\tan^8(x) - 16\tan^6(x) - 6\tan^4(x) + 8\tan(x)^2 + 4} which is zero at… ahem. The expression factors as

{\displaystyle p = -2(\tan^2 x+1)(3\tan^6(x) + 5\tan^4(x) - 2\tan^2(x) - 2)}

Writing {u = \tan^2(x)} we can get a cubic equation in {u}, but it is not a nice one. Or we could do some trigonometry and reduce {p=0} to the equation {8 - 41\cos 2x + \cos 6x =0}. Either way, a numerical solution is called for: {x \approx \pm 0.6937} (and {+\pi n} for other periods).

tan
y = tan(x)

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