This curve, described with the inverse hyperbolic secant $\text{arsech}(x) = \text{arcosh}(1/x)$,

$$y = \text{arsech}\frac x2 - \sqrt{4-x^2}$$

has its curvature equal to the linear coordinate $x$:

$$\frac{dy}{dx} = \frac{-2}{x\sqrt{4-x^2}} - \frac{-x}{\sqrt{4-x^2}}\quad = \frac{x^2-2}{x\sqrt{4-x^2}}$$

$$\frac{d^2y}{dx^2} = \frac{2x}{x\sqrt{4-x^2}} + \frac{-(x^2-2)}{x^2\sqrt{4-x^2}} + \frac{(x^2-2)x}{x\sqrt{4-x^2\;}^3}\quad = \frac{8}{x^2\sqrt{4-x^2\;}^3}$$

$$k = \frac{\frac{d^2y}{dx^2}}{\sqrt{1+\Big(\frac{dy}{dx}\Big)^2\;}^3} = \frac{\bigg(\frac{8x}{x^3\sqrt{4-x^2\;}^3}\bigg)}{\sqrt{\frac{x^2(4-x^2)+(x^4-4x^2+4)}{x^2(4-x^2)}\;}^3}\quad = x$$

Here's a graph. (This curve can be extended by reflecting it across the $x$ and $y$ axes, and the relation $k = x$ holds, depending on sign conventions.)

Is there a surface in 3D with this property?

Obviously, the (generalized) cylinder based on the above curve would work. What about a surface of revolution?

After some calculations in cylindrical coordinates (with $r$ and $z$ functions of $t$, independent of $\theta$), I find the mean curvature:

$$2H = \frac{(r'^2+z'^2)z' + (r'z''-r''z')r}{r\sqrt{r'^2+z'^2\;}^3}\quad \overset{?}{=} z$$

If the parameter $t = r$ or $t = z$, then this simplifies to

$$\frac{(1+z'^2)z' + rz''}{r\sqrt{1+z'^2\;}^3} = z$$

($z$ is a function of $r$)


$$\frac{1+r'^2 - rr''}{r\sqrt{1+r'^2\;}^3} = z$$

($r$ is a function of $z$).

Are these equations solvable? in terms of elementary functions? in terms of elliptic functions?


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