Function equal to its own arclength.

Question

Does there exist a differentiable function $f(x)$ such that

$$f(x) = \int_0^x \sqrt{1+[f'(t)]^2}dt?$$

I'm quite interested in finding if there is such a function, as its value at any point $t$ would be the length of the curve from $x=0$ to $x=t$ and I think such a function must necessarily be beautiful in some way. Any help is appreciated!

Answer 1

Suppose there was such a function $f$. Fix an $x > 0$. Since the arclength of $f$ between $0$ and $x$ is at least the length of the straight line between $(0,f(0)) = (0,0)$ and $(x,f(x))$, we have $$f(x) = \int_0^x\sqrt{1+f'(t)^2}\,dt \ge \sqrt{x^2+f(x)^2} > f(x),$$ a contradiction. Thus, no such function $f$ exists.

Answer 2

No, there's no such function. If there was, you could take the $x$-derivative of both sides, then square both sides, then you'd obtain $0=1$. That's wrong, therefore no such function exists.