Give an analytic definition for a function $f(x)$ defined on $(0,\infty)$ for which the arclength from $0$ to $d$ is always $(d+1)^2$.
I’m generally pretty competent with calculus, and I think that my setup ought to be $$(d+1)^2=\lim_{a\to 0^+}\int_a^d \sqrt{1+f’(x)^2}dx$$ But I’m kind of lost as to how to actually evaluate this. For one thing, I know that the arclength integral usually has no “nice” antiderivative, but I don’t think that’s a problem because I’m just asking about a case in which it does. Next, I think the function has some sort of vertical asymptote at $0$ (because at $x=\varepsilon$ for $\varepsilon$ arbitrarily close to $0$, the arclength is already $(1+\varepsilon)^2>1$, and in particular $>\varepsilon$). Otherwise, I’m pretty much stumped. I could maybe take the derivative of both sides? But then I don’t know how to show that the derivative necessarily commutes inside the limit, and I’m not sure what to do with what would come out of that anyway.
I got this question by modifying a similar one asked on a math LinkedIn group that asked about arclength $d^2$, for which I noted that the arclength would by shorter than that of a straight line for $0<x<1$, so no such function could exist. But I’m pretty sure no similar argument would work for this function because $(x+1)^2>x$ (and in particular $(x+1)^2\not<x$) for all real $x$.
Any help or even hints would be appreciated. Thanks!
Pick $a>0$. Then for all $t>a$, we need $$ (t+1)^2=(a+1)^2+\int_a^t \sqrt{1+f’(x)^2}dx.$$ Apply $\frac{\mathrm d}{\mathrm dt}$ to both sides: $$ 2(t+1)=\sqrt{1+f'(t)^2}.$$ Solve for $f'(t)$: $$ f'(t)=\sqrt{(2(t+1))^2-1}=\sqrt{4t^2+8t+3}.$$ Conclude that $$ f(x)=f(0)+\int_0^x\sqrt{4t^2+8t+3}\,\mathrm dt.$$