= Solution
{c}
Let $g>0$ be the gravitational acceleration, and let $y$ denote positive downward depth. For a frictionless particle released from rest, <conservation of energy> gives $v^2=2gy$. The <arc length> element of $(x,-y(x))$ is $ds=\sqrt{1+y'^2}\,dx$, so the <travel time> and length are
$$
T[y]=\frac1{\sqrt{2g}}\int_0^1\frac{\sqrt{1+y'^2}}{\sqrt y}\,dx,\qquad S[y]=\int_0^1\sqrt{1+y'^2}\,dx.
$$
For a finite-time single arch, require $y>0$ in $0<x<1$, $y(0)=y(1)=0$, and interpret the endpoint integrals improperly. At a regular constrained minimizer with $S[y]=L$, the <Lagrange multiplier> rule for an <isoperimetric constraint> makes $\sqrt{2g}\,T+\lambda(S-L)$ stationary. Dropping the constant $-\lambda L$ gives
$$
\boxed{I[y]=\int_0^1(y^{-1/2}+\lambda)\sqrt{1+y'^2}\,dx,\qquad y(0)=y(1)=0.}
$$
For $L>1$ the constraint is regular under interior <variations>: if its first variation vanished for every compactly supported variation, the <Euler-Lagrange equation> for $S$ would force $y'$ constant, hence $y\equiv0$ and $S=1$, a contradiction. Thus the multiplier is justified for a nondegenerate arch.
\b[The printed case $L=1$ is degenerate.] Since $\sqrt{1+y'^2}\ge1$, equality of the length with the horizontal separation forces $y'=0$ and $y=0$. A particle initially at rest then has no tangential gravitational acceleration and cannot complete the trip. Its <travel time> is infinite. The finite-time calculation therefore concerns $L>1$, with endpoint regularity qualified below.
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