minimal and shortest curve How can we find the shortest curve in the xy plane that joints the 2 given points (0,a) and (1,b) and that has a given area A below it (above the x-axis and between x = 0 and x = 1) a and b are positive
Thanks
 A: I assume you are looking for a curve that can be written as a graph of a function $y=f(x)$. (Otherwise, it becomes a little unclear what area you are talking about.) Then the problem is to minimize
$$ L(f) = \int_0^1 \sqrt{1+f'(x)^2} \, dx $$
under the conditions
$$ f(0)=a, \quad f(1) = b, \quad \text{ and } \quad \int_0^1 f(x) \, dx = A. $$
To set up a variation, consider functions $v:[0,1] \to \mathbb{R}$ with
$$ v(0)=v(1)= 0, \quad \text{ and } \quad \int_0^1 v(x) \, dx = 0, $$
and let $f_t(x) = f(x) + tv(x)$. If $f$ is admissible for the variational problem, then so is $f_t$. Assume now that $f$ is the minimizer, and let
$$ L_t = \int_0^1 \sqrt{1+f_t'(x)^2} \, dx. $$
Since $L_t$ has a minimum at $t=0$, we have
$$ 0 = \left.\frac{d}{dt}\right|_{t=0} L_t  = \int_0^1\frac{f'(x)}{\sqrt{1+f'(x)^2}} v'(x) \, dx.$$
Now if $V:[0,1] \to \infty$ is any smooth function with compact support in $(0,1)$, then $v(x) = V'(x)$ is an admissible variation. Integrating by parts twice in the last integral gives
$$ 0 = \int_0^1 \frac{d^2}{dx^2} \left[\frac{f'(x)}{\sqrt{1+f'(x)^2}}\right] V(x) \, dx $$
for any compactly supported smooth $V$. This finally gives you the Euler-Lagrange equation
$$
\frac{d^2}{dx^2} \left[\frac{f'(x)}{\sqrt{1+f'(x)^2}}\right] = 0,
$$
which has the general solution
$$
\frac{f'(x)}{\sqrt{1+f'(x)^2}} = \alpha x + \beta
$$
with some constants $\alpha,\beta \in \mathbb{R}$. From here on it is an easy integration to see that the graph has to be an arc of a circle or a line segment.
There are a few issues if $A$ is too large, so that there is no arc of a circle through the given points with area $A$ under it (then you would probably want a curve made up of a graph and one or two vertical segments at the endpoints), or if $A$ is too small so that the arc actually dips below the $x$-axis. However, what this method shows is that if there is a smooth minimizer $f$, then its graph has to be an arc of a circle or a line segment.
