Shortest distance between two points via calculus of variations This problem might be trivial but when solving it using calculus of variations it's not so stupid.  
Suppose we have a fixed boundary condition $f(a) = f(b) = 0$ and we want to find the shortest distance between two points, so we choose $f$ to minimize the functional
$$
I(f) = \int_a^b \sqrt{1+f'(x)^2} dx
$$
we solved this in class and got $f'' = 0$ which is a straight line.  However, could we have just solved the problem
$$
\min I(f) = \int_a^b f'(x)^2 dx
$$
since it is a monotonic transformation of the original function?  I know this works in calculus but not sure about when dealing with functionals.  It gives me the same answer ($f'' = 0$) but is this just coincidental?  The second problem involves significantly less work. Thanks!
 A: I would say that the result is coincidental.
Consider minimization of the functional:
$$
I(f)=\int_a^b\sqrt{{f(x)}^2+{f'(x)}^2}\;dx.\tag1
$$
If I correctly understand your point, the alternative way in this case would be the minimization of
$$
I^*(f)=\int_a^b\left({f(x)}^2+{f'(x)}^2\right)\;dx.\tag2
$$
But these two functionals lead to completely different Euler-Lagrange equations:
$$\begin{align}
f(f^2+2f'^2-f''f)=0;\tag{1'}\\
f-f''=0,\tag{2'}
\end{align}$$
which have (except for the case $f(a)=f(b)=0$) completely different solutions.
A: Consider the general functional
$$\int_a^b\Phi(f'(x))\,dx.$$
The Euler-Lagrange equations read
$$\frac{d}{dx}\frac\partial{\partial f'}\Phi(f')=0$$
and
$$\frac{d}{dx}\Phi'(f')=f''\Phi''(f')=0.$$
Hence $f''(x)=0$ is always a solution, as is $f'(x)=\Phi''^{-1}(0)$, a constant.
A: This is an extended comment, not an answer, just to share my thoughts.
You mention that monotone transformations of functions yield equivalent optimization problems. This is true because, taking $g(x)$ to be monotone increasing so that $g'(x)>0$ for all $x$ in the domain, $$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)=0\iff f'(x)=0$$ since $g'(x)$ is never zero.
But I simply cannot find a way to extend this argument to the case of variational problems. If we have a solution to the transformed problem with $g(f(x,y,y'))$ (in your case $g(t)=t^2-1$ and the domain is $t\ge 0$) then we know that $$\frac{\partial }{\partial y}g\big(f(x,y,y')\big)=\frac{d}{dx}\frac{\partial }{\partial y'}g\big(f(x,y,y')\big)$$ which expands to $$g'(f)\frac{\partial f}{\partial y}=g''(f)\frac{df}{dx}\frac{\partial f}{\partial y'}+g'(f)\frac{d}{dx}\frac{\partial f}{\partial y'}.$$ We want this to imply that $$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}$$ so that we can conclude (under suitable conditions) that we have optimized the original problem. But I can find no way to realize that implication, given only that $g'(t)>0$.
Clearly the proof is done if we have $g''(t)\equiv 0$, that is, if $g$ is a linear function. Unfortunately yours is quadratic.
A: Yes, it is expected, not coincidental. That is how it should be as per calculus of variations concepts verifiable using Euler-Lagrange. The effect is the same in the two cases you mention. However the cause is different, it is not from monotonic changes from the functional. It is a class (or type) of functionals dealt with Euler-Lagrange through this EL uniform procedure. Let us see how.
Application of Euler-Lagrange on functionals 
$$ \sqrt{1+ f^{'2}(x)}, f^{'2}(x),\dfrac{1- f^{'2}(x)}{\sin(1+ f^{'7}(x))},\;...$$ 
all lead to the very same solution viz., $ y'=$  a constant, the straight lines.
This happens due to the appearance of a single derivative variable $y'$ dependent pure functional.
For example when you take another  hypothetical 
$$ \int y' (1- y^{'2})dx$$
after extremization the solution should be the same again, you can verify with EL Equn.
For that matter in general any $f(y')$ pure functional (without x) would lead to the same straight lines for a solution, viz., $ y^{'}=const,\;y^{''}=0.$ 
I am supplying the proof here that any  functional dependent purely on  $y'$ would lead to straight line solutions.
$$ L= \int f(y') dx  \tag1$$
Beltrami's formula when devoid of $x$ or its functions gives:
$$ f(y') -y'\cdot \dfrac{df(y')}{dy'}=C_1 \tag2 $$
(Full derivative as for single variable, $y$ terms treated constant when differentiating wrt $y'$).
For symbolic brevity let $u= y'$  operating as independent variable.
$$ f(u)-u\cdot\dfrac{d f(u)}{du}=C_1\tag3$$
$$\dfrac{df(u)}{f(u)-C_1}=\dfrac{du}{u} \tag4$$
Integrating,
$$ log [f(u)-C_1]= log \;u + log\;C_2 \tag5 $$
$$ f(u)= C_2 \;u +C_1\tag6 $$
$$ f(y')= C_2\; y'+ C_1\tag7 $$
Differentiating with respect to $x$
$$\dfrac{df(y')}{dx}= C_2\; y^{''} \tag8$$
The LHS should vanish as we began with the premise/assumption that $f'(u)$ does not involve (is independent of)  $x$ explicitly.
$$ y''=0 \rightarrow y= C_3 x + C_4 \tag9$$
are straight lines in the plane for all such functionals as a class.
