@Rouben Rostamian  

Yes, I made a mistake, unfortunately I did not check the statement about uniquness. y(1-x) is a path joining the points (0,0) and  (1,0) but not an admissible one (the boat will not have a constant speed here).
Also, because -Pi/2 < alpha, psi < Pi/2, it is not difficult to see that psi-alpha = Pi - arcsin(k)   cannot occur.
Sorry, my excuse is that I'm on holyday :-)

The optimal path is not unique in general: if y(x) is an optimal path, then y(1-x) is optimal too because the boat can go back on the same path in the same amount of time.

Furthermore, when V(x) is symmetric wrt x=1/2 i.e. V(x)=V(1-x) (as in the provided parabolic example) then it seems that  y(x) = - y(1-x).

[In the worksheet, from sin(psi-alpha)=k ==> psi-alpha = arcsin(k) OR  psi-alpha = Pi - arcsin(k) ...].

[Edited]

@Rouben Rostamian  

Excellent presentation. Congratulations!

A bare definition appears here.

I think that a good idea would be to have an easy access to a list (database) of known bugs + workarounds (not necessarily patches) if available.

The definition is contained in the worksheet (formula before (3))
such that anyone can see whether it suits his/her needs (and adapt it if necessary).
Otherwise it could have been given directly as:

L:=z->(sqrt(2)*exp(-z^2/2)+z*sqrt(Pi)*(erf(z*sqrt(2)/2)-1))/(2*sqrt(Pi));

On the other side, it would be nice a higher level of politeness in comments.

@tomleslie 

Probably (only?) acer can do it :-)

Please try to formulate mathematically the problem (forget about Maple for the moment).
For example:

Find the C^1 functions f(t,x,p) defined on R^3 such that

   f(t,x(t),x'(t)) = x'(t) * (∂f/∂p)(t,x(t),x'(t))

for any C^1 function x(t) defined on R.

For this problem the solution would be

f(t,x,p) = p * C(t,x),  where C is an arbitrary C^1 function on R^2.

Now, try to formulate in this manner your problem.

@asa12 

You must define/explain it. Use quantifiers.

@mikemeson 

In the above example the Array was obtained of course by a simple procedure, but you may define it as complicated as you want and then use surfdata. I don't understand your objection.

@tomleslie 

I posted once a modification like yours including the modified proc() and the post was deleded because of "Copyright" problems.
I did not understand why it was considered so, but this was the fact. A patch like mine (even almost  identical to yours) seems to be acceptable.

@miguelbravo 

Rouben's method is fine for parallel projections of plot curves and plot3d surfaces. It will not work for your dodecahedron.
Using images here it is:

@miguelbravo 

Have you read the last line of my answer? Probably not.

@Adam Ledger 

33..127  are the ASCII codes of "regular" characters.

Note that "!" (ASCII 33) also corresponds to an operator (factorial) but it is posfix and has a special treatment

( op(0, x!)  returns factorial).

@Seb1123 

The ode is solved symbolically; using the initial conditions y(0)=0, D(y)(0)=1;
the solution (Y above) depends on two constants _C3, _C4.

Computing Y' (Y1 above) it is easy to see (even without Maple) that the limit at infinity
exists only when _C3 = _C4 = 0.