TITLE: Beach Path math question
QUESTION [6 upvotes]: Anyone who has walked on the beach knows that walking speed is dependent upon how far away from the ocean one walks. If you walk on the wet sand you can walk much more quickly than if you walked on the dry sand. I have a question that discusses this principal.
On a Cartesian $xy$ plane limited by: Domain: {$x|0 \le x \le 1$} Range: {$y|0 \le y \le 1$} you start at the point ($0,0$) and you would like to travel on a defined path to ($1,1$) in the shortest amount of time. This sounds simple just take the path $y=x$ because it is the shortest path so it will take the shortest amount of time, but there is a catch. Your forward speed $\frac{dS}{dt}$ is equal to ($1−\frac{3}{4}y$). With this constraint in mind the path $y=x$ would not be the fastest path. What is the fastest path. I am open to questions about the problem itself if I have not been clear. Thank you.

REPLY [2 votes]: This is a very cute problem.  Here is a solution using hyperbolic geometry which explains the suggestion you got at Mathoverflow. Take the formula
$$
T(\gamma)=\int_0^1 \frac{\sqrt{1+(y')^2}}{1- \frac{3}{4}y}dx
$$
as in user7530s answer. Now, let's make the change of variables:
$$
u= \frac{3}{4}x, v= 1- \frac{3}{4}y. 
$$
Note that this change of variables is a Euclidean similarity (it contracts all distances by the same amount). The integral becomes
$$
T(\gamma)= \frac{3}{4}\int_0^{3/4} \frac{\sqrt{1+(v')^2}}{v}du
$$
Declare $t=u$ (time parameter). Then this integral (I am ignoring the constant factor in front) is the hyperbolic arc-length of the curve $\gamma$ with respect to the Riemannian metric
$$
\frac{du^2 + dv^2}{v^2} 
$$
on the upper half-plane $H=\{(u,v): v>0\}$. 
Your goal is to minimize $T$ over all paths $\gamma$ connecting the two given points. It is well-known that minimizers of lengths are hyperbolic geodesics which are arcs of circles perpendicular to the $v$-axis. See for instance J.Anderson "Hyperbolic Geometry". 
Converting back to $xy$-coordinates, we see that the minimizing path is arc of circle through the points $(1,1), (0,0)$ and with center $p$ on the line $y=\frac{4}{3}$. Elementary computation then gives you that $p=(-\frac{1}{3}, \frac{4}{3})$ and the radius of the circle is $\frac{\sqrt{17}}{3}$.