speediest inclined plane

We set the problem, how great must be the difference in altitude of the top and the bottom of an inclined plane in that a little ball would frictionlessly roll the whole length of the plane as soon as possible(cf. the brachistochrone problem (http://planetmath.org/CalculusOfVariations)).  It is assumed that the http://planetmath.org/node/9475projection of the length on a horizontal plane has a given value $b$.

Using notations of mechanics, we can write

$$F=ma=mg\sin \alpha =m\frac{gx}{\sqrt{x^2+b^2}},$$

$$\sqrt{x^2+b^2}=s=\frac{1}{2}{t}^{2}a=\frac{t^2}{2}\cdot \frac{gx}{\sqrt{x^2+b^2}}.$$

Thus we get the function

$$t^2=\frac{2}{g}\cdot \frac{x^2+b^2}{x}=:f(x)\mathit{\hspace{1em}\hspace{1em}}(x>0),$$

the absolute minimum point of which is to be found.  This function is differentiable, and its derivative is

$$f^{\prime }(x)=\frac{2}{g}\cdot \frac{x^2-b^2}{x^2}.$$

The only zero of $f^{\prime }(x)$ is  $x=b$,  where the sign changes from minus to plus as $x$ increases.  It means that  $x=b$  is the searched minimum point.  The difference in altitude is thus equal to the http://planetmath.org/node/11642base, and the inclination $\alpha$ must be ${45}^{\circ }$.