# 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

 $t^{2}\;=\;\frac{2}{g}\!\cdot\!\frac{x^{2}\!+\!b^{2}}{x}\;=:\;f(x)\qquad(x>0),$

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 $45^{\circ}$.

Title speediest inclined plane SpeediestInclinedPlane 2013-03-22 19:19:11 2013-03-22 19:19:11 pahio (2872) pahio (2872) 9 pahio (2872) Example msc 26A09 msc 26A06 Extremum CalculusOfVariations BrachistochroneCurve