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Homespeediest inclined plane

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# 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 order that a little ball would frictionlessly roll the whole length of the plane as soon as possible (cf. the brachistochrone problem). It is assumed that the projection 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 base, and the inclination $\alpha$ must be $45^{\circ}$.

## Mathematics Subject Classification

26A09*no label found*26A06

*no label found*

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