# 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

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

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

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 ${45}^{\circ}$.

Title | speediest inclined plane |
---|---|

Canonical name | SpeediestInclinedPlane |

Date of creation | 2013-03-22 19:19:11 |

Last modified on | 2013-03-22 19:19:11 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 9 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 26A09 |

Classification | msc 26A06 |

Related topic | Extremum^{} |

Related topic | CalculusOfVariations |

Related topic | BrachistochroneCurve |