gtp | if dg/dt = 0 then df/dg DNE that's all I acn prove to you that f is a differentiable function of f from the equation he gave me f is a differentiable function of g* f=0 g=0 sorry df/dt=0 dg/dt=0 so that takes care of the "exception" |

Kiryn | hey whats a vertex perpendicular |

TRWBW | Kiryn: is that the start of a joke? i don't know, what's a vertex perpendicular... |

Kiryn | err i meant perpendicular |

gtp | @jinc there's an infinite set of solutions |

mbot | Maybe you meant: join kind |

Kiryn | sorry i copied and pasted it from a tutorial |

TRWBW | Kiryn: there is a vertex bisector, a line that bisects the angle at a vertex of a polygon |

Kiryn | TRWBW, I just meant perpendicular. not vertex perpendicular |

gtp | df/dg = (cos t + b)/ (a - sin t), that's all you can get, there's an infinite set of functions of f and g various forms |

TRWBW | Kiryn: perpendicular to a line is another line that meets it at right angles. Kiryn: a more precise definition depends on the type of geometry you are doing |

demio | % D [Arctan[x/y], x] |

mbot | demio: Derivative[1][Arctan][x/y]/y |

Kiryn | TRWBW, http://www.riemers.net/eng/Tutorials/XNA/Csharp/Series1/Lighting_basics.php |

demio | wtf |

Kiryn | thats where im reading about it |

demio | % D [Atan[x/y], x] |

mbot | demio: Derivative[1][Atan][x/y]/y |

demio | >_> % D [Arctan[x/y], x] |

mbot | demio: Derivative[1][Arctan][x/y]/y |

gtp | demio derivative of arctan(x) is 1/(1+x^2) |

TRWBW | Kiryn: do you know differential geometry? |

Steve|Office | demio: http://www.brainyquote.com/quotes/quotes/a/alberteins133991.html |

demio | what about arctan(x/y) |

Kiryn | i have no idea what that means |

demio | dx |

gtp | so derivative of artan(x/y) is (1/(1+x^2))*(1/y) |

TRWBW | Kiryn: how about vector calculus? |

gtp | chain rule opps |

Kiryn | never did calculus |

TRWBW | Kiryn: linear algebra? |