mk_ | but how do I figure out the distance? |

TRWBW | mk_: c(t)=c0+t*v, define the line with a unit normal to it, get a quadradic |

mk_ | since there are an infinite number of points on the line unit normal to it? |

TRWBW | mk_: define the line as ax+by=c, or {a,b}.{x,y}=c |

mk_ | alright |

TRWBW | mk_: {a,b} is a normal vector to the line |

anubiss | i have some problems with trig identities : i have to integrate sqrt(1 -sin(t) ) |

TRWBW | mk_: if it's a unit normal, then |{a,b}.{x,y}-c| is distance to the line |

Eclipsor | substitute, anubiss ? |

mk_ | what is the {a,b}.{x,y} notation? |

Eclipsor | and remember that cos is an evil derivative because it turns sin negative while sin is all peaceful and turns into positive cosine :) |

anubiss | i can use the conjugate to obtain: sqrt( 1 - sin(t) ) * ( ( 1+sin(t) ) / ( ( 1+ sin(t) ) ) |

TRWBW | mk_: i use . for dot mk_: raise it up half a line height and it looks like dot product |

mk_ | ...multiplication? |

anubiss | and the answer book gives after this step : sqrt( cos^2(t) / sqrt(1+ sin(t) and i dont undetstand |

droptothetop | In relation to existence and uniqueness of a solution to a differential equation, if it's determined that there is no unique solution given a certain initial condition how can you show that there is more than one solution? One way would be to solve the equation. |

anubiss | first i dont understand how you can multiply 1 - sin(t) by 1+ sin(t) and second the answer cos^2 .. |

droptothetop | But how to show that there are two? |

TRWBW | droptothetop: how about f+1=f, that's a differential equation with no unique solution |

Olathe | anubiss: (1 - sin)(1 + sin) = 1 - sin^2. cos^2 + sin^2 = 1. |

anubiss | ok first i dont understand how you can multiply sqrt(1 - sin(t)) by sqrt(1+ sin(t)).. they are different! |

Olathe | anubiss: (1 - sin)(1 + sin)/(1 + sin) |

droptothetop | TRWBW, well, I am speaking specifically about ones that are given initial conditions, and at those initial conditions there is discontinuity in f(y,t) in y'=f(y,t) |

anubiss | hum, well my problem boils down to the sqrt's |

Olathe | anubiss: a = ab/b |

anubiss | why do you remove them |

TRWBW | droptothetop: f=1+f, f(0)=1 |

droptothetop | or in d f(y,t) /dy |

Olathe | anubiss: Where are they removed ? |

mk_ | ok, I see what the dot product is. I use a and b from the line, and x and y from the circle... what's c then? |

anubiss | i see so you dont care about the sqrt's you just multiply whats inside kk |

TRWBW | mk_: that lets you shift the line. |