Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Replies: 38   Last Post: Apr 13, 2013 11:57 PM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 Dono Posts: 368 Registered: 11/6/06
MTW Exercise 8.5 (c) - Crotchm the poser puts his foot in his mouth,
big time

Posted: Apr 13, 2013 10:03 AM
 Plain Text Reply

On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:
> > which again with [1] gives
> >     r r''' + 3 r'' r' = 0

>
> > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > you're done. Easy. Take
> >     r(t) = A t + B        [3]

>
> Yes, thats a "trivial" solution  and as you hinted, there are others.
> The general solution to this is
> r = s(At2 + Bt  + C), which can be ~ at + b if its a perfect square.
>

No, pretentious imbecile, what you put down is not a solution since r"
isn't null for all t (while r"'=0). You are not only a pretender, you
are an outright idiot. And to think that the other french idiot, YBM,
rushes to suck your cock and lick your ass every time you post one of
your patented idiocies.

© The Math Forum at NCTM 1994-2018. All Rights Reserved.