Almost Only Counts in Horseshoes by Bryan
I almost did it. Rather I almost didn't do it. And all for a class I'm not even in.
So like Mitra pointed out in her blog this week, this week MIT celebrated just about every week it could.
These included Career Fair Week, Libraries Week, and MISTI Week.
Knowing that this week would be packed with activities and stuff that I’d want to do, I decided to get all my homework done early and take my free Monday off to get ahead. And so I did.
Cue the fun part of the week.
“Attend the tale of Sweeney Todd…”
MIT Musical Theatre Guild (MTG) performed “Sweeney Todd” this week. My friend Tim, played Tobias. Little Kresge was outfitted with a barber shop atop a bakery where most of the action took place.
Surgeon General’s Warning: This musical is not for the weak stomached. It involves mild acts of cannibalism and bad haircuts.
Wednesday night, I braved the long lines to wait for advance screening tickets to Corpse Bride courtesy of LSC. Of what I got to see, it was an awesome movie. I definitely suggest checking it out just for the art alone.
Thursday was originally going to be exciting as well. I hoped to get free tickets to Carmen from the Council for the Arts, but they ran out.
(Bryan talking to himself in 3rd person)
Bryan: So pray tell, what did you do Thursday night?
Bryan: I did 18.03.
Bryan: I thought you already took that class?
Bryan: I did.
I didn’t realize how hard math is sometimes.
So Thursday night, at the house, a lot of the freshmen were working on problem sets, and I volunteered to help some people on their math homework. (8 pm)
I went to sleep at 4 am, and I still didn’t finish the problem. I would have stayed up longer trying to finish the problem, but I have made a contract with myself to NEVER pull an all-nighter.
Instead, I just woke up at 7 am to finish the problem.
Here’s the problem, try it on for size.
Almost only counts in horseshoes…and hand grenades. That’s what my third grade teacher always said. Now that I think of it, that’s kind of sketchy… Haha
Hi, seeing as it’s rather lateish at our place and I’ve just come home from a soccer-match I just glanced at the problem and I’m actually not sure I got it right, but here’s how I understood (and maybe solved) it:
If the dog is supposed to meet the mailman _at_ the gate then the mailman needs to cover ‘a’ in the same time the dog needs to cover a*pi/2, right? If that is the case then you find v = a/t_m for the mailman’s speed and u = (a*pi/2)/t_d for the dog’s speed. Seeing as we’re looking at the scene for a certain time (until the mailman reaches the gate) t_m = t_d. This would mean that we can write that the ratio k = v/u = a/(a*pi*1/2). I’m very unsure of this, but I thought I might aswell give it a try.
use the condition that the dot product of the position vector pointing from D(x,y) to M(t) with the velocity vector of D is equal to the product of their magnitudes, since they are parallel. then write t as an integral of the differential dl, and dy/dt = dy/dx * dx/dt. differentiate the first equation with respect to x and use the relations for t and dy/dt to write a differential equation for y’, y”, and x. solve for y(x) to give the trajectory of D and find necessary conditions so that y(a) = a.
BRYAN! Are you going to blog about the soiree, or should I?
we BOTH will!
Dinyar, that solution isn’t correct… even though it’s also the answer that I got (2/pi) initially.
i’m still trying to work out another solution… and Anon’s post seems to be helping a great deal..