October [censored] thread - Page 10

dazraf69 · #91 10-16-2007, 12:44 AM

A canoeist is traveling upstream. 2 miles later he comes across a log going the opposite way.He travels for another 1 hour then turns around. He gets to his starting point the same time the log does. What is the speed of the stream?

This word problem has me stumped. Any ideas?

James282 · #92 10-16-2007, 01:05 AM

The problem doesn't make any sense. He shouldn't catch up to the log at all, assuming they are just floating along without paddling.

James

dazraf69 · #93 10-16-2007, 01:13 AM

[ QUOTE ]
The problem doesn't make any sense. He shouldn't catch up to the log at all, assuming they are just floating along without paddling.

James

[/ QUOTE ]

2 miles in 2 hours.

James282 · #94 10-16-2007, 01:36 AM

We need to figure out the speed that the paddler needs to travel to go 1 hour north and then turn around and cover his previous distance + 2 miles in the time that the stream carries the log 2 miles. You figure that if the canoeist is using the same force in both directions, his speed would obviously differ depending on which way he was travelling. If we assume he'd travel a certain speed in still water, he'd theoretically travel that speed - the stream's speed when heading upstream, and that speed + the stream's speed heading downstream.

So if the stream's speed is 1 mph - the paddler's speed would need to be 3 mph...he would go 2 miles upstream in the first hour(3mph - 1mph), and then in the next hour he would come back 4 miles to catch the log - having traveled 2 miles upstream and 4 miles downstream over the course of these 2 hours. The log would travel the 2 miles over the course of 2 hours(at 1mph)

James

James282 · #95 10-16-2007, 01:38 AM

ha, looks like you snuck the answer in as i was typing mine. your answer is correct, i presume you figured out the course to solve it as well!

dazraf69 · #96 10-16-2007, 01:47 AM

[ QUOTE ]
ha, looks like you snuck the answer in as i was typing mine. your answer is correct, i presume you figured out the course to solve it as well!

[/ QUOTE ]

I cant take credit one of the guys in the math forum came up with it. I was over thinking the problem when taking into consideration the effect of the stream speed on the canoe. Nice explanation though.

Phone Booth · #97 10-16-2007, 09:55 AM

[ QUOTE ]
We need to figure out the speed that the paddler needs to travel to go 1 hour north and then turn around and cover his previous distance + 2 miles in the time that the stream carries the log 2 miles. You figure that if the canoeist is using the same force in both directions, his speed would obviously differ depending on which way he was travelling. If we assume he'd travel a certain speed in still water, he'd theoretically travel that speed - the stream's speed when heading upstream, and that speed + the stream's speed heading downstream.

So if the stream's speed is 1 mph - the paddler's speed would need to be 3 mph...he would go 2 miles upstream in the first hour(3mph - 1mph), and then in the next hour he would come back 4 miles to catch the log - having traveled 2 miles upstream and 4 miles downstream over the course of these 2 hours. The log would travel the 2 miles over the course of 2 hours(at 1mph)

James

[/ QUOTE ]

I don't follow - as far as I can see, the paddler's speed is an indeterminate (assuming that the problem doesn't specify how long it took him to go the first three miles - the wording is unclear - if it does, it's a bad question).

Let paddler's speed be p and stream speed be s. After 2 miles *and* one hour, paddler's position is at 2 + p - s from the starting position. The log is at 2 - s from the paddler's starting position. The paddler's going at the speed of p + s and the log at the speed of s. The ratio of their speeds must be equal to the ratio of their distance away from the starting position.

(2 + p - s) / (2 - s) = (p + s) / s

which is same as

(p + (2 - s)) / (2 - s) = (p + s) / s

which, assuming p > 0 and s > 0, implies 2 - s = s, thus s = 1, leaving p an indeterminate.

Another way to look at it is that, if s = 1, the log takes two hours to cover two miles. It doesn't matter what the paddler's speed is, if he goes upstream for 1 hour and downstream for one hour, he will end up where the stream takes him, because he's cancelled out his own paddling.

James282 · #98 10-16-2007, 11:42 AM

Right, I was just using a hypothetical speed to make it easier to understand for the poster who asked the question. Of course, speed is relative and if he paddles away from the log for an hour, he will paddle for an hour in the other direction to catch it(the stream's speed will cancel itself out) - so all we need to determine is a speed over which the log will travel 2 miles in 2 hours....which is 1 mph obv.

James

ahnuld · #99 10-16-2007, 11:47 AM

nice work james. Yeah the answer is going to be a multiple of the speed of the current, it doesnt really matter what speed it flows. Now some brags:

Brag: got 95% on my investment portfolio midterm, I could teach this [censored] class.

Brag: got my fake Patrick Bateman business cards to finalize my halloween costume.

ahnuld · #**100** 10-16-2007, 11:51 AM

Only question I got wrong is this, could someone explain it to me?

Q: The global minimum variance portfolio formed from two risky securities will be riskless when the correlation coeffcient between the two securities is..

A) 0
B) 1.0
C) .5
D) -1
E) negative

I answered E, the correct answer is d? I assumed if you balance the weights and can go over 100% in one security and short the other as long as there is a neg correlation we can reduce variance to zero.