Need help for a homework

[Khris]

I could be horribly wrong but I guess, no matter what the angle, he'll never make it.
My approach was assuming he makes it and calculating the angle from there.
But then sin α * cos α would have to be 1,962 which is impossible since it can't be more than 0.5.

[Scarab]

a)

as_m =  120π
as_s = 7200π

b)

x² = 8² + 10² - 2*8*10 cosθ

x² = 164 - 160 cosθ

c)

x = (164-160cosθ)^0.5
dx = 0.5*(164-160*1)^-0.5
dx = 0.5*4^-0.5
dx = 1/2 * 1/(4^0.5)
dx = 1/(2*2)

dx = 0.25

(a) incorrect
(b) correct
(c) incorrect

This one's still open guys!

peace
Scarab

p.s. @ Khris, I sent you a p.m.

[Bulbapuck]

I could be horribly wrong but I guess, no matter what the angle, he'll never make it.
My approach was assuming he makes it and calculating the angle from there.
But then sin α * cos α would have to be 1,962 which is impossible since it can't be more than 0.5.

You're correct. We got this question in class and had to solve it in 5 mins using no calculation
In a class of 150, where everyone is excellent at math, 2 were able to solve it, in different ways too:
1)People who compete in long jump run at approx. 10 m/s when they jump and can't reach 10m.
2)If you shoot straight up he's in the air for approx. 2 sec. He needs more time than that at any angle that isn't straight.

It would be cool to have a dedicated thread for solving physics/maths/chemistry problems. Although I do not think that's in too much demand really.

Have at ye!


A watch has a minute hand that is 8mm long and a second hand 10 mm long.

(a) Determine the angular speeds of rotation of the minute and seconds hands in radians per hour.

[2]

(c) At exactly 7.20pm, how fast is the distance between the tips of the hands chcnging? Give your answer in mm/s correct to 2 decimal places.

[6]

a)
2*pi
(1/6)*pi

c)
The speeds are constant and at 7:20 the distance decreaces:

(1/6)*pi - 2*pi = ((1 - 12)/6)*pi = (-11/6)*pi
pi ~ 3.14
-11/6 ~ -1.83
(-1.83)*3.14 = -5.66142
.... I think, I don't have a calculator on me so that could be seriously wrong but my money's on:
Ans: -5.66
Wait.. crap.. I read wrong

10*(1/6)*pi - 8*2*pi = ((5 - 8*6)/3)*pi = (-43/3)*pi
I just remembered that the computer has a calculator so I cheated , sue me
Ans: -45.03

[Scarab]

a)
2*pi
(1/6)*pi

c)
The speeds are constant and at 7:20 the distance decreaces:

(1/6)*pi - 2*pi = ((1 - 12)/6)*pi = (-11/6)*pi
pi ~ 3.14
-11/6 ~ -1.83
(-1.83)*3.14 = -5.66142
.... I think, I don't have a calculator on me so that could be seriously wrong but my money's on:
Ans: -5.66
Wait.. crap.. I read wrong

10*(1/6)*pi - 8*2*pi = ((5 - 8*6)/3)*pi = (-43/3)*pi
I just remembered that the computer has a calculator so I cheated , sue me
Ans: -45.03

Not quite,
Still open!
Focus on the units people!

when I did this question, it needed 14 lines of working, so I think it needs about 10 minimum.

peace
scarab

[Kweepa]

(c)
x^2 = 164 - 160*cosθ
x = sqrt(164 - 160*cosθ)
dx/dt = dx/dθ * dθ/dt
dx/dθ = 40*sinθ/sqrt(41 - 40*cosθ)
dθ/dt = pi/30 - pi/1800 = 59*pi/1800
dx/dt = (59*pi/1800)*40*sinθ/sqrt(41 - 40*cosθ)
dx/dt = (59*pi/45)*sinθ/sqrt(41 - 40*cosθ)
Angle between them is 120. Plug in to get
dx/dt = (59*pi/45)*(sqrt(3)/2)/sqrt(41 + 20)
dx/dt = (59/90)*pi*sqrt(3/61), or approximately 0.46 mm/s

[Scarab]

***DING!DING!DING!DING!***
We have a winner. The answer is really -0.457, because dθ/dt = -59/1800 pi, but all the working was there.

Peace
Scarab

[Bulbapuck]

I believe this math question has not yet been solved, but since I don't know how to approach it at all I decided to post it for some insight. There are many variations of the question, this is just one.

You're the participant at a game show and you've won a prize. But which prize? There are two boxes filled with money, you have no clue as to how much. However, the host tells you that one box contains twice the amount of money then the other one.
You pick one of the boxes. Now the host asks you if you want to change box. Do you?

Logically it doesn't matter, right? However, the twist to the problem is this:
You get to open the box.
So now you have a certain amount of money. Say N dollars. Changing the box means a 50% chance that you end up with N/2 dollars, and a 50% chance that you end up with 2N dollars. Meaning you bet N/2 dollars on a 50-50 game and can win 3N/2 dollars.

But this quite obviously means you should change the box.



Where in my reasoning did I go wrong?

[Lufia]

Some kind of variation on the "three doors" problem? What I understand from such problems is that conditional probabilities are counter-intuitive.

(What do you mean "not helping"?)

[Scarab]

Logically it doesn't matter, right? However, the twist to the problem is this:
You get to open the box.
So now you have a certain amount of money. Say N dollars. Changing the box means a 50% chance that you end up with N/2 dollars, and a 50% chance that you end up with 2N dollars. Meaning you bet N/2 dollars on a 50-50 game and can win 3N/2 dollars.

But this quite obviously means you should change the box.

Where in my reasoning did I go wrong?

Why are you betting N/2 dollars? As far as I can see you arre betting N dollars with a 50/50 chance of going double or 'half'. (Please correct me if I've misunderstood)

The fact that you get to open the box changes nothing because the amount is arbitrary.

The Monty Hall Problem (or The Three Doors as Luf said it) is counter intuitive because it appears that the third outcome is removed, leaving a 50/50, when actually it is not. This problem just seems perfectly straightforward to me. (however I remember saying something similar when doing the Monty Hall Problem the first time as well  )

Diagram:


So I think whether you know the contents or not is irrelevant, and is just thrown in there to throw us off.

[Bulbapuck]

Well, I said "bet N/2" because there's no scenario where you end up with less than N/2. Hence you can only lose N/2 dollars. And also, if you don't know the amount of money there is a simple solution a mathematician came up with that states that you can't use probability theorems like that when you don't know the amount. But you're right, I don't really think it matters either.

Your diagram shows my first point perfectly, it doesn't matter.

However, to show my second point: Imagine this fair game:

You bet N dollars on a coin toss to come up heads. If the coin comes up heads, you get 2N dollars. If it comes up tails, you lose the N dollars.

This game is obviously fair. Acording to probability theory your projected winnings are as much as you bet:
(1/2)*0 + (1/2)*2N = N

Now imagine this one:

You bet N dollars on a coin toss to come up heads. If the coin comes up heads, you get 2N dollars. If it comes up tails, you lose N/2 dollars.

Acording to probability theory, this game is in your favour:
(1/2)*(N/2) + (1/2)*2N = 5N/4 > N
Hence you should play the game.

[Khris]

Well, betting N/2 isn't the same as winning at least N/2.

Here's one I particularly like (rather a riddle than a real math problem though):
You work at a place that deals with counterfeit money. Your specialty is coins.
Just before your shift ends, eight big boxes full of coins come in. The coins all look exactly the same, and the real ones are all supposed to have a weight of 10 grams.

Your supervisor explains that
a) one box contains exclusively counterfeit coins while the other seven hold only real ones.
b) all of the bad coins are either 1 gram lighter or 1 gram heavier (not mixed!).

Just as you want to start weighing the coins you notice that your scales are running out of power. You're too lazy to get new batteries and want to go home five minutes ago, so: can you find out the bad box and the weight of the bad coins while using the scales only once? And how?

[Bulbapuck]

Well, betting N/2 isn't the same as winning at least N/2.

But in this case, the outcome is exactly the same, right? So I don't see how that matters.

Here's one I particularly like (rather a riddle than a real math problem though):
You work at a place that deals with counterfeit money. Your specialty is coins.
Just before your shift ends, eight big boxes full of coins come in. The coins all look exactly the same, and the real ones are all supposed to have a weight of 10 grams.

Your supervisor explains that
a) one box contains exclusively counterfeit coins while the other seven hold only real ones.
b) all of the bad coins are either 1 gram lighter or 1 gram heavier (not mixed!).

Just as you want to start weighing the coins you notice that your scales are running out of power. You're too lazy to get new batteries and want to go home five minutes ago, so: can you find out the bad box and the weight of the bad coins while using the scales only once? And how?

I numbered the boxes from 1 to 8. Take one coin from the first box, two from the second, three from the fourth and so on. Now weigh those coins together. If all coins were real this would be 360g. However, since they're not all real there will be a difference of Xg. That means that it's box number X. And to figure out the weight you simply see if the scale is giving you a value above or below 360g.

Code: ags

if(below){badCoinsWeight = 9g;}
else if(above){badCoinsWeight = 11g;}
else{Display("damn lying boss..");} // :P

[Khris]

Yep, correct

[Bulbapuck]

Here's one:
A hotel has an infinite amount of rooms, and every room is occupied. Suddenly a new guest shows up. Can they find him a room?

[Wonkyth]

Oo, this one has so many angles it can be taken from!

OK, if I'm allowed to quote from Wikipedia:
"The Isha Upanishad of the Yajurveda (c. 4th to 3rd century BC) states that "if you remove a part from infinity or add a part to infinity, still what remains is infinity"."

does that answer?

[Calin Leafshade]

Ah Hilbert's Hotel :p

I would argue that yes they can.

There are an infinite number of odd numbers.
There are an infinite number of even numbers.
There are an infinite number of even and odd numbers combined.

So even if you have an infinite number of rooms and an infinite number of guests the two cannot be equal since infinity is not a value but rather a concept.

[Wonkyth]

Yes.
It reminds me of the a=b problem...

[Bulbapuck]

Ah Hilbert's Hotel :p

I would argue that yes they can.

There are an infinite number of odd numbers.
There are an infinite number of even numbers.
There are an infinite number of even and odd numbers combined.

So even if you have an infinite number of rooms and an infinite number of guests the two cannot be equal since infinity is not a value but rather a concept.

Actually, they are equal! The amount of hotel rooms is the same as the amount of odd numbers which is the same as the number of rational numbers.
I know that sounds odd, but there are actually only two infinities. One that's countable, and one that isn't.

Oo, this one has so many angles it can be taken from!

OK, if I'm allowed to quote from Wikipedia:
"The Isha Upanishad of the Yajurveda (c. 4th to 3rd century BC) states that "if you remove a part from infinity or add a part to infinity, still what remains is infinity"."

does that answer?

Not really, you can't add a new hotel room... Think like you're the keeper for the hotel. How would you go about actually doing it?

This one is still open!

[Calin Leafshade]

If you're just looking for the classic hilberts hotel answer;

move the occupant of room 1 to room 2 and the occupant of room 2 to room 3 and so on.

thus room 1 is now free.

[Bulbapuck]

If you're just looking for the classic hilberts hotel answer;

move the occupant of room 1 to room 2 and the occupant of room 2 to room 3 and so on.

thus room 1 is now free.

Yes, correct.

I was looking for any variation on the solution, and wonkyth's answer didn't seem too practical. It's not like the hotel keeper could've said: "just add a room" to the guy.

FOLLOWUP QUESTION:
And those who know the solution, please don't answer

If everyone at said hotel was bringing in a freind to stay in a different room, how can you put each individual in a room of his/her own?