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Can you solve the maths question for Singapore schoolkids that went viral?

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If the current is the same speed as the boat it cannot make progress up stream.  To travel on from 6 o'clock requires it to move upstream.

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Here’s my thinking. I assumed that the ship maintains its maximum speed, 2pi, which of course is relative to the moving water. It adjusts its direction such that the resultant course is along the circle. Using the cosine rule, you get a quadratic equation for the speed at whatever point you’re on the circular course. It turns out that the speed is cos x + sqrt(15+cos^2 x) times pi/2. Integrate the reciprocal of this over -pi to pi and you get your answer. If the current cancelled out overall it would be 1 hour. I got it to be just over, corroborating China’s initial suspicion that it doesn’t. 

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5 hours ago, wrighty said:

Here’s my thinking. I assumed that the ship maintains its maximum speed, 2pi, which of course is relative to the moving water. It adjusts its direction such that the resultant course is along the circle. Using the cosine rule, you get a quadratic equation for the speed at whatever point you’re on the circular course. It turns out that the speed is cos x + sqrt(15+cos^2 x) times pi/2. Integrate the reciprocal of this over -pi to pi and you get your answer. If the current cancelled out overall it would be 1 hour. I got it to be just over, corroborating China’s initial suspicion that it doesn’t. 

Is ship speed measured relative to water using log or to land, using GPS?

Does it need to be specified.

Or does it make a difference in the problem?

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I'm still working my way through it.

John, the circle it is sailing around is fixed on the infinite plane it is sailing on (I probably am going to have to look at the effects of the curvature of the earth at some point but not now!).

The speed of the ship is its speed through the water and the water is moving in all cases steadily from north to south.

I've put the circle at the origin, called the Ship's speed S, the Water speed W and said the ship is at an angle of theta from the center of the circle on the Radius R.

It sounds like I've done something similar to Wrighty - I've had the ship sail at a course (phi) where the vectors of its speed and the water speed result in a vector T which is tangential to the circle.

The cosine rule produces a quadratic which solves for T = W cos(theta) + square root(S^2 -W^2x(sin(theta))^2)

That seems to be different from Wighty's, I may have made a mistake, but when I graph it I get a shape which seems right - a sort of sine wave, but flattened at the bottom.  If I make W=S it goes to zero in a shape which looks right.  I'll keep going on integrating this numerically, I agree with Wrighty it looks a bugger to do it algebraically and it is probably impossible!

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1 hour ago, Chinahand said:

I'm still working my way through it.

John, the circle it is sailing around is fixed on the infinite plane it is sailing on (I probably am going to have to look at the effects of the curvature of the earth at some point but not now!).

The speed of the ship is its speed through the water and the water is moving in all cases steadily from north to south.

I've put the circle at the origin, called the Ship's speed S, the Water speed W and said the ship is at an angle of theta from the center of the circle on the Radius R.

It sounds like I've done something similar to Wrighty - I've had the ship sail at a course (phi) where the vectors of its speed and the water speed result in a vector T which is tangential to the circle.

The cosine rule produces a quadratic which solves for T = W cos(theta) + square root(S^2 -W^2x(sin(theta))^2)

That seems to be different from Wighty's, I may have made a mistake, but when I graph it I get a shape which seems right - a sort of sine wave, but flattened at the bottom.  If I make W=S it goes to zero in a shape which looks right.  I'll keep going on integrating this numerically, I agree with Wrighty it looks a bugger to do it algebraically and it is probably impossible!

I suspect we’re the same - I put the values in for W and S and simplified to remove fractions and pi, and you’ve got 1-sin^2 instead of cos^2. Be interested to see the result of your numerical integration. 

Don’t even try to include the earth’s curvature - we all know the earth is really flat anyway:P - what would a uniform current even mean on the surface of a sphere? For a 1 mile radius on the earth the effect would be negligible. 

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23 hours ago, wrighty said:

.  1.0498 hours is my revised answer.

I concur!!  That's a relief!  

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3 hours ago, Chinahand said:
On ‎03‎/‎02‎/‎2018 at 4:32 PM, wrighty said:

.  1.0498 hours is my revised answer.

I concur!!  That's a relief!  

I think you're right.

I get the same answer.:lol:

Trying hard to bite my tongue here.

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Ha, a colleague who is a far better mathematician than I has concerns but I cannot check his analysis on my phone and so it will have to wait till later. 

Ho hum is there an error? I’m a little surprised- it seems to give the right answers for zero and equal water current speeds, but I’d be rash to ignore my colleague’s concern!

 

Edited by Chinahand

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42 minutes ago, the stinking enigma said:

Sorry chinahand. Had visitors, will ring you back shortly.

Best post in the whole thread. :lol:

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Can you solve Bertrand's paradox:

You have three boxes in front of you:

1) a box containing two gold coins,

2) a box containing two silver coins,

3) a box containing one gold coin and a silver coin.

What is the probability, after choosing a box at random and withdrawing one coin at random, which is found to be a gold coin, that the next coin drawn from the same box is also a gold coin?

I'm sad to say when I first thought about it I got it wrong and fell into the trap the problem deliberately sets.  Think a bit harder about it and I hope you won't make the same mistake as me!

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Hm-mm, is there a definitive answer to this paradox? 2/3, surely? 

Searching google... 

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2 hours ago, Chinahand said:

Can you solve Bertrand's paradox:

You have three boxes in front of you:

1) a box containing two gold coins,

2) a box containing two silver coins,

3) a box containing one gold coin and a silver coin.

What is the probability, after choosing a box at random and withdrawing one coin at random, which is found to be a gold coin, that the next coin drawn from the same box is also a gold coin?

I'm sad to say when I first thought about it I got it wrong and fell into the trap the problem deliberately sets.  Think a bit harder about it and I hope you won't make the same mistake as me!

This is not Bertrand’s paradox, at least according to Wiki, and is more akin to the Monty Hall problem. 

If you draw a decision tree of the six possible outcomes, given that you picked a gold coin, go back a node from those 3 outcomes and 2 indicate that the other coin in the box is gold. Hence 2/3. 

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Lovely example of how seemingly disparate areas of maths are in fact connected. 

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