Solved

Posted on 2011-04-26

The original question and the work I did on it is attached (the work I did is a bit short-cut since I've verified my answer until that point).

My problem is with part B, when I solve for it, I get x = 4/3, which would technically make this traveler travel a bit further than needed to get to the destination, which would make no sense, unless he just really loves to row/walk...

Any guidance on the matter?

And to the mods and experts who will be complaining about providing direct solutions, this is no different than me posting C++ code and having other experts rip it apart and show me better ways of doing the same thing, which does not violate any rules.

My problem is with part B, when I solve for it, I get x = 4/3, which would technically make this traveler travel a bit further than needed to get to the destination, which would make no sense, unless he just really loves to row/walk...

Any guidance on the matter?

And to the mods and experts who will be complaining about providing direct solutions, this is no different than me posting C++ code and having other experts rip it apart and show me better ways of doing the same thing, which does not violate any rules.

19 Comments

http://zenoture.net/math_p

http://zenoture.net/math_p

Where in the domain set is the travel time the least in part B?

then with trigonometry

BP = 1 / cos x = sec x

PT = 1 - OP = 1 - tan x

then

time,

t = (1/3) sec x + (1/5) - (1/5) tan x

dt/dx = (1/3) sec x tan x - (1/5) sec^2(x)

need

(1/3) tanx - (1/5) sec x = 0

leads to

sin x = (3/5)

tan(x) = sin x / cos x = (3/5) / sqrt(1 - (3/5)^2) = (3/5)/ sqrt(16/25) = 3/4

so OP works out as 3/4

think you inverted it somewhere

sin(x) = (4/5)

tan(x) = (4/5) / [1 - (4/5)^2] = (4/5) / 3/5 = 4 / 3

I've got the same problem as you

----------------------

I've seen what's wrong, the walk from P to T would take 'negative time' in the formula if x>1, but it would take positive time in reality, you need a different formula for x> pi/4 - as ozo said though already.

General reasoning says that if he rows straight there, then it has to take longer to row to any point beyond P and walk back (as you have correctly reasoned), for the purpose of the question you know that there is no minimum or maximum on 0 to 1, so if you show that rowing straight there is faster than rowing to 0 then walking a mile, you have proved that rowing straight there has to be fastest (and rowing to 0 and walking a mile is slowest)

No, I'd just simply call the point at the right angle O, and the distance OP x and solve quite simply.

Isn't it interesting how in nicely constructed puzzles like this how 3,4 and 5 turn up?

In fact f'(x) is discontinuous and flips from being negative to positive when the rower heads straight for the target - but if you draw a graph of time taken by landing point, it will have a minimum, but not a minimum where the first derivative is continuous

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