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tricky math problem

Posted on 2011-04-26
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Last Modified: 2012-05-11
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.
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Question by:Zenoture
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19 Comments
 

Author Comment

by:Zenoture
ID: 35464985
Oops, forgot to attach the image. Here it is.
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Author Comment

by:Zenoture
ID: 35465007
Upload script isn't working for me for some reason... Here are links to the images:

http://zenoture.net/math_prob.png
http://zenoture.net/math_prob_work.png
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LVL 9

Expert Comment

by:sshah254
ID: 35465010
No attachment yet :-)

Ss
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Author Comment

by:Zenoture
ID: 35465015
Yea, I can't get it to for some reason, regardless, the links work! =)
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LVL 85

Expert Comment

by:ozo
ID: 35465034
So how do you get x = 4/3?
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Author Comment

by:Zenoture
ID: 35465045
taking the derivative of sqrt(1+x^2)/4 + 1-x/5 results in x = 4/3, sorry I should have clarified that better, i thought is was implied where I wrote "3 or 4"
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LVL 85

Accepted Solution

by:
ozo earned 1000 total points
ID: 35465057
In the real world, walk time would be abs|1-x|/5, not (1-x)/5
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Author Comment

by:Zenoture
ID: 35465062
true, but a domain is set regardless, we know based on common sense we would have results greater than or equal to 0, and no greater than or equal to 1, since both given distances are 1 mile each.
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LVL 85

Assisted Solution

by:ozo
ozo earned 1000 total points
ID: 35465068
Either way, you can reject  x = 4/3 as an answer
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Author Comment

by:Zenoture
ID: 35465112
So then x = 4/3 is the only possible answer that will work? But that only solves part A, does that mean that there is no reasonable answer for part B?
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Expert Comment

by:ozo
ID: 35465232
A reasonable answer for part B would be 0<=x<=1
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Author Comment

by:Zenoture
ID: 35465333
And since the critical number from the first derivative of that function does not fall within the domain set, we can sat that x = 3/4 is the point at which the time traveled to destination will be the least, correct?
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Author Comment

by:Zenoture
ID: 35465335
is the only point*
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LVL 85

Expert Comment

by:ozo
ID: 35465354
In part A, there are no points within the domain set for which the travel time is less than at  x = 3/4.
Where in the domain set is the travel time the least in part B?
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LVL 18

Assisted Solution

by:deighton
deighton earned 1000 total points
ID: 35465757
how about calling 'x' the angle the boat makes with the vertical, angle  OBP

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



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LVL 18

Expert Comment

by:deighton
ID: 35465852
sorry, for the second,

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)



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LVL 27

Expert Comment

by:BigRat
ID: 35465876
>>how about calling 'x' the angle the boat makes with the vertical, angle  OBP

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?
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LVL 18

Assisted Solution

by:deighton
deighton earned 1000 total points
ID: 35465880
the educative point of the question is that the solution might not be where f'(x) = 0.  

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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LVL 18

Expert Comment

by:deighton
ID: 35465924
BigRat - it's equally as easy to get the second part wrong using trigonometry - that's what I found.

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