17th Jan 2017, 23:26  #1 
Gazing at the stars
Join Date: Mar 2004
Location: Staffordshire
Posts: 1,454

Maths Problem  stumper
I've been knocking this one around for a little while now but can't seem to reduce it to something simple. It's a relative velocity problem:
_________________________________ A river flows at 5ms1 from west to east between parallel banks which are at a distance 300m apart. A man rows a boat at speed of 3ms1 in still water. Find the direction in which the boat must be steered in order to cross the river from the southern bank to the northern bank by the shortest possible route. Find the time taken and the actual distance covered by the boat for this crossing. _________________________________ Treating it as a vector problem where the 3ms1 velocity is at x degrees to the north (upstream) and splitting it into components I came up with a function which I could graph giving an optimal solution of x=36.9 degrees but I need to find a graphical/algebraic solution. Any ideas?
__________________
Exploring the outer atmosphere with Project Horizon 
18th Jan 2017, 00:02  #2 
Mod Master
Join Date: Mar 2008
Posts: 2,107

__________________
"Power without love is reckless and abusive; Love without power is anemic and sentimental" ~ Dr. Martin Luther King

18th Jan 2017, 00:32  #3 
Gazing at the stars
Join Date: Mar 2004
Location: Staffordshire
Posts: 1,454

Thanks for the link. Unfortunately all the problems and examples assume a perpendicular velocity to the bank. In the problem I've posed, I think both the boat's velocity and it's velocity relative to the river are at angles to the bank and that's what makes this so tough to pick apart.
__________________
Exploring the outer atmosphere with Project Horizon 
18th Jan 2017, 00:44  #4 
I Mod, Therefore I Own
Join Date: Apr 2009
Location: Leicestershire
Posts: 7,368

The two vectors 5ms and 3ms go togethor into a right angle triangle, the shortest route is the hypotenuse of that triangle.
angle is x opposite = 3ms adjacent = 5ms tan(x) = 3/5 tan^1(3/4) = x x = 31 degrees I would assume by direction it means heading, so that would be You could then use either sin or cos to work out the hypotenuse to get that velocity, then using the 300m and that velocity work out the time taken.
__________________
Steam: Cerberus90  i34160 3.6GHz  MSI Z97S SLI Krait  16GiB HyperX Fury 1866 256GB Samsung 830   Strix GTX980 WC'd  Thermaltake Armor  
18th Jan 2017, 01:01  #5  
Gazing at the stars
Join Date: Mar 2004
Location: Staffordshire
Posts: 1,454

Quote:
I've gone back to my component formula: If the boat's velocity, relative to the river, makes an angle of x degrees with the perpendicular to the bank upstream then the components of velocity are 3sin(x) upstream and 3cos(x) perpendicular to the bank. The time taken to cross the river is given by t=300/(3cos(x))=100/cos(x). The drift is given by (53sin(x)) multiplied by time. The resultant distance squared (by Pythagoras' Theorem) is 300^2+(((53sin(x))*100)/(cos(x)))^2 I'm trying a differentiation of the distance squared with respect to x and then solving equal to 0 in order to find the optimal angle between 0 and 90 degrees. Fingers crossed.
__________________
Exploring the outer atmosphere with Project Horizon 

18th Jan 2017, 01:49  #6 
Gazing at the stars
Join Date: Mar 2004
Location: Staffordshire
Posts: 1,454

It worked! Optimal angle is 36.9 degrees upstream of the perpendicular to the bank (53.1 degrees above the upstream bank). Total time is 125s and total distance is 500m.
Solved using quotient rule and chain rule to differentiate the distance squared. It produced a horrific mess of trigonometric terms but by factorising I obtained a slightly less messy fraction which I could then set equal to zero to determine the minimum point for the domain of 0<x<90. Setting the numerator equal to zero helped me discard some solutions which were not feasible (sin x>1) and find the only feasible solution in the domain. A victory cup of tea is in order. How the heck will I explain it tomorrow as it looks hideous. Must be a simpler method but I can't spot anything obvious with the geometry. Simpler solution still up for grabs vs my somewhat inelegant brute force approach.
__________________
Exploring the outer atmosphere with Project Horizon 
18th Jan 2017, 09:51  #7 
[DELETE] means [DELETE]
Join Date: May 2012
Posts: 4,738

I think the optimal angle is just 90 degrees. Which will give some some shallower angle of inverse tan of 3/5 due to the water.
Since the vertical distance is 300 metres you get an equivalent tiangle with horizontal component of 500 metres and a total distance travelled of 583.09 metres using pythagoras. 
18th Jan 2017, 11:38  #8  
I Mod, Therefore I Own
Join Date: Apr 2009
Location: Leicestershire
Posts: 7,368

Quote:
If it's pointing more downstream, then the river would push you further than required, the boat would end up going faster, but it would take longer to go 300m north. Anything pointing upstream would get you to a point more perpendicular to where you started, but would take longer as the 5ms would cancel out some of the 3ms speed of the boat. *EDIT* Getting stuck now trying to work it all the way through. *EDIT 2* So changed tack, Vector addition on 3ms and 5ms using pythagoras gives 5.8ms1, still at 31 degrees NorthEast. Using Sin then comes up with 576m for the total distance, and 99 seconds at 5.8ms. Here's another way to think about it. That 3ms that the boat can be rowed at, all of that speed must be put towards going north. If you row upstream, some of that 3ms is used in fighting the current, so you won't get across as quick, downstream, you're just propelling yourself further downstream. I think the time taken can actually be worked out right from the start, just by using 3ms and 300m, which comes out to 100 seconds, which is the same as my 99 seconds above accounting for rounding errors.
__________________
Steam: Cerberus90  i34160 3.6GHz  MSI Z97S SLI Krait  16GiB HyperX Fury 1866 256GB Samsung 830   Strix GTX980 WC'd  Thermaltake Armor  Last edited by Cerberus90; 18th Jan 2017 at 12:20. 

18th Jan 2017, 17:04  #10 
Bunned
Join Date: Jan 2008
Posts: 5,135

Unless I'm misreading the question, surely you just row towards the bank?
You have a horizontal component of whatever the river is flowing at, but that will never make any difference to the vertical speed of the rower. I mean, imagine it's the banks that are moving at 5m/s horizontally instead, but then they might as well be staying still because the rower doesn't need to get to any particular point on the bank, just to the bank. it makes a difference for the distance, but then you just do a simple pythagoras calc. 
18th Jan 2017, 17:09  #11  
I Mod, Therefore I Own
Join Date: Apr 2009
Location: Leicestershire
Posts: 7,368

Quote:
__________________
Steam: Cerberus90  i34160 3.6GHz  MSI Z97S SLI Krait  16GiB HyperX Fury 1866 256GB Samsung 830   Strix GTX980 WC'd  Thermaltake Armor  

18th Jan 2017, 17:15  #12  
Downwind from the bloodhounds
Join Date: Apr 2008
Location: Devon, England
Posts: 2,277

Quote:
__________________
Wolfticket: Ambitious but rubbish "A facility for quotation covers the absence of original thought." Dorothy L. Sayers (1893  1957) 

18th Jan 2017, 17:39  #13 
[DELETE] means [DELETE]
Join Date: May 2012
Posts: 4,738

If that was the case you would travel at as near as possible to zero degrees( the opposite direction to the water). That would minimise the effect of the water while still retaining a tiny vertical motion component to actually move you to the other side.
That doesn't really fit in as a solution for this kind of question. 
18th Jan 2017, 17:54  #14 
Mod Master
Join Date: Mar 2008
Posts: 2,107

I think he means that the shortest possible route is in reference to time, rather than distance. Also, the current is able to add momentum to the boat, which can be steered, midtrip, to balance momentum gained from the current with angle of traversal to optimise travel time.
__________________
"Power without love is reckless and abusive; Love without power is anemic and sentimental" ~ Dr. Martin Luther King

18th Jan 2017, 21:12  #15 
Surfacing sucks!
Join Date: Mar 2008
Posts: 2,438

Surely its a curve, as you row out perpendicular from the bank you use the rivers speed to assist you.
Or you could get a motor on the damn thing and not care.
__________________
I <3 Aquatuning & Enermax long time!

19th Jan 2017, 00:12  #16  
I Mod, Therefore I Own
Join Date: Apr 2009
Location: Leicestershire
Posts: 7,368

Quote:
But wait, if you can't row faster than the speed of the river then there's no point rowing against the current is there? Quote:
I still think it's 3ms perpendicular to the bank, then for every second, you get pushed 5ms downstream, but you'd move 3m across, with a total of 576m distance. Quote:
Using that time and distance, that means that a speed of 4ms is required, which when pointing upstream against the current of 5ms I don't think is possible if the boat can only be rowed at 3ms, if you pointed the boat perpendicular to the flow of water you could only travel at 2ms, so any direction pointing into the current cannot be faster than 2ms.
__________________
Steam: Cerberus90  i34160 3.6GHz  MSI Z97S SLI Krait  16GiB HyperX Fury 1866 256GB Samsung 830   Strix GTX980 WC'd  Thermaltake Armor  Last edited by Cerberus90; 19th Jan 2017 at 00:34. 

19th Jan 2017, 01:55  #17 
dreams are made of this
Join Date: Apr 2009
Posts: 3,018

thanks to this thread, I am going back through maths to get back to a decent state of affairs, thanks for the jolt (and nice space pix's).

19th Jan 2017, 15:26  #18 
Bunned
Join Date: Jan 2008
Posts: 5,135

Nah, you can do it in 500m at roughly 40 degrees.
EDIT: Basically what awoken said, except I used differentiation by Wolfram. 
19th Jan 2017, 20:43  #19  
I Mod, Therefore I Own
Join Date: Apr 2009
Location: Leicestershire
Posts: 7,368

So, I asked my dad what he thought (he's a chartered civil engineer), he came up with the same solution as me, row straight across.
300m at 3ms which takes 100s, that gives a drift downstream of 500m, which using pythagoras, gives a total distance of 583m. So the direction the boat must be steered is 0 degrees, which makes the resultant vector of the boat 5.83ms at an angle of 30.97 degrees to horizontal, which would be ~59 degree heading. I worked out that at 45 degrees upstream, you would have 1.5ms north component and 3.5ms downstream component (5ms1.5ms), giving a travel time of 200s, which makes 300m north and 700m drift downstream distances, which works out to be 684m total distance at 26 degrees from horitzontal, which is a 64 degree heading. Which is longer than the 583m distance that we worked out. What level/grade of maths is this by the way? Quote:
Using all of that and plugging it back in to get components, accounting for the river flowing east at 5ms, the boat would have to travel at 7.35ms in the upstream/east velocity component to make that work. As with a 36 degree angle, and the distance of 500m at 125s, which gives a resultant of 4ms, that gives a North velocity component of 3.2ms and an East velocity component of 2.35ms, but because of the river at 5ms, the east velocity must be 7.35ms to work, which isn't possible. Make sure you get what the correct answer is and how it's worked out and post it up here, I need to know otherwise it'll be bugging me for ages,
__________________
Steam: Cerberus90  i34160 3.6GHz  MSI Z97S SLI Krait  16GiB HyperX Fury 1866 256GB Samsung 830   Strix GTX980 WC'd  Thermaltake Armor  Last edited by Cerberus90; 19th Jan 2017 at 21:00. 

21st Jan 2017, 01:34  #20 
Gazing at the stars
Join Date: Mar 2004
Location: Staffordshire
Posts: 1,454

Hi All,
I worked through my solution the following morning and at the end saw two numbers staring me in the face: 5 and 3 My initial values...s**t. After a little fiddling with the vectors it became obvious. In order to obtain the shortest possible distance, the boats velocity and it's velocity relative to the river had to be perpendicular. Point the relative velocity upsteam (at an angle to the upstream bank) and the actual velocity, relative to its starting position, would point downstream (perpendicular to the velocity of the boat relative to the velocity). From there it just took a couple of minutes to realise that this whole thing could be reduced to a little bit of trig (cos x = 3/5) and some suvat calculations. With the optimal angle being 53.1 degrees (to 3sf) upstream (measured from the river bank), you can calculate the velocity of the boat, the minimum total distance travelled is therefore 500m and the time is 125s. Thanks to everyone for giving this a shot, there were some interesting ideas that led me to a better understanding of this. The work on my original, blunt & complex solution was not wasted as it turned up some interesting results such as the plot of the function I created for the distance dependent upon the angle. For those interested, this is Mechanics 3 from the AQA ALevel Further Maths syllabus. I've still not mastered this topic of Relative Motion but I'm getting closer every day. This has been, without a doubt, the toughest of all the topics so far (I'm looking at you Vertical Circular Motion from M2 and nth roots of unity from FP2). The M3 textbook is a free download online. M1 and M2 are relatively inexpensive but I'd recommend that you were confident with double angle formulae and your differentiation/integration rules before you progress on to M2 and later M3.
__________________
Exploring the outer atmosphere with Project Horizon Last edited by Awoken; 21st Jan 2017 at 01:41. 
Thread Tools  
View Mode  

